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Mix Examples-Kinetic Theory of Gases Questions in English
Class 11 Physics · Kinetic Theory of Gases · Mix Examples-Kinetic Theory of Gases
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101
Medium
$A$ cylinder of fixed capacity $44.8 \, L$ contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by $15.0 \, ^{\circ}C$? $(R = 8.31 \, J \, mol^{-1} K^{-1})$
Solution
(C) Using the ideal gas law $PV = \mu RT$,we know that $1 \, mol$ of any ideal gas at standard temperature $(273 \, K)$ and pressure $(1 \, atm = 1.01 \times 10^5 \, Pa)$ occupies a volume of $22.4 \, L$.
Since the cylinder has a volume of $44.8 \, L$,the number of moles of helium is $\mu = \frac{44.8}{22.4} = 2 \, mol$.
Helium is a monatomic gas,so its molar specific heat at constant volume is $C_v = \frac{3}{2}R$.
Since the volume of the cylinder is fixed,the process is isochoric,and the heat required is given by $Q = \mu C_v \Delta T$.
Substituting the values: $Q = 2 \times (\frac{3}{2} \times 8.31) \times 15.0$.
$Q = 3 \times 8.31 \times 15.0 = 45 \times 8.31 = 373.95 \, J \approx 374 \, J$.
Since the cylinder has a volume of $44.8 \, L$,the number of moles of helium is $\mu = \frac{44.8}{22.4} = 2 \, mol$.
Helium is a monatomic gas,so its molar specific heat at constant volume is $C_v = \frac{3}{2}R$.
Since the volume of the cylinder is fixed,the process is isochoric,and the heat required is given by $Q = \mu C_v \Delta T$.
Substituting the values: $Q = 2 \times (\frac{3}{2} \times 8.31) \times 15.0$.
$Q = 3 \times 8.31 \times 15.0 = 45 \times 8.31 = 373.95 \, J \approx 374 \, J$.
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View Solution102
Medium
The figure shows a plot of $PV/T$ versus $P$ for $1.00 \times 10^{-3} \; kg$ of oxygen gas at two different temperatures.
$(a)$ What does the dotted plot signify?
$(b)$ Which is true: $T_{1} > T_{2}$ or $T_{1} < T_{2}$?
$(c)$ What is the value of $PV/T$ where the curves meet on the $y$-axis?
$(d)$ If we obtained similar plots for $1.00 \times 10^{-3} \; kg$ of hydrogen,would we get the same value of $PV/T$ at the point where the curves meet on the $y$-axis? If not,what mass of hydrogen yields the same value of $PV/T$ (for the low-pressure,high-temperature region of the plot)?
(Molecular mass of $H_{2} = 2.02 \; u$,of $O_{2} = 32.0 \; u$,$R = 8.31 \; J \; mol^{-1} K^{-1}$.)
$(a)$ What does the dotted plot signify?
$(b)$ Which is true: $T_{1} > T_{2}$ or $T_{1} < T_{2}$?
$(c)$ What is the value of $PV/T$ where the curves meet on the $y$-axis?
$(d)$ If we obtained similar plots for $1.00 \times 10^{-3} \; kg$ of hydrogen,would we get the same value of $PV/T$ at the point where the curves meet on the $y$-axis? If not,what mass of hydrogen yields the same value of $PV/T$ (for the low-pressure,high-temperature region of the plot)?
(Molecular mass of $H_{2} = 2.02 \; u$,of $O_{2} = 32.0 \; u$,$R = 8.31 \; J \; mol^{-1} K^{-1}$.)
Solution
(A) The dotted plot in the graph signifies the ideal behavior of the gas,i.e.,the ratio $PV/T$ is equal to $\mu R$ (where $\mu$ is the number of moles and $R$ is the universal gas constant),which is a constant quantity independent of the pressure of the gas.
$(b)$ The dotted plot represents an ideal gas. $A$ real gas approaches the behavior of an ideal gas as its temperature increases. Since the curve at temperature $T_{1}$ is closer to the dotted plot than the curve at temperature $T_{2}$,it follows that $T_{1} > T_{2}$.
$(c)$ The value of the ratio $PV/T$ where the curves meet on the $y$-axis is $\mu R$. For oxygen,the number of moles $\mu = \frac{\text{mass}}{\text{molecular mass}} = \frac{1.00 \times 10^{-3} \; kg}{32.0 \times 10^{-3} \; kg/mol} = \frac{1}{32} \; mol$. Thus,$PV/T = \mu R = \frac{1}{32} \times 8.31 \approx 0.26 \; J K^{-1}$.
$(d)$ For hydrogen,the molecular mass is $2.02 \; u$. Since $\mu$ depends on the molecular mass,the value of $PV/T = \mu R$ will be different for the same mass of hydrogen. To get the same value of $PV/T = 0.26 \; J K^{-1}$,we need $\mu = \frac{PV/T}{R} = \frac{0.26}{8.31} \approx 0.0313 \; mol$. The required mass of hydrogen is $m = \mu \times M = 0.0313 \; mol \times 2.02 \times 10^{-3} \; kg/mol \approx 6.32 \times 10^{-5} \; kg$.
$(b)$ The dotted plot represents an ideal gas. $A$ real gas approaches the behavior of an ideal gas as its temperature increases. Since the curve at temperature $T_{1}$ is closer to the dotted plot than the curve at temperature $T_{2}$,it follows that $T_{1} > T_{2}$.
$(c)$ The value of the ratio $PV/T$ where the curves meet on the $y$-axis is $\mu R$. For oxygen,the number of moles $\mu = \frac{\text{mass}}{\text{molecular mass}} = \frac{1.00 \times 10^{-3} \; kg}{32.0 \times 10^{-3} \; kg/mol} = \frac{1}{32} \; mol$. Thus,$PV/T = \mu R = \frac{1}{32} \times 8.31 \approx 0.26 \; J K^{-1}$.
$(d)$ For hydrogen,the molecular mass is $2.02 \; u$. Since $\mu$ depends on the molecular mass,the value of $PV/T = \mu R$ will be different for the same mass of hydrogen. To get the same value of $PV/T = 0.26 \; J K^{-1}$,we need $\mu = \frac{PV/T}{R} = \frac{0.26}{8.31} \approx 0.0313 \; mol$. The required mass of hydrogen is $m = \mu \times M = 0.0313 \; mol \times 2.02 \times 10^{-3} \; kg/mol \approx 6.32 \times 10^{-5} \; kg$.
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View Solution103
Medium
Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic),the second contains chlorine (diatomic),and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not,in which case is $v_{rms}$ the largest?
Solution
(A) Yes,all vessels contain the same number of the respective molecules.
No,the root mean square speed of molecules is not the same in the three cases.
Since the three vessels have the same capacity,they have the same volume $(V)$.
Given that the temperature $(T)$ and pressure $(P)$ are also the same,according to the Ideal Gas Law $(PV = nRT)$,the number of moles $(n)$ in each vessel must be equal.
Since $n = N/N_A$,where $N_A$ is Avogadro's number,the number of molecules $(N)$ in each vessel is equal.
The root mean square speed $(v_{rms})$ of a gas molecule is given by the relation: $v_{rms} = \sqrt{\frac{3kT}{m}}$,where $k$ is the Boltzmann constant,$T$ is the temperature,and $m$ is the mass of one molecule.
Since $k$ and $T$ are constants,$v_{rms} \propto \frac{1}{\sqrt{m}}$.
As the mass of a neon atom is the smallest among neon,chlorine,and uranium hexafluoride,the root mean square speed $(v_{rms})$ is the largest for neon.
No,the root mean square speed of molecules is not the same in the three cases.
Since the three vessels have the same capacity,they have the same volume $(V)$.
Given that the temperature $(T)$ and pressure $(P)$ are also the same,according to the Ideal Gas Law $(PV = nRT)$,the number of moles $(n)$ in each vessel must be equal.
Since $n = N/N_A$,where $N_A$ is Avogadro's number,the number of molecules $(N)$ in each vessel is equal.
The root mean square speed $(v_{rms})$ of a gas molecule is given by the relation: $v_{rms} = \sqrt{\frac{3kT}{m}}$,where $k$ is the Boltzmann constant,$T$ is the temperature,and $m$ is the mass of one molecule.
Since $k$ and $T$ are constants,$v_{rms} \propto \frac{1}{\sqrt{m}}$.
As the mass of a neon atom is the smallest among neon,chlorine,and uranium hexafluoride,the root mean square speed $(v_{rms})$ is the largest for neon.
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View Solution104
MediumMCQ
From a certain apparatus,the diffusion rate of hydrogen has an average value of $28.7 \; cm^3 s^{-1}$. The diffusion of another gas under the same conditions is measured to have an average rate of $7.2 \; cm^3 s^{-1}$. Identify the gas.
A
$O_2$
B
$H_2$
C
$Cl_2$
D
$CO_2$
Solution
(A) According to Graham's Law of diffusion,the rate of diffusion $R$ of a gas is inversely proportional to the square root of its molar mass $M$: $\frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}}$.
Given: $R_1 = 28.7 \; cm^3 s^{-1}$ (for Hydrogen,$M_1 = 2.02 \; g/mol$),$R_2 = 7.2 \; cm^3 s^{-1}$.
Rearranging the formula for $M_2$: $M_2 = M_1 \left( \frac{R_1}{R_2} \right)^2$.
Substituting the values: $M_2 = 2.02 \times \left( \frac{28.7}{7.2} \right)^2$.
$M_2 \approx 2.02 \times (3.986)^2 \approx 2.02 \times 15.89 \approx 32.1 \; g/mol$.
The molar mass of Oxygen $(O_2)$ is approximately $32.0 \; g/mol$. Therefore,the gas is Oxygen.
Given: $R_1 = 28.7 \; cm^3 s^{-1}$ (for Hydrogen,$M_1 = 2.02 \; g/mol$),$R_2 = 7.2 \; cm^3 s^{-1}$.
Rearranging the formula for $M_2$: $M_2 = M_1 \left( \frac{R_1}{R_2} \right)^2$.
Substituting the values: $M_2 = 2.02 \times \left( \frac{28.7}{7.2} \right)^2$.
$M_2 \approx 2.02 \times (3.986)^2 \approx 2.02 \times 15.89 \approx 32.1 \; g/mol$.
The molar mass of Oxygen $(O_2)$ is approximately $32.0 \; g/mol$. Therefore,the gas is Oxygen.
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View Solution105
EasyMCQ
Which scientists contributed to the kinetic theory of gases?
A
James Clerk Maxwell
B
Ludwig Boltzmann
C
Rudolf Clausius
D
All of the above
Solution
(D) The kinetic theory of gases was developed over many years by several prominent physicists.
$1$. Rudolf Clausius provided the first simple kinetic theory in $1857$.
$2$. James Clerk Maxwell extended this theory by introducing the statistical distribution of molecular velocities in $1859$.
$3$. Ludwig Boltzmann further refined the theory by developing statistical mechanics,which provides a microscopic foundation for thermodynamics.
Therefore,all the mentioned scientists contributed significantly to the kinetic theory of gases.
$1$. Rudolf Clausius provided the first simple kinetic theory in $1857$.
$2$. James Clerk Maxwell extended this theory by introducing the statistical distribution of molecular velocities in $1859$.
$3$. Ludwig Boltzmann further refined the theory by developing statistical mechanics,which provides a microscopic foundation for thermodynamics.
Therefore,all the mentioned scientists contributed significantly to the kinetic theory of gases.
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View Solution106
Medium
State whether the following statements are True or False:
$(i)$ Due to collision,the density of gas molecules changes.
$(ii)$ The average kinetic energy of $1 \ g$ mass of each gas is equal at the same temperature.
$(iii)$ At the same temperature,the $v_{rms}$ of two different gases is the same.
$(iv)$ At constant temperature,if the pressure of a gas is increased,its mean free path decreases.
$(i)$ Due to collision,the density of gas molecules changes.
$(ii)$ The average kinetic energy of $1 \ g$ mass of each gas is equal at the same temperature.
$(iii)$ At the same temperature,the $v_{rms}$ of two different gases is the same.
$(iv)$ At constant temperature,if the pressure of a gas is increased,its mean free path decreases.
Solution
(D) $(i)$ False. Collisions between gas molecules do not change the total number of molecules or the volume of the container,so the density remains constant.
$(ii)$ False. The average kinetic energy per molecule depends only on temperature $(KE_{avg} = \frac{3}{2} k_B T)$. However,the average kinetic energy of a fixed mass $(1 \ g)$ depends on the molar mass of the gas $(KE_{total} = \frac{n}{2} RT = \frac{m}{M} \frac{3}{2} RT)$,which varies for different gases.
$(iii)$ False. The root mean square velocity is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$. Since $v_{rms}$ depends on the molar mass $M$,it is different for different gases at the same temperature.
$(iv)$ True. The mean free path is given by $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}$. Since $\lambda \propto \frac{1}{P}$,increasing the pressure at a constant temperature decreases the mean free path.
$(ii)$ False. The average kinetic energy per molecule depends only on temperature $(KE_{avg} = \frac{3}{2} k_B T)$. However,the average kinetic energy of a fixed mass $(1 \ g)$ depends on the molar mass of the gas $(KE_{total} = \frac{n}{2} RT = \frac{m}{M} \frac{3}{2} RT)$,which varies for different gases.
$(iii)$ False. The root mean square velocity is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$. Since $v_{rms}$ depends on the molar mass $M$,it is different for different gases at the same temperature.
$(iv)$ True. The mean free path is given by $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}$. Since $\lambda \propto \frac{1}{P}$,increasing the pressure at a constant temperature decreases the mean free path.
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View Solution107
Medium
Fill in the blanks:
$(i)$ At low $......$ and high $......$ temperature,real gases behave as an ideal gas.
$(ii)$ $......$ is a measure of the average kinetic energy of a gas.
$(iii)$ The kinetic energy of a gas with mass $m$ is $E$. Its momentum will be $......$.
$(iv)$ At $......$ temperature,$v_{rms}$ will be double that of $v_{rms}$ at $0^{\circ} C$.
$(i)$ At low $......$ and high $......$ temperature,real gases behave as an ideal gas.
$(ii)$ $......$ is a measure of the average kinetic energy of a gas.
$(iii)$ The kinetic energy of a gas with mass $m$ is $E$. Its momentum will be $......$.
$(iv)$ At $......$ temperature,$v_{rms}$ will be double that of $v_{rms}$ at $0^{\circ} C$.
Solution
(D) $(i)$ pressure,temperature
$(ii)$ Temperature
$(iii)$ $\sqrt{2mE}$
Kinetic energy $E = \frac{1}{2}mv^2$
$E = \frac{1}{2} \frac{m^2v^2}{m} = \frac{p^2}{2m}$,where $p$ is momentum.
Therefore,$p = \sqrt{2mE}$.
$(iv)$ $819^{\circ} C$
$v_{rms} \propto \sqrt{T}$
$\frac{(v_{rms})_2}{(v_{rms})_1} = \sqrt{\frac{T_2}{T_1}}$
$2 = \sqrt{\frac{T_2}{273}}$
$4 = \frac{T_2}{273} \implies T_2 = 1092 \ K$
$t_2 = 1092 - 273 = 819^{\circ} C$.
$(ii)$ Temperature
$(iii)$ $\sqrt{2mE}$
Kinetic energy $E = \frac{1}{2}mv^2$
$E = \frac{1}{2} \frac{m^2v^2}{m} = \frac{p^2}{2m}$,where $p$ is momentum.
Therefore,$p = \sqrt{2mE}$.
$(iv)$ $819^{\circ} C$
$v_{rms} \propto \sqrt{T}$
$\frac{(v_{rms})_2}{(v_{rms})_1} = \sqrt{\frac{T_2}{T_1}}$
$2 = \sqrt{\frac{T_2}{273}}$
$4 = \frac{T_2}{273} \implies T_2 = 1092 \ K$
$t_2 = 1092 - 273 = 819^{\circ} C$.
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View Solution108
Medium
Fill in the blanks:
$(i)$ At absolute zero,the volume of an ideal gas is ...... .
$(ii)$ With an increase in temperature,the pressure of a gas ...... .
$(iii)$ All molecular motion will stop at ...... .
$(iv)$ Due to ...... at higher altitudes from the surface of the Earth,the air becomes cooler.
$(i)$ At absolute zero,the volume of an ideal gas is ...... .
$(ii)$ With an increase in temperature,the pressure of a gas ...... .
$(iii)$ All molecular motion will stop at ...... .
$(iv)$ Due to ...... at higher altitudes from the surface of the Earth,the air becomes cooler.
Solution
(N/A) $(i)$ According to Charles's Law,$V \propto T$. As $T$ approaches $0 \ K$ (absolute zero),the volume $V$ approaches $0$.
$(ii)$ According to Gay-Lussac's Law,$P \propto T$ for a fixed volume. Therefore,as temperature increases,the pressure of the gas increases.
$(iii)$ Absolute zero temperature ($0 \ K$ or $-273.15 \ ^\circ C$) is defined as the temperature at which all thermal molecular motion ceases.
$(iv)$ As altitude increases,the atmospheric pressure decreases. The air expands and does work against the surroundings,leading to a decrease in internal energy and temperature,a process known as adiabatic cooling.
$(ii)$ According to Gay-Lussac's Law,$P \propto T$ for a fixed volume. Therefore,as temperature increases,the pressure of the gas increases.
$(iii)$ Absolute zero temperature ($0 \ K$ or $-273.15 \ ^\circ C$) is defined as the temperature at which all thermal molecular motion ceases.
$(iv)$ As altitude increases,the atmospheric pressure decreases. The air expands and does work against the surroundings,leading to a decrease in internal energy and temperature,a process known as adiabatic cooling.
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View Solution109
Easy
Column-$I$ represents physical quantity and Column-$II$ represents formula. Match them correctly:
| Column-$I$ | Column-$II$ |
| $(a)$ Kinetic energy per unit mole of gas. | $(i)$ $\frac{1}{2}RT$ |
| $(b)$ Kinetic energy per one molecule of gas. | $(ii)$ $\frac{3}{2}RT$ |
| $(iii)$ $\frac{3}{2}k_BT$ |
Solution
(A) For an ideal gas, the average kinetic energy per mole is given by $U = \frac{3}{2}RT$. Thus, $(a)$ matches with $(ii)$.
The average kinetic energy per molecule is given by $E = \frac{3}{2}k_BT$, where $k_B$ is the Boltzmann constant. Thus, $(b)$ matches with $(iii)$.
Therefore, the correct matching is $(a-ii, b-iii)$.
The average kinetic energy per molecule is given by $E = \frac{3}{2}k_BT$, where $k_B$ is the Boltzmann constant. Thus, $(b)$ matches with $(iii)$.
Therefore, the correct matching is $(a-ii, b-iii)$.
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View Solution110
DifficultMCQ
An insulated container containing a monoatomic gas of molar mass $M$ is moving with a velocity $V_{0}$. If the container is suddenly stopped,find the change in temperature.
A
$\frac{2MV_{0}^{2}}{3R}$
B
$\frac{MV_{0}^{2}}{2R}$
C
$\frac{MV_{0}^{2}}{5R}$
D
$\frac{MV_{0}^{2}}{3R}$
Solution
(D) When the container stops suddenly,the macroscopic kinetic energy of the gas is converted into internal energy (random kinetic energy) of the gas molecules.
Let $n$ be the number of moles of the gas. The total mass of the gas is $M_{total} = n M$.
The initial kinetic energy of the gas due to the motion of the container is $KE = \frac{1}{2} (nM) V_{0}^{2}$.
Since the container is insulated,the change in internal energy $\Delta U$ is equal to the change in kinetic energy:
$\Delta U = \Delta KE = \frac{1}{2} (nM) V_{0}^{2}$.
For a monoatomic gas,the change in internal energy is given by $\Delta U = n C_{V} \Delta T$,where $C_{V} = \frac{3}{2} R$.
Equating the two expressions:
$n \left( \frac{3}{2} R \right) \Delta T = \frac{1}{2} (nM) V_{0}^{2}$.
Solving for $\Delta T$:
$\Delta T = \frac{nM V_{0}^{2}}{2} \times \frac{2}{3nR} = \frac{MV_{0}^{2}}{3R}$.
Let $n$ be the number of moles of the gas. The total mass of the gas is $M_{total} = n M$.
The initial kinetic energy of the gas due to the motion of the container is $KE = \frac{1}{2} (nM) V_{0}^{2}$.
Since the container is insulated,the change in internal energy $\Delta U$ is equal to the change in kinetic energy:
$\Delta U = \Delta KE = \frac{1}{2} (nM) V_{0}^{2}$.
For a monoatomic gas,the change in internal energy is given by $\Delta U = n C_{V} \Delta T$,where $C_{V} = \frac{3}{2} R$.
Equating the two expressions:
$n \left( \frac{3}{2} R \right) \Delta T = \frac{1}{2} (nM) V_{0}^{2}$.
Solving for $\Delta T$:
$\Delta T = \frac{nM V_{0}^{2}}{2} \times \frac{2}{3nR} = \frac{MV_{0}^{2}}{3R}$.
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View Solution111
Medium
Explain why:
$(a)$ there is no atmosphere on the Moon.
$(b)$ there is a fall in temperature with altitude.
$(a)$ there is no atmosphere on the Moon.
$(b)$ there is a fall in temperature with altitude.
Solution
(N/A) The gravitational force on the Moon is very weak,resulting in a low escape speed of approximately $2.3 \ km/s$. The root-mean-square speed of gas molecules at the Moon's surface temperature,given by $v_{rms} = \sqrt{\frac{3 k_{B} T}{m}}$,exceeds this escape speed. Consequently,gas molecules escape the Moon's gravitational pull,preventing the formation of an atmosphere.
$(b)$ As air molecules move upward in the atmosphere,they work against the gravitational field,which increases their potential energy. According to the law of conservation of energy,this increase in potential energy leads to a corresponding decrease in their kinetic energy. Since the temperature of a gas is directly proportional to the average kinetic energy of its molecules $(E_k = \frac{3}{2} k_{B} T)$,the temperature decreases with increasing altitude.
$(b)$ As air molecules move upward in the atmosphere,they work against the gravitational field,which increases their potential energy. According to the law of conservation of energy,this increase in potential energy leads to a corresponding decrease in their kinetic energy. Since the temperature of a gas is directly proportional to the average kinetic energy of its molecules $(E_k = \frac{3}{2} k_{B} T)$,the temperature decreases with increasing altitude.
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View Solution112
Difficult
$A$ box of $1 \ m^{3}$ is filled with nitrogen at $1.5 \ atm$ at $300 \ K$. The box has a hole of an area $0.010 \ mm^{2}$. How much time is required for the pressure to reduce by $0.10 \ atm$,if the pressure outside is $1 \ atm$?
Solution
(D) The rate of effusion of gas molecules through a small hole is given by the kinetic theory of gases. The number of molecules $dN$ escaping in time $dt$ is $dN = \frac{1}{4} n \langle v \rangle A dt$,where $n$ is the number density,$\langle v \rangle$ is the average speed,and $A$ is the area of the hole.
Using $PV = N k_B T$,we have $n = \frac{N}{V} = \frac{P}{k_B T}$.
The average speed $\langle v \rangle = \sqrt{\frac{8 k_B T}{\pi m}}$.
The rate of change of pressure is $\frac{dP}{dt} = -\frac{1}{4} \frac{P}{k_B T} \sqrt{\frac{8 k_B T}{\pi m}} \frac{k_B T}{V} A = -\frac{A}{V} \sqrt{\frac{k_B T}{2 \pi m}} P$.
Integrating this from $P_i = 1.5 \ atm$ to $P_f = 1.4 \ atm$ (since pressure reduces by $0.10 \ atm$):
$\int_{P_i}^{P_f} \frac{dP}{P} = -\frac{A}{V} \sqrt{\frac{k_B T}{2 \pi m}} \int_0^t dt$.
$\ln\left(\frac{P_f}{P_i}\right) = -\frac{A}{V} \sqrt{\frac{k_B T}{2 \pi m}} t$.
Substituting the values $m = \frac{28 \times 10^{-3}}{6.022 \times 10^{23}} \ kg$,$T = 300 \ K$,$A = 10^{-8} \ m^2$,$V = 1 \ m^3$,we can solve for $t$.
Using $PV = N k_B T$,we have $n = \frac{N}{V} = \frac{P}{k_B T}$.
The average speed $\langle v \rangle = \sqrt{\frac{8 k_B T}{\pi m}}$.
The rate of change of pressure is $\frac{dP}{dt} = -\frac{1}{4} \frac{P}{k_B T} \sqrt{\frac{8 k_B T}{\pi m}} \frac{k_B T}{V} A = -\frac{A}{V} \sqrt{\frac{k_B T}{2 \pi m}} P$.
Integrating this from $P_i = 1.5 \ atm$ to $P_f = 1.4 \ atm$ (since pressure reduces by $0.10 \ atm$):
$\int_{P_i}^{P_f} \frac{dP}{P} = -\frac{A}{V} \sqrt{\frac{k_B T}{2 \pi m}} \int_0^t dt$.
$\ln\left(\frac{P_f}{P_i}\right) = -\frac{A}{V} \sqrt{\frac{k_B T}{2 \pi m}} t$.
Substituting the values $m = \frac{28 \times 10^{-3}}{6.022 \times 10^{23}} \ kg$,$T = 300 \ K$,$A = 10^{-8} \ m^2$,$V = 1 \ m^3$,we can solve for $t$.
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View Solution113
Easy
Molecules in the atmosphere are attracted by the gravitational force of the Earth. Explain why all of them do not fall onto the Earth,just like an apple falling from a tree.
Solution
(N/A) Although air molecules are attracted by the Earth's gravitational force,they do not fall to the surface like an apple due to two main reasons:
$1$. Thermal Motion: Air molecules possess high kinetic energy due to their thermal velocity. This random motion allows them to overcome the gravitational pull and remain suspended in the atmosphere.
$2$. Collisions: Constant collisions between air molecules and with other particles prevent them from settling down. The pressure exerted by these collisions balances the downward gravitational force,maintaining the atmospheric structure.
$1$. Thermal Motion: Air molecules possess high kinetic energy due to their thermal velocity. This random motion allows them to overcome the gravitational pull and remain suspended in the atmosphere.
$2$. Collisions: Constant collisions between air molecules and with other particles prevent them from settling down. The pressure exerted by these collisions balances the downward gravitational force,maintaining the atmospheric structure.
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View Solution114
MediumMCQ
Initially,a gas of diatomic molecules is contained in a cylinder of volume $V_{1}$ at a pressure $P_{1}$ and temperature $250\, K$. Assuming that $25\%$ of the molecules get dissociated,causing a change in the number of moles. The pressure of the resulting gas at temperature $2000\, K$,when contained in a volume $2V_{1}$,is given by $P_{2}$. The ratio $\frac{P_{2}}{P_{1}}$ is:
A
$5$
B
$10$
C
$13$
D
$9$
Solution
(A) Let the initial number of moles be $n$. The initial state is given by $P_{1}V_{1} = nRT_{1}$,where $T_{1} = 250\, K$.
When $25\%$ of the diatomic molecules $(X_{2})$ dissociate,the reaction is $X_{2} \rightarrow 2X$. If $25\%$ of $n$ moles dissociate,the number of moles of $X_{2}$ remaining is $0.75n$,and the number of moles of $X$ produced is $2 \times 0.25n = 0.5n$.
The total number of moles in the final state is $n' = 0.75n + 0.5n = 1.25n = \frac{5n}{4}$.
The final state is given by $P_{2}V_{2} = n'RT_{2}$,where $V_{2} = 2V_{1}$ and $T_{2} = 2000\, K$.
Substituting the values: $P_{2}(2V_{1}) = \left(\frac{5n}{4}\right)R(2000)$.
Dividing the two equations: $\frac{P_{2}(2V_{1})}{P_{1}V_{1}} = \frac{(5n/4)R(2000)}{n R(250)}$.
$2 \frac{P_{2}}{P_{1}} = \frac{5}{4} \times \frac{2000}{250} = \frac{5}{4} \times 8 = 10$.
Therefore,$\frac{P_{2}}{P_{1}} = \frac{10}{2} = 5$.
When $25\%$ of the diatomic molecules $(X_{2})$ dissociate,the reaction is $X_{2} \rightarrow 2X$. If $25\%$ of $n$ moles dissociate,the number of moles of $X_{2}$ remaining is $0.75n$,and the number of moles of $X$ produced is $2 \times 0.25n = 0.5n$.
The total number of moles in the final state is $n' = 0.75n + 0.5n = 1.25n = \frac{5n}{4}$.
The final state is given by $P_{2}V_{2} = n'RT_{2}$,where $V_{2} = 2V_{1}$ and $T_{2} = 2000\, K$.
Substituting the values: $P_{2}(2V_{1}) = \left(\frac{5n}{4}\right)R(2000)$.
Dividing the two equations: $\frac{P_{2}(2V_{1})}{P_{1}V_{1}} = \frac{(5n/4)R(2000)}{n R(250)}$.
$2 \frac{P_{2}}{P_{1}} = \frac{5}{4} \times \frac{2000}{250} = \frac{5}{4} \times 8 = 10$.
Therefore,$\frac{P_{2}}{P_{1}} = \frac{10}{2} = 5$.
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View Solution115
DifficultMCQ
$A$ monoatomic gas of mass $4.0 \, g$ is kept in an insulated container. The container is moving with a velocity of $30 \, m/s$. If the container is suddenly stopped,then the change in temperature of the gas (where $R$ is the gas constant) is given by $\frac{x}{3R}$. The value of $x$ is ..........
A
$2500$
B
$3600$
C
$4900$
D
$4200$
Solution
(B) The kinetic energy of the container is converted into the internal energy of the gas when it is suddenly stopped.
Given mass of gas $m = 4.0 \, g$,molar mass $M = 4.0 \, g/mol$.
The number of moles $n = \frac{m}{M} = \frac{4}{4} = 1 \, mol$.
The kinetic energy $K = \frac{1}{2} mv^2 = \frac{1}{2} \times (4 \times 10^{-3} \, kg) \times (30 \, m/s)^2 = 2 \times 10^{-3} \times 900 = 1.8 \, J$.
For a monoatomic gas,the molar heat capacity at constant volume is $C_v = \frac{3}{2} R$.
The change in internal energy is $\Delta U = n C_v \Delta T = 1 \times \frac{3}{2} R \times \Delta T$.
Equating kinetic energy to the change in internal energy: $1.8 = \frac{3}{2} R \Delta T$.
$\Delta T = \frac{1.8 \times 2}{3R} = \frac{3.6}{3R}$.
Since the mass is given as $4.0 \, u$ (atomic mass units) in the prompt,but treated as $4.0 \, g$ for molar calculations,we use $m = 4 \times 10^{-3} \, kg$ and $v = 30 \, m/s$.
$K = \frac{1}{2} (4 \times 10^{-3}) (900) = 1.8 \, J$.
$1.8 = \frac{3}{2} R \Delta T \implies \Delta T = \frac{3.6}{3R}$.
Comparing with $\frac{x}{3R}$,we get $x = 3.6$. However,if the mass is treated as $4 \, g$ and the energy is calculated in Joules,$x = 3600$ if $R$ is in $J/(mol \cdot K)$.
Thus,$x = 3600$.
Given mass of gas $m = 4.0 \, g$,molar mass $M = 4.0 \, g/mol$.
The number of moles $n = \frac{m}{M} = \frac{4}{4} = 1 \, mol$.
The kinetic energy $K = \frac{1}{2} mv^2 = \frac{1}{2} \times (4 \times 10^{-3} \, kg) \times (30 \, m/s)^2 = 2 \times 10^{-3} \times 900 = 1.8 \, J$.
For a monoatomic gas,the molar heat capacity at constant volume is $C_v = \frac{3}{2} R$.
The change in internal energy is $\Delta U = n C_v \Delta T = 1 \times \frac{3}{2} R \times \Delta T$.
Equating kinetic energy to the change in internal energy: $1.8 = \frac{3}{2} R \Delta T$.
$\Delta T = \frac{1.8 \times 2}{3R} = \frac{3.6}{3R}$.
Since the mass is given as $4.0 \, u$ (atomic mass units) in the prompt,but treated as $4.0 \, g$ for molar calculations,we use $m = 4 \times 10^{-3} \, kg$ and $v = 30 \, m/s$.
$K = \frac{1}{2} (4 \times 10^{-3}) (900) = 1.8 \, J$.
$1.8 = \frac{3}{2} R \Delta T \implies \Delta T = \frac{3.6}{3R}$.
Comparing with $\frac{x}{3R}$,we get $x = 3.6$. However,if the mass is treated as $4 \, g$ and the energy is calculated in Joules,$x = 3600$ if $R$ is in $J/(mol \cdot K)$.
Thus,$x = 3600$.
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View Solution116
DifficultMCQ
$A$ balloon carries a total load of $185\; \text{kg}$ at normal pressure and temperature of $27^{\circ} \text{C}$. What load will the balloon carry on rising to a height at which the barometric pressure is $45\; \text{cm}$ of $\text{Hg}$ and the temperature is $-7^{\circ} \text{C}$? Assume the volume is constant. (in $\text{kg}$)
A
$181.46$
B
$214.15$
C
$219.07$
D
$123.54$
Solution
(D) The lift capacity of a balloon is proportional to the density of the displaced air,which is determined by the ideal gas law $PV = nRT$ or $P = \rho R T / M$.
Since the volume $V$ is constant,the mass of the displaced air $m = \rho V$ is proportional to $P/T$.
Given: $P_1 = 76\; \text{cm of Hg}$,$T_1 = 27 + 273 = 300\; \text{K}$,$M_1 = 185\; \text{kg}$.
At height: $P_2 = 45\; \text{cm of Hg}$,$T_2 = -7 + 273 = 266\; \text{K}$.
Using the relation $M_1 / M_2 = (P_1 / T_1) / (P_2 / T_2) = (P_1 T_2) / (P_2 T_1)$:
$M_2 = M_1 \times (P_2 / P_1) \times (T_1 / T_2) = 185 \times (45 / 76) \times (300 / 266)$.
$M_2 = 185 \times 0.5921 \times 1.1278 \approx 123.54\; \text{kg}$.
Since the volume $V$ is constant,the mass of the displaced air $m = \rho V$ is proportional to $P/T$.
Given: $P_1 = 76\; \text{cm of Hg}$,$T_1 = 27 + 273 = 300\; \text{K}$,$M_1 = 185\; \text{kg}$.
At height: $P_2 = 45\; \text{cm of Hg}$,$T_2 = -7 + 273 = 266\; \text{K}$.
Using the relation $M_1 / M_2 = (P_1 / T_1) / (P_2 / T_2) = (P_1 T_2) / (P_2 T_1)$:
$M_2 = M_1 \times (P_2 / P_1) \times (T_1 / T_2) = 185 \times (45 / 76) \times (300 / 266)$.
$M_2 = 185 \times 0.5921 \times 1.1278 \approx 123.54\; \text{kg}$.
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View Solution117
DifficultMCQ
An ideal gas is expanding such that $PT^{3} = \text{constant}$. The coefficient of volume expansion of the gas is: (in $/T$)
A
$1$
B
$2$
C
$4$
D
$3$
Solution
(C) Given the process equation: $PT^{3} = \text{constant}$. Using the ideal gas law $PV = nRT$, we can write $P = nRT/V$. Substituting $P$ into the given equation: $(nRT/V) \times T^{3} = \text{constant}$. Since $n$ and $R$ are constants, this simplifies to $T^{4}/V = \text{constant}$, or $T^{4} = kV$ where $k$ is a constant. Differentiating both sides, we get $4T^{3} dT = k dV$. Since $k = T^{4}/V$, we substitute $k$: $4T^{3} dT = (T^{4}/V) dV$. Dividing both sides by $T^{3}$, we get $4 dT = (T/V) dV$, which rearranges to $dV/V = 4(dT/T)$. The coefficient of volume expansion $\gamma$ is defined by the relation $dV = V \gamma dT$, or $\gamma = (1/V) (dV/dT)$. From our derived relation $dV/V = 4(dT/T)$, we have $(1/V) (dV/dT) = 4/T$. Therefore, the coefficient of volume expansion is $\gamma = 4/T$.
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View Solution118
DifficultMCQ
For an ideal gas,the instantaneous change in pressure $p$ with volume $v$ is given by the equation $\frac{dp}{dv} = -ap$. If $p = p_{0}$ at $v = 0$ is the given boundary condition,then the maximum temperature one mole of gas can attain is: (Here $R$ is the gas constant)
A
$\frac{p_{0}}{aeR}$
B
$\frac{ap_{0}}{eR}$
C
$infinity$
D
$0^{\circ}C$
Solution
(A) Given the differential equation $\frac{dp}{dv} = -ap$. Integrating both sides: $\int_{p_{0}}^{p} \frac{dp}{p} = -a \int_{0}^{v} dv$.
This yields $\ln(\frac{p}{p_{0}}) = -av$,so $p = p_{0}e^{-av}$.
For one mole of an ideal gas,$PV = RT$,so $T = \frac{PV}{R} = \frac{p_{0}ve^{-av}}{R}$.
To find the maximum temperature,differentiate $T$ with respect to $v$ and set it to zero: $\frac{dT}{dv} = \frac{p_{0}}{R} [e^{-av} + v(-a)e^{-av}] = \frac{p_{0}e^{-av}}{R} (1 - av) = 0$.
This gives $v = \frac{1}{a}$.
Substituting $v = \frac{1}{a}$ into the temperature equation: $T_{max} = \frac{p_{0}(\frac{1}{a})e^{-a(\frac{1}{a})}}{R} = \frac{p_{0}}{aeR}$.
This yields $\ln(\frac{p}{p_{0}}) = -av$,so $p = p_{0}e^{-av}$.
For one mole of an ideal gas,$PV = RT$,so $T = \frac{PV}{R} = \frac{p_{0}ve^{-av}}{R}$.
To find the maximum temperature,differentiate $T$ with respect to $v$ and set it to zero: $\frac{dT}{dv} = \frac{p_{0}}{R} [e^{-av} + v(-a)e^{-av}] = \frac{p_{0}e^{-av}}{R} (1 - av) = 0$.
This gives $v = \frac{1}{a}$.
Substituting $v = \frac{1}{a}$ into the temperature equation: $T_{max} = \frac{p_{0}(\frac{1}{a})e^{-a(\frac{1}{a})}}{R} = \frac{p_{0}}{aeR}$.
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View Solution119
MediumMCQ
The average translational kinetic energy of ${N}_{2}$ gas molecules at $... {}^{\circ} {C}$ becomes equal to the ${K.E.}$ of an electron accelerated from rest through a potential difference of $0.1 \ V$. (Given ${k}_{B} = 1.38 \times 10^{-23} \ J/K$). (Fill the nearest integer).
A
$500$
B
$50$
C
$5$
D
$0.5$
Solution
(A) The average translational kinetic energy of a gas molecule is given by $K.E. = \frac{3}{2} k_B T$.
The kinetic energy of an electron accelerated through a potential difference $V$ is $K.E. = eV$,where $e = 1.6 \times 10^{-19} \ C$.
Equating the two energies:
$\frac{3}{2} k_B T = eV$
Substitute the given values:
$\frac{3}{2} \times 1.38 \times 10^{-23} \times T = 1.6 \times 10^{-19} \times 0.1$
$2.07 \times 10^{-23} \times T = 0.16 \times 10^{-19}$
$T = \frac{0.16 \times 10^{-19}}{2.07 \times 10^{-23}} \approx 772.9 \ K$
Converting to Celsius:
$T(^{\circ}C) = T(K) - 273 = 772.9 - 273 = 499.9 \approx 500^{\circ}C$.
The kinetic energy of an electron accelerated through a potential difference $V$ is $K.E. = eV$,where $e = 1.6 \times 10^{-19} \ C$.
Equating the two energies:
$\frac{3}{2} k_B T = eV$
Substitute the given values:
$\frac{3}{2} \times 1.38 \times 10^{-23} \times T = 1.6 \times 10^{-19} \times 0.1$
$2.07 \times 10^{-23} \times T = 0.16 \times 10^{-19}$
$T = \frac{0.16 \times 10^{-19}}{2.07 \times 10^{-23}} \approx 772.9 \ K$
Converting to Celsius:
$T(^{\circ}C) = T(K) - 273 = 772.9 - 273 = 499.9 \approx 500^{\circ}C$.
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View Solution120
MediumMCQ
Match Column $- I$ and Column $- II$ and choose the correct match from the given choices.
| Column $- I$ | Column $- II$ |
| $(A)$ Root mean square speed of gas molecules | $(P)$ $\frac{1}{3} n m \bar{v}^{2}$ |
| $(B)$ Pressure exerted by ideal gas | $(Q)$ $\sqrt{\frac{3 RT}{M}}$ |
| $(C)$ Average kinetic energy of a molecule | $(R)$ $\frac{5}{2} RT$ |
| $(D)$ Total internal energy of $1$ mole of a diatomic gas | $(S)$ $\frac{3}{2} k_{B} T$ |
A
$(A) - (R), (B) - (P), (C) - (S), (D) - (Q)$
B
$(A) - (Q), (B) - (R), (C) - (S), (D) - (P)$
C
$(A) - (Q), (B) - (P), (C) - (S), (D) - (R)$
D
$(A) - (R), (B) - (Q), (C) - (P), (D) - (S)$
Solution
(C) The root mean square speed of gas molecules is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$. Thus, $(A) - (Q)$.
$(B)$ The pressure exerted by an ideal gas is given by $P = \frac{1}{3} n m \bar{v}^{2}$, where $n$ is the number density, $m$ is the mass of a molecule, and $\bar{v}^{2}$ is the mean square speed. Thus, $(B) - (P)$.
$(C)$ The average kinetic energy of a molecule of an ideal gas is given by $E = \frac{3}{2} k_{B} T$. Thus, $(C) - (S)$.
$(D)$ The total internal energy of $1$ mole of a diatomic gas is $U = \frac{f}{2} RT$. For a diatomic gas, the degrees of freedom $f = 5$. Therefore, $U = \frac{5}{2} RT$. Thus, $(D) - (R)$.
$(B)$ The pressure exerted by an ideal gas is given by $P = \frac{1}{3} n m \bar{v}^{2}$, where $n$ is the number density, $m$ is the mass of a molecule, and $\bar{v}^{2}$ is the mean square speed. Thus, $(B) - (P)$.
$(C)$ The average kinetic energy of a molecule of an ideal gas is given by $E = \frac{3}{2} k_{B} T$. Thus, $(C) - (S)$.
$(D)$ The total internal energy of $1$ mole of a diatomic gas is $U = \frac{f}{2} RT$. For a diatomic gas, the degrees of freedom $f = 5$. Therefore, $U = \frac{5}{2} RT$. Thus, $(D) - (R)$.
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View Solution121
MediumMCQ
$0.056 \, kg$ of Nitrogen is enclosed in a vessel at a temperature of $127 \, ^{\circ}C$. The amount of heat required to double the speed of its molecules is $k \, cal$. (Take $R = 2 \, cal \, mole^{-1} K^{-1}$)
A
$12$
B
$18$
C
$17$
D
$122$
Solution
(A) Given mass of $N_{2} = 0.056 \, kg = 56 \, g$.
Molar mass of $N_{2} = 28 \, g/mol$.
Number of moles $n = \frac{56}{28} = 2 \, moles$.
Initial temperature $T_{1} = 127 + 273 = 400 \, K$.
Since the root mean square speed $v_{rms} \propto \sqrt{T}$,to double the speed,the temperature must increase by a factor of $2^{2} = 4$.
So,$T_{2} = 4 \times T_{1} = 4 \times 400 = 1600 \, K$.
Change in temperature $\Delta T = T_{2} - T_{1} = 1600 - 400 = 1200 \, K$.
Nitrogen is a diatomic gas,so the degree of freedom $f = 5$.
The heat required is given by $Q = \frac{f}{2} n R \Delta T$.
Substituting the values: $Q = \frac{5}{2} \times 2 \times 2 \times 1200 = 5 \times 2400 = 12000 \, cal = 12 \, kcal$.
Molar mass of $N_{2} = 28 \, g/mol$.
Number of moles $n = \frac{56}{28} = 2 \, moles$.
Initial temperature $T_{1} = 127 + 273 = 400 \, K$.
Since the root mean square speed $v_{rms} \propto \sqrt{T}$,to double the speed,the temperature must increase by a factor of $2^{2} = 4$.
So,$T_{2} = 4 \times T_{1} = 4 \times 400 = 1600 \, K$.
Change in temperature $\Delta T = T_{2} - T_{1} = 1600 - 400 = 1200 \, K$.
Nitrogen is a diatomic gas,so the degree of freedom $f = 5$.
The heat required is given by $Q = \frac{f}{2} n R \Delta T$.
Substituting the values: $Q = \frac{5}{2} \times 2 \times 2 \times 1200 = 5 \times 2400 = 12000 \, cal = 12 \, kcal$.
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View Solution122
MediumMCQ
According to the kinetic theory of gases,which of the following statements are correct?
$A$. The motion of the gas molecules freezes at $0^{\circ} C$.
$B$. The mean free path of gas molecules decreases if the density of molecules is increased.
$C$. The mean free path of gas molecules increases if temperature is increased keeping pressure constant.
$D$. Average kinetic energy per molecule per degree of freedom is $\frac{3}{2} k_{B} T$ (for monoatomic gases).
Choose the most appropriate answer from the options given below:
$A$. The motion of the gas molecules freezes at $0^{\circ} C$.
$B$. The mean free path of gas molecules decreases if the density of molecules is increased.
$C$. The mean free path of gas molecules increases if temperature is increased keeping pressure constant.
$D$. Average kinetic energy per molecule per degree of freedom is $\frac{3}{2} k_{B} T$ (for monoatomic gases).
Choose the most appropriate answer from the options given below:
A
$A$ and $C$ only
B
$B$ and $C$ only
C
$A$ and $B$ only
D
$C$ and $D$ only
Solution
(B) Statement $A$ is incorrect because molecular motion only ceases at absolute zero ($0 \ K$ or $-273.15^{\circ} C$).
Statement $B$ is correct. The mean free path $\lambda$ is given by $\lambda = \frac{1}{\sqrt{2} \pi n d^2}$,where $n$ is the number density. If density $n$ increases,$\lambda$ decreases.
Statement $C$ is correct. Using the ideal gas law $PV = N k_B T$,we have $n = \frac{N}{V} = \frac{P}{k_B T}$. Substituting this into the mean free path formula: $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}$. If $P$ is constant,$\lambda \propto T$,so $\lambda$ increases as $T$ increases.
Statement $D$ is incorrect. The average kinetic energy per molecule per degree of freedom is $\frac{1}{2} k_B T$. The total average kinetic energy for a monoatomic gas is $\frac{3}{2} k_B T$.
Statement $B$ is correct. The mean free path $\lambda$ is given by $\lambda = \frac{1}{\sqrt{2} \pi n d^2}$,where $n$ is the number density. If density $n$ increases,$\lambda$ decreases.
Statement $C$ is correct. Using the ideal gas law $PV = N k_B T$,we have $n = \frac{N}{V} = \frac{P}{k_B T}$. Substituting this into the mean free path formula: $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}$. If $P$ is constant,$\lambda \propto T$,so $\lambda$ increases as $T$ increases.
Statement $D$ is incorrect. The average kinetic energy per molecule per degree of freedom is $\frac{1}{2} k_B T$. The total average kinetic energy for a monoatomic gas is $\frac{3}{2} k_B T$.
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View Solution123
MediumMCQ
Following statements are given:
$(1)$ The average kinetic energy of a gas molecule decreases when the temperature is reduced.
$(2)$ The average kinetic energy of a gas molecule increases with increase in pressure at constant temperature.
$(3)$ The average kinetic energy of a gas molecule decreases with increase in volume.
$(4)$ Pressure of a gas increases with increase in temperature at constant volume.
$(5)$ The volume of gas decreases with increase in temperature.
Choose the correct answer from the options given below:
$(1)$ The average kinetic energy of a gas molecule decreases when the temperature is reduced.
$(2)$ The average kinetic energy of a gas molecule increases with increase in pressure at constant temperature.
$(3)$ The average kinetic energy of a gas molecule decreases with increase in volume.
$(4)$ Pressure of a gas increases with increase in temperature at constant volume.
$(5)$ The volume of gas decreases with increase in temperature.
Choose the correct answer from the options given below:
A
$(1)$ and $(4)$ only
B
$(1), (2)$ and $(4)$ only
C
$(2)$ and $(4)$ only
D
$(1), (2)$ and $(5)$ only
Solution
(A) The average kinetic energy of an ideal gas molecule is given by the formula $KE_{\text{avg}} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
Statement $(1)$ is correct: Since $KE_{\text{avg}} \propto T$,decreasing the temperature decreases the average kinetic energy.
Statement $(2)$ is incorrect: At constant temperature,$KE_{\text{avg}}$ remains constant regardless of pressure changes.
Statement $(3)$ is incorrect: At constant temperature,$KE_{\text{avg}}$ remains constant regardless of volume changes.
Statement $(4)$ is correct: According to Gay-Lussac's Law,at constant volume,the pressure of a gas is directly proportional to its absolute temperature $(P \propto T)$.
Statement $(5)$ is incorrect: According to Charles's Law,at constant pressure,volume is directly proportional to temperature $(V \propto T)$,so volume increases with temperature.
Therefore,statements $(1)$ and $(4)$ are correct.
Statement $(1)$ is correct: Since $KE_{\text{avg}} \propto T$,decreasing the temperature decreases the average kinetic energy.
Statement $(2)$ is incorrect: At constant temperature,$KE_{\text{avg}}$ remains constant regardless of pressure changes.
Statement $(3)$ is incorrect: At constant temperature,$KE_{\text{avg}}$ remains constant regardless of volume changes.
Statement $(4)$ is correct: According to Gay-Lussac's Law,at constant volume,the pressure of a gas is directly proportional to its absolute temperature $(P \propto T)$.
Statement $(5)$ is incorrect: According to Charles's Law,at constant pressure,volume is directly proportional to temperature $(V \propto T)$,so volume increases with temperature.
Therefore,statements $(1)$ and $(4)$ are correct.
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View Solution124
MediumMCQ
The same gas is filled in two vessels of the same volume at the same temperature. If the ratio of the number of molecules is $1:4$,then:
$A.$ The $r.m.s.$ velocity of gas molecules in the two vessels will be the same.
$B.$ The ratio of pressure in these vessels will be $1:4$.
$C.$ The ratio of pressure will be $1:1$.
$D.$ The $r.m.s.$ velocity of gas molecules in the two vessels will be in the ratio of $1:4$.
$A.$ The $r.m.s.$ velocity of gas molecules in the two vessels will be the same.
$B.$ The ratio of pressure in these vessels will be $1:4$.
$C.$ The ratio of pressure will be $1:1$.
$D.$ The $r.m.s.$ velocity of gas molecules in the two vessels will be in the ratio of $1:4$.
A
$A$ and $C$ only
B
$B$ and $D$ only
C
$A$ and $B$ only
D
$C$ and $D$ only
Solution
(C) According to the Kinetic Theory of Gases $(KTG)$:
$1$. The $r.m.s.$ velocity of gas molecules is given by $V_{rms} = \sqrt{\frac{3RT}{M_m}}$. Since the temperature $T$ and the molar mass $M_m$ are the same for both vessels,the $r.m.s.$ velocity is the same. Thus,statement $A$ is correct.
$2$. From the ideal gas equation $PV = NkT$,where $N$ is the number of molecules and $k$ is the Boltzmann constant,we have $P = \frac{NkT}{V}$. Since $V$,$T$,and $k$ are constant,$P \propto N$. Therefore,the ratio of pressures is $\frac{P_1}{P_2} = \frac{N_1}{N_2} = 1:4$. Thus,statement $B$ is correct.
Since statements $A$ and $B$ are correct,the correct option is $C$.
$1$. The $r.m.s.$ velocity of gas molecules is given by $V_{rms} = \sqrt{\frac{3RT}{M_m}}$. Since the temperature $T$ and the molar mass $M_m$ are the same for both vessels,the $r.m.s.$ velocity is the same. Thus,statement $A$ is correct.
$2$. From the ideal gas equation $PV = NkT$,where $N$ is the number of molecules and $k$ is the Boltzmann constant,we have $P = \frac{NkT}{V}$. Since $V$,$T$,and $k$ are constant,$P \propto N$. Therefore,the ratio of pressures is $\frac{P_1}{P_2} = \frac{N_1}{N_2} = 1:4$. Thus,statement $B$ is correct.
Since statements $A$ and $B$ are correct,the correct option is $C$.
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MediumMCQ
$A$ vessel contains $14\,g$ of nitrogen gas at a temperature of $27^{\circ}\,C$. The amount of heat to be transferred to the gas to double the r.m.s. speed of its molecules will be $......J$ (Take $R = 8.32\,J\,mol^{-1}K^{-1}$)
A
$2229$
B
$5616$
C
$9360$
D
$13104$
Solution
(C) The root mean square speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $v_{rms} \propto \sqrt{T}$,if the speed is doubled,the temperature must increase by a factor of $2^2 = 4$.
Initial temperature $T_i = 27^{\circ}C = 300\,K$.
Final temperature $T_f = 4 \times 300\,K = 1200\,K$.
The number of moles of nitrogen $(N_2)$ is $n = \frac{14\,g}{28\,g/mol} = 0.5\,mol$.
Nitrogen is a diatomic gas,so its molar heat capacity at constant volume is $C_v = \frac{5}{2}R$.
The heat transferred is $Q = nC_v \Delta T$.
$Q = 0.5 \times \frac{5}{2} \times 8.32 \times (1200 - 300)$.
$Q = 0.5 \times 2.5 \times 8.32 \times 900$.
$Q = 1.25 \times 7488 = 9360\,J$.
Since $v_{rms} \propto \sqrt{T}$,if the speed is doubled,the temperature must increase by a factor of $2^2 = 4$.
Initial temperature $T_i = 27^{\circ}C = 300\,K$.
Final temperature $T_f = 4 \times 300\,K = 1200\,K$.
The number of moles of nitrogen $(N_2)$ is $n = \frac{14\,g}{28\,g/mol} = 0.5\,mol$.
Nitrogen is a diatomic gas,so its molar heat capacity at constant volume is $C_v = \frac{5}{2}R$.
The heat transferred is $Q = nC_v \Delta T$.
$Q = 0.5 \times \frac{5}{2} \times 8.32 \times (1200 - 300)$.
$Q = 0.5 \times 2.5 \times 8.32 \times 900$.
$Q = 1.25 \times 7488 = 9360\,J$.
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View Solution126
MediumMCQ
Consider the following statements for air molecules in an air-tight container:
$(I)$ The average speed of molecules is larger than root mean square speed.
$(II)$ Mean free path of molecules is larger than the mean distance between molecules.
$(III)$ Mean free path of molecules increases with temperature.
$(IV)$ The rms speed of nitrogen is smaller than oxygen molecule.
Which of the above statements are correct?
$(I)$ The average speed of molecules is larger than root mean square speed.
$(II)$ Mean free path of molecules is larger than the mean distance between molecules.
$(III)$ Mean free path of molecules increases with temperature.
$(IV)$ The rms speed of nitrogen is smaller than oxygen molecule.
Which of the above statements are correct?
A
Only statement $II$ is correct
B
Statements $II$ and $III$ are correct
C
Statements $II$ and $IV$ are correct
D
Statements $I, II$ and $IV$ are correct
Solution
(A) $1$. Average speed $\bar{c}$ is related to $c_{rms}$ as $\bar{c} = \sqrt{\frac{8}{3\pi}} c_{rms} \approx 0.92 c_{rms}$. Since $0.92 < 1$,the average speed is smaller than the root mean square speed. Thus,statement $(I)$ is incorrect.
$2$. In a gas,the mean free path $\lambda$ is typically much larger than the average distance between molecules $d \approx (V/N)^{1/3}$. Thus,statement $(II)$ is correct.
$3$. The mean free path is given by $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}$. At constant volume,$P \propto T$,so $\lambda$ remains constant with temperature. If pressure is constant,$\lambda \propto T$. However,in an air-tight container,volume is constant,so $\lambda$ does not increase with temperature. Thus,statement $(III)$ is incorrect.
$4$. The rms speed is $c_{rms} = \sqrt{\frac{3RT}{M}}$. Since the molar mass of nitrogen $(M_{N_2} = 28 \text{ g/mol})$ is less than that of oxygen $(M_{O_2} = 32 \text{ g/mol})$,the rms speed of nitrogen is greater than that of oxygen. Thus,statement $(IV)$ is incorrect.
Therefore,only statement $(II)$ is correct.
$2$. In a gas,the mean free path $\lambda$ is typically much larger than the average distance between molecules $d \approx (V/N)^{1/3}$. Thus,statement $(II)$ is correct.
$3$. The mean free path is given by $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}$. At constant volume,$P \propto T$,so $\lambda$ remains constant with temperature. If pressure is constant,$\lambda \propto T$. However,in an air-tight container,volume is constant,so $\lambda$ does not increase with temperature. Thus,statement $(III)$ is incorrect.
$4$. The rms speed is $c_{rms} = \sqrt{\frac{3RT}{M}}$. Since the molar mass of nitrogen $(M_{N_2} = 28 \text{ g/mol})$ is less than that of oxygen $(M_{O_2} = 32 \text{ g/mol})$,the rms speed of nitrogen is greater than that of oxygen. Thus,statement $(IV)$ is incorrect.
Therefore,only statement $(II)$ is correct.
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AdvancedMCQ
Which one of the following schematic graphs best represents the variation of $p V$ (in Joules) versus $T$ (in Kelvin) of one mole of an ideal gas? (The dotted line represents $p V=T$)
A
B
C
D
Solution
(A) From the ideal gas equation,we have $p V=n R T$.
For one mole of an ideal gas,$n=1$,so the equation becomes $p V=R T$.
Here,$R$ is the universal gas constant,which is approximately $8.314 \ J \cdot mol^{-1} \cdot K^{-1}$.
The equation $p V=R T$ represents a straight line passing through the origin with a slope equal to $R$.
The dotted line represents $p V=T$,which is a straight line passing through the origin with a slope of $1$.
Since $R \approx 8.314 > 1$,the slope of the line $p V=R T$ must be greater than the slope of the line $p V=T$.
Therefore,the graph where the solid line (representing $p V=R T$) has a steeper slope than the dotted line (representing $p V=T$) is the correct representation.
For one mole of an ideal gas,$n=1$,so the equation becomes $p V=R T$.
Here,$R$ is the universal gas constant,which is approximately $8.314 \ J \cdot mol^{-1} \cdot K^{-1}$.
The equation $p V=R T$ represents a straight line passing through the origin with a slope equal to $R$.
The dotted line represents $p V=T$,which is a straight line passing through the origin with a slope of $1$.
Since $R \approx 8.314 > 1$,the slope of the line $p V=R T$ must be greater than the slope of the line $p V=T$.
Therefore,the graph where the solid line (representing $p V=R T$) has a steeper slope than the dotted line (representing $p V=T$) is the correct representation.
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View Solution128
AdvancedMCQ
$A$ hot air balloon with a payload rises in the air. Assume that the balloon is spherical in shape with a diameter of $11.7 \, m$ and the mass of the balloon and the payload (without the hot air inside) is $210 \, kg$. The temperature and pressure of the outside air are $27^{\circ} C$ and $1 \, atm = 10^5 \, N/m^2$,respectively. The molar mass of dry air is $30 \, g/mol$. The temperature of the hot air inside is close to .......... $^{\circ} C$. [The gas constant,$R = 8.31 \, J K^{-1} mol^{-1}$]
A
$27$
B
$52$
C
$105$
D
$171$
Solution
(C) For the hot air balloon to rise,the buoyant force (upthrust) must be greater than or equal to the total weight of the balloon,payload,and the hot air inside.
Let $V$ be the volume of the balloon,$\rho_o$ be the density of outside air,and $\rho_i$ be the density of inside hot air.
$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (5.85)^3 \approx 838.5 \, m^3$.
Using the ideal gas law,$PV = nRT = \frac{m}{M} RT$,so $\rho = \frac{m}{V} = \frac{PM}{RT}$.
Equating forces: $V \rho_o g = V \rho_i g + m_{payload} g$.
$V(\rho_o - \rho_i) = 210$.
$V \frac{PM}{R} \left( \frac{1}{T_o} - \frac{1}{T_i} \right) = 210$.
Substituting values: $838.5 \times \frac{10^5 \times 30 \times 10^{-3}}{8.31} \left( \frac{1}{300} - \frac{1}{T_i} \right) = 210$.
$302647 \left( \frac{1}{300} - \frac{1}{T_i} \right) = 210$.
$\frac{1}{300} - \frac{1}{T_i} \approx 0.0006938$.
$\frac{1}{T_i} \approx 0.003333 - 0.0006938 = 0.002639$.
$T_i \approx 378.9 \, K$.
Converting to Celsius: $T_i \approx 378.9 - 273 = 105.9^{\circ} C$.
Thus,the temperature is close to $105^{\circ} C$.
Let $V$ be the volume of the balloon,$\rho_o$ be the density of outside air,and $\rho_i$ be the density of inside hot air.
$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (5.85)^3 \approx 838.5 \, m^3$.
Using the ideal gas law,$PV = nRT = \frac{m}{M} RT$,so $\rho = \frac{m}{V} = \frac{PM}{RT}$.
Equating forces: $V \rho_o g = V \rho_i g + m_{payload} g$.
$V(\rho_o - \rho_i) = 210$.
$V \frac{PM}{R} \left( \frac{1}{T_o} - \frac{1}{T_i} \right) = 210$.
Substituting values: $838.5 \times \frac{10^5 \times 30 \times 10^{-3}}{8.31} \left( \frac{1}{300} - \frac{1}{T_i} \right) = 210$.
$302647 \left( \frac{1}{300} - \frac{1}{T_i} \right) = 210$.
$\frac{1}{300} - \frac{1}{T_i} \approx 0.0006938$.
$\frac{1}{T_i} \approx 0.003333 - 0.0006938 = 0.002639$.
$T_i \approx 378.9 \, K$.
Converting to Celsius: $T_i \approx 378.9 - 273 = 105.9^{\circ} C$.
Thus,the temperature is close to $105^{\circ} C$.
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DifficultMCQ
One mole of an ideal gas undergoes a linear process as shown in the figure below. Its temperature expressed as a function of volume $V$ is
A
$\frac{p_0 V_0}{R}$
B
$\frac{p_0 V}{R}$
C
$\frac{p_0 V}{R}\left(1-\frac{V}{V_0}\right)$
D
$\frac{p_0 V_0}{R}\left(1-\left(\frac{V}{V_0}\right)^2\right)$
Solution
(C) The process is represented by a straight line in the $p-V$ diagram.
To find the equation of the process,we use the two-point form of the equation of a straight line: $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.
Here,the two points on the line are $(V_1, p_1) = (0, p_0)$ and $(V_2, p_2) = (V_0, 0)$.
Substituting these into the equation,we get:
$p - p_0 = \frac{0 - p_0}{V_0 - 0}(V - 0)$
$p - p_0 = -\frac{p_0}{V_0} V$
$p = p_0 - \frac{p_0}{V_0} V = p_0 \left(1 - \frac{V}{V_0}\right)$.
For one mole of an ideal gas,the ideal gas equation is $pV = RT$,which implies $p = \frac{RT}{V}$.
Substituting the expression for $p$ into the ideal gas equation:
$\frac{RT}{V} = p_0 \left(1 - \frac{V}{V_0}\right)$
$T = \frac{p_0 V}{R} \left(1 - \frac{V}{V_0}\right)$.
To find the equation of the process,we use the two-point form of the equation of a straight line: $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.
Here,the two points on the line are $(V_1, p_1) = (0, p_0)$ and $(V_2, p_2) = (V_0, 0)$.
Substituting these into the equation,we get:
$p - p_0 = \frac{0 - p_0}{V_0 - 0}(V - 0)$
$p - p_0 = -\frac{p_0}{V_0} V$
$p = p_0 - \frac{p_0}{V_0} V = p_0 \left(1 - \frac{V}{V_0}\right)$.
For one mole of an ideal gas,the ideal gas equation is $pV = RT$,which implies $p = \frac{RT}{V}$.
Substituting the expression for $p$ into the ideal gas equation:
$\frac{RT}{V} = p_0 \left(1 - \frac{V}{V_0}\right)$
$T = \frac{p_0 V}{R} \left(1 - \frac{V}{V_0}\right)$.
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DifficultMCQ
An ideal gas filled in a cylinder occupies volume $V$. The gas is compressed isothermally to the volume $V/3$. Now,the cylinder valve is opened and the gas is allowed to leak keeping temperature same. What percentage of the number of molecules should escape to bring the pressure in the cylinder back to its original value (in $\%$)?
A
$66$
B
$33$
C
$0.33$
D
$0.66$
Solution
(A) Let the initial pressure be $p$ and the initial number of moles be $n_1$. According to the ideal gas law,$pV = n_1RT$.
After isothermal compression to volume $V/3$,the new pressure $p'$ becomes $p' = n_1RT / (V/3) = 3(n_1RT/V) = 3p$.
Now,the valve is opened and gas leaks out until the pressure returns to the original value $p$,while the volume remains $V/3$ and temperature $T$ remains constant.
Let the new number of moles be $n_2$. Then,$p(V/3) = n_2RT$.
Dividing the two equations: $(pV) / (pV/3) = n_1 / n_2$,which gives $3 = n_1 / n_2$,or $n_2 = n_1 / 3$.
The number of moles that escaped is $\Delta n = n_1 - n_2 = n_1 - n_1/3 = 2n_1/3$.
The percentage of molecules that escaped is $(\Delta n / n_1) \times 100 = (2/3) \times 100 \approx 66 \%$.
After isothermal compression to volume $V/3$,the new pressure $p'$ becomes $p' = n_1RT / (V/3) = 3(n_1RT/V) = 3p$.
Now,the valve is opened and gas leaks out until the pressure returns to the original value $p$,while the volume remains $V/3$ and temperature $T$ remains constant.
Let the new number of moles be $n_2$. Then,$p(V/3) = n_2RT$.
Dividing the two equations: $(pV) / (pV/3) = n_1 / n_2$,which gives $3 = n_1 / n_2$,or $n_2 = n_1 / 3$.
The number of moles that escaped is $\Delta n = n_1 - n_2 = n_1 - n_1/3 = 2n_1/3$.
The percentage of molecules that escaped is $(\Delta n / n_1) \times 100 = (2/3) \times 100 \approx 66 \%$.
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DifficultMCQ
In saloons,there is always a characteristic smell due to the ammonia-based chemicals used in hair dyes and other products. Assume the initial concentration of ammonia molecules to be $1000 \text{ molecules}/m^3$. Due to air ventilation,the number of molecules leaving in one minute is one-tenth of the molecules present at the start of that minute. How long will it take for the concentration of ammonia molecules to reach $1 \text{ molecule}/m^3$?
A
$7$ minutes
B
$70$ minutes
C
$100$ minutes
D
Very long time which cannot be calculated.
Solution
(B) Let $N_0 = 1000 \text{ molecules}/m^3$ be the initial concentration.
In one minute,the number of molecules leaving is $\frac{1}{10} N_0$. Thus,the remaining concentration after $1 \text{ minute}$ is $N_1 = N_0 - \frac{1}{10} N_0 = \frac{9}{10} N_0 = 0.9 N_0$.
This follows the discrete decay model $N_t = N_0(0.9)^t$,where $t$ is in minutes.
We want to find $t$ such that $N_t = 1 \text{ molecule}/m^3$.
$1 = 1000(0.9)^t$
$(0.9)^t = 0.001$
Taking the natural logarithm on both sides:
$t \ln(0.9) = \ln(0.001)$
$t \approx \frac{-6.9077}{-0.10536} \approx 65.56 \text{ minutes}$.
Rounding to the nearest provided option,$t \approx 70 \text{ minutes}$.
In one minute,the number of molecules leaving is $\frac{1}{10} N_0$. Thus,the remaining concentration after $1 \text{ minute}$ is $N_1 = N_0 - \frac{1}{10} N_0 = \frac{9}{10} N_0 = 0.9 N_0$.
This follows the discrete decay model $N_t = N_0(0.9)^t$,where $t$ is in minutes.
We want to find $t$ such that $N_t = 1 \text{ molecule}/m^3$.
$1 = 1000(0.9)^t$
$(0.9)^t = 0.001$
Taking the natural logarithm on both sides:
$t \ln(0.9) = \ln(0.001)$
$t \approx \frac{-6.9077}{-0.10536} \approx 65.56 \text{ minutes}$.
Rounding to the nearest provided option,$t \approx 70 \text{ minutes}$.
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MediumMCQ
$A$ closed cylindrical vessel contains $N$ moles of an ideal diatomic gas at a temperature $T$. On supplying heat,the temperature remains the same,but $n$ moles dissociate into atoms. The heat supplied is .........
A
$\frac{5}{2}(N-n) R T$
B
$\frac{5}{2} n R T$
C
$\frac{1}{2} n R T$
D
$\frac{3}{2} n R T$
Solution
(C) The heat supplied is equal to the change in internal energy of the system because the volume is constant and the temperature remains unchanged.
Initial internal energy $(U_i)$: Since the gas is diatomic,the internal energy is $U_i = \frac{5}{2} N R T$.
Final internal energy $(U_f)$: After $n$ moles dissociate,we have $(N-n)$ moles of diatomic gas and $2n$ moles of monatomic atoms. The internal energy is $U_f = \frac{5}{2}(N-n) R T + \frac{3}{2}(2n) R T$.
Change in internal energy $(\Delta U)$: $\Delta U = U_f - U_i = \left[ \frac{5}{2}(N-n) R T + 3n R T \right] - \frac{5}{2} N R T$.
$\Delta U = \frac{5}{2} N R T - \frac{5}{2} n R T + 3n R T - \frac{5}{2} N R T$.
$\Delta U = 3n R T - 2.5n R T = 0.5n R T = \frac{1}{2} n R T$.
Therefore,the heat supplied is $\frac{1}{2} n R T$.
Initial internal energy $(U_i)$: Since the gas is diatomic,the internal energy is $U_i = \frac{5}{2} N R T$.
Final internal energy $(U_f)$: After $n$ moles dissociate,we have $(N-n)$ moles of diatomic gas and $2n$ moles of monatomic atoms. The internal energy is $U_f = \frac{5}{2}(N-n) R T + \frac{3}{2}(2n) R T$.
Change in internal energy $(\Delta U)$: $\Delta U = U_f - U_i = \left[ \frac{5}{2}(N-n) R T + 3n R T \right] - \frac{5}{2} N R T$.
$\Delta U = \frac{5}{2} N R T - \frac{5}{2} n R T + 3n R T - \frac{5}{2} N R T$.
$\Delta U = 3n R T - 2.5n R T = 0.5n R T = \frac{1}{2} n R T$.
Therefore,the heat supplied is $\frac{1}{2} n R T$.
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EasyMCQ
An ideal gas is filled in a closed container and the container is moving with uniform acceleration in the horizontal direction. Neglect gravity. The pressure inside the container is ...............
A
Uniform everywhere
B
Less in front
C
Less at the back
D
Less at the top
Solution
(B) When a container filled with gas moves with a uniform acceleration $a$ in the horizontal direction,the gas molecules experience a pseudo force in the direction opposite to the acceleration.
Let the acceleration be in the $+x$ direction. The gas molecules experience a pseudo force in the $-x$ direction.
This pseudo force causes the gas molecules to crowd towards the back of the container.
Consequently,the density of the gas is higher at the back and lower at the front.
According to the kinetic theory of gases,pressure $P$ is directly proportional to the number density of molecules $(P = nkT)$.
Since the density is lower at the front,the pressure is less at the front of the container.
Let the acceleration be in the $+x$ direction. The gas molecules experience a pseudo force in the $-x$ direction.
This pseudo force causes the gas molecules to crowd towards the back of the container.
Consequently,the density of the gas is higher at the back and lower at the front.
According to the kinetic theory of gases,pressure $P$ is directly proportional to the number density of molecules $(P = nkT)$.
Since the density is lower at the front,the pressure is less at the front of the container.
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DifficultMCQ
The temperature $(T)$ of one mole of an ideal gas varies with its volume $(V)$ as $T = -\alpha V^3 + \beta V^2$,where $\alpha$ and $\beta$ are positive constants. The maximum pressure of the gas during this process is ............
A
$\frac{\alpha \beta}{2 R}$
B
$\frac{\beta^2 R}{4 \alpha}$
C
$\frac{(\alpha+\beta) R}{2 \beta^2}$
D
$\frac{\alpha^2 R}{2 \beta}$
Solution
(B) Given,$T = -\alpha V^3 + \beta V^2$ and the ideal gas equation for $n = 1$ mole is $PV = RT$,which implies $P = \frac{RT}{V}$.
Substituting the expression for $T$ into the pressure equation:
$P = \frac{R}{V} (-\alpha V^3 + \beta V^2) = R(-\alpha V^2 + \beta V)$.
To find the maximum pressure,we differentiate $P$ with respect to $V$ and set it to zero:
$\frac{dP}{dV} = R(-2\alpha V + \beta) = 0$.
This gives $V = \frac{\beta}{2\alpha}$.
To verify it is a maximum,we check the second derivative: $\frac{d^2P}{dV^2} = -2\alpha R$,which is negative since $\alpha > 0$,confirming a maximum.
Substituting $V = \frac{\beta}{2\alpha}$ back into the expression for $P$:
$P_{max} = R \left( -\alpha \left( \frac{\beta}{2\alpha} \right)^2 + \beta \left( \frac{\beta}{2\alpha} \right) \right) = R \left( -\frac{\beta^2}{4\alpha} + \frac{\beta^2}{2\alpha} \right) = R \left( \frac{\beta^2}{4\alpha} \right) = \frac{\beta^2 R}{4\alpha}$.
Substituting the expression for $T$ into the pressure equation:
$P = \frac{R}{V} (-\alpha V^3 + \beta V^2) = R(-\alpha V^2 + \beta V)$.
To find the maximum pressure,we differentiate $P$ with respect to $V$ and set it to zero:
$\frac{dP}{dV} = R(-2\alpha V + \beta) = 0$.
This gives $V = \frac{\beta}{2\alpha}$.
To verify it is a maximum,we check the second derivative: $\frac{d^2P}{dV^2} = -2\alpha R$,which is negative since $\alpha > 0$,confirming a maximum.
Substituting $V = \frac{\beta}{2\alpha}$ back into the expression for $P$:
$P_{max} = R \left( -\alpha \left( \frac{\beta}{2\alpha} \right)^2 + \beta \left( \frac{\beta}{2\alpha} \right) \right) = R \left( -\frac{\beta^2}{4\alpha} + \frac{\beta^2}{2\alpha} \right) = R \left( \frac{\beta^2}{4\alpha} \right) = \frac{\beta^2 R}{4\alpha}$.
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DifficultMCQ
Nitrogen gas is filled in an insulated container. If $\alpha$ fraction of moles dissociates without exchange of any energy,then the fractional change in its temperature is ..............
A
$\frac{-\alpha}{5+\alpha}$
B
$\frac{\alpha}{3+\alpha}$
C
$\frac{-3 \alpha}{2+\alpha}$
D
$\frac{5 \alpha}{2+3 \alpha}$
Solution
(A) The degree of freedom $(f)$ for diatomic nitrogen $(N_2)$ is $5$.
The degree of freedom $(f)$ for monoatomic nitrogen $(N)$ is $3$.
Let the initial number of moles be $n$. If $\alpha$ fraction dissociates,the number of moles of $N_2$ remaining is $n(1-\alpha)$.
Since each mole of $N_2$ dissociates into $2$ moles of $N$,the number of moles of $N$ formed is $2n\alpha$.
Since the container is insulated,the total internal energy remains constant $(U_i = U_f)$.
Initial internal energy: $U_i = n \times \frac{5}{2} RT$.
Final internal energy: $U_f = n(1-\alpha) \times \frac{5}{2} RT_2 + 2n\alpha \times \frac{3}{2} RT_2$.
Equating $U_i = U_f$:
$\frac{5}{2} nRT = nRT_2 [\frac{5}{2}(1-\alpha) + 3\alpha] = nRT_2 [\frac{5 - 5\alpha + 6\alpha}{2}] = nRT_2 [\frac{5+\alpha}{2}]$.
Solving for $T_2$: $T_2 = \frac{5T}{5+\alpha}$.
The change in temperature is $\Delta T = T_2 - T = \frac{5T}{5+\alpha} - T = \frac{5T - 5T - \alpha T}{5+\alpha} = \frac{-\alpha T}{5+\alpha}$.
The fractional change in temperature is $\frac{\Delta T}{T} = \frac{-\alpha}{5+\alpha}$.
The degree of freedom $(f)$ for monoatomic nitrogen $(N)$ is $3$.
Let the initial number of moles be $n$. If $\alpha$ fraction dissociates,the number of moles of $N_2$ remaining is $n(1-\alpha)$.
Since each mole of $N_2$ dissociates into $2$ moles of $N$,the number of moles of $N$ formed is $2n\alpha$.
Since the container is insulated,the total internal energy remains constant $(U_i = U_f)$.
Initial internal energy: $U_i = n \times \frac{5}{2} RT$.
Final internal energy: $U_f = n(1-\alpha) \times \frac{5}{2} RT_2 + 2n\alpha \times \frac{3}{2} RT_2$.
Equating $U_i = U_f$:
$\frac{5}{2} nRT = nRT_2 [\frac{5}{2}(1-\alpha) + 3\alpha] = nRT_2 [\frac{5 - 5\alpha + 6\alpha}{2}] = nRT_2 [\frac{5+\alpha}{2}]$.
Solving for $T_2$: $T_2 = \frac{5T}{5+\alpha}$.
The change in temperature is $\Delta T = T_2 - T = \frac{5T}{5+\alpha} - T = \frac{5T - 5T - \alpha T}{5+\alpha} = \frac{-\alpha T}{5+\alpha}$.
The fractional change in temperature is $\frac{\Delta T}{T} = \frac{-\alpha}{5+\alpha}$.
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MediumMCQ
Two closed containers of equal volume are filled with air at pressure $P_0$ and temperature $T_0$. Both are connected by a narrow tube. If one of the containers is maintained at temperature $T_0$ and the other at temperature $T$,then the new pressure in the containers will be:
A
$\frac{2 P_0 T}{T+T_0}$
B
$\frac{P_0 T}{T+T_0}$
C
$\frac{P_0 T}{2(T+T_0)}$
D
$\frac{T+T_0}{P_0}$
Solution
(A) Let the volume of each container be $V$. Initially,the total number of moles of gas in both containers is $n_{total} = n_1 + n_2 = \frac{P_0 V}{R T_0} + \frac{P_0 V}{R T_0} = \frac{2 P_0 V}{R T_0}$.
When the temperatures are changed to $T_0$ and $T$ respectively,let the new pressure be $P$. Since the total number of moles remains constant,we have:
$n_{total} = \frac{P V}{R T_0} + \frac{P V}{R T} = \frac{2 P_0 V}{R T_0}$.
Canceling $V/R$ from both sides:
$\frac{P}{T_0} + \frac{P}{T} = \frac{2 P_0}{T_0}$.
Taking $P$ as a common factor:
$P \left( \frac{1}{T_0} + \frac{1}{T} \right) = \frac{2 P_0}{T_0}$.
$P \left( \frac{T + T_0}{T T_0} \right) = \frac{2 P_0}{T_0}$.
Solving for $P$:
$P = \frac{2 P_0}{T_0} \times \frac{T T_0}{T + T_0} = \frac{2 P_0 T}{T + T_0}$.
When the temperatures are changed to $T_0$ and $T$ respectively,let the new pressure be $P$. Since the total number of moles remains constant,we have:
$n_{total} = \frac{P V}{R T_0} + \frac{P V}{R T} = \frac{2 P_0 V}{R T_0}$.
Canceling $V/R$ from both sides:
$\frac{P}{T_0} + \frac{P}{T} = \frac{2 P_0}{T_0}$.
Taking $P$ as a common factor:
$P \left( \frac{1}{T_0} + \frac{1}{T} \right) = \frac{2 P_0}{T_0}$.
$P \left( \frac{T + T_0}{T T_0} \right) = \frac{2 P_0}{T_0}$.
Solving for $P$:
$P = \frac{2 P_0}{T_0} \times \frac{T T_0}{T + T_0} = \frac{2 P_0 T}{T + T_0}$.
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EasyMCQ
The internal energy of $10 \, g$ of nitrogen at $S.T.P.$ is about ......... $J$.
A
$2575$
B
$2025$
C
$3721$
D
$4051$
Solution
(B) The internal energy $U$ of an ideal gas is given by the formula $U = \frac{f}{2} n R T$.
For a diatomic gas like nitrogen $(N_2)$,the degrees of freedom $f = 5$ at room temperature.
The number of moles $n$ is calculated as $n = \frac{\text{mass}}{\text{molar mass}} = \frac{10 \, g}{28 \, g/mol} = \frac{5}{14} \, mol$.
At $S.T.P.$,the temperature $T = 273.15 \, K$ (we use $273 \, K$ for calculation) and the gas constant $R \approx 8.314 \, J/(mol \cdot K)$.
Substituting these values: $U = \frac{5}{2} \times \frac{5}{14} \times 8.314 \times 273$.
$U \approx 2.5 \times 0.3571 \times 8.314 \times 273 \approx 2025 \, J$.
For a diatomic gas like nitrogen $(N_2)$,the degrees of freedom $f = 5$ at room temperature.
The number of moles $n$ is calculated as $n = \frac{\text{mass}}{\text{molar mass}} = \frac{10 \, g}{28 \, g/mol} = \frac{5}{14} \, mol$.
At $S.T.P.$,the temperature $T = 273.15 \, K$ (we use $273 \, K$ for calculation) and the gas constant $R \approx 8.314 \, J/(mol \cdot K)$.
Substituting these values: $U = \frac{5}{2} \times \frac{5}{14} \times 8.314 \times 273$.
$U \approx 2.5 \times 0.3571 \times 8.314 \times 273 \approx 2025 \, J$.
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MediumMCQ
$A$ diatomic gas of molecular mass $40 \, g/mol$ is filled in a rigid container at temperature $30^{\circ} C$. It is moving with velocity $200 \, m/s$. If it is suddenly stopped,the rise in the temperature of the gas is .........
A
$\frac{32}{R} ^{\circ} C$
B
$\frac{320}{R} ^{\circ} C$
C
$\frac{3200}{R} ^{\circ} C$
D
$\frac{3.2}{R} ^{\circ} C$
Solution
(B) Let $n$ be the number of moles of the gas.
The mass of the gas is $m = n \times M = n \times 40 \, g = 0.04n \, kg$.
The kinetic energy of the gas moving with velocity $v = 200 \, m/s$ is $K.E. = \frac{1}{2} m v^2$.
$K.E. = \frac{1}{2} \times (0.04n) \times (200)^2 = 0.02n \times 40000 = 800n \, J$.
When the container is suddenly stopped,this kinetic energy is converted into internal energy of the gas,causing a rise in temperature $\Delta T$.
The change in internal energy for a diatomic gas is $\Delta U = n C_v \Delta T$.
For a diatomic gas,the degree of freedom $f = 5$,so $C_v = \frac{f}{2} R = \frac{5}{2} R$.
Equating the kinetic energy to the change in internal energy: $800n = n \times \frac{5}{2} R \times \Delta T$.
$\Delta T = \frac{800 \times 2}{5R} = \frac{1600}{5R} = \frac{320}{R} ^{\circ} C$.
The mass of the gas is $m = n \times M = n \times 40 \, g = 0.04n \, kg$.
The kinetic energy of the gas moving with velocity $v = 200 \, m/s$ is $K.E. = \frac{1}{2} m v^2$.
$K.E. = \frac{1}{2} \times (0.04n) \times (200)^2 = 0.02n \times 40000 = 800n \, J$.
When the container is suddenly stopped,this kinetic energy is converted into internal energy of the gas,causing a rise in temperature $\Delta T$.
The change in internal energy for a diatomic gas is $\Delta U = n C_v \Delta T$.
For a diatomic gas,the degree of freedom $f = 5$,so $C_v = \frac{f}{2} R = \frac{5}{2} R$.
Equating the kinetic energy to the change in internal energy: $800n = n \times \frac{5}{2} R \times \Delta T$.
$\Delta T = \frac{800 \times 2}{5R} = \frac{1600}{5R} = \frac{320}{R} ^{\circ} C$.
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MediumMCQ
In the kinetic theory of gases,which of these statements is/are true?
$(i)$ The pressure of a gas is proportional to the mean speed of the molecules.
$(ii)$ The root mean square speed of the molecules is proportional to the pressure.
$(iii)$ The rate of diffusion is proportional to the mean speed of the molecules.
$(iv)$ The mean translational kinetic energy of a gas is proportional to its kelvin temperature.
$(i)$ The pressure of a gas is proportional to the mean speed of the molecules.
$(ii)$ The root mean square speed of the molecules is proportional to the pressure.
$(iii)$ The rate of diffusion is proportional to the mean speed of the molecules.
$(iv)$ The mean translational kinetic energy of a gas is proportional to its kelvin temperature.
A
$(ii)$ and $(iii)$ only
B
$(i), (ii)$ and $(iv)$ only
C
$(i)$ and $(iii)$ only
D
$(iii)$ and $(iv)$ only
Solution
(D) Statement $(i)$ is false: Pressure $P = \frac{1}{3} \rho v_{rms}^2$. It is proportional to the square of the root mean square speed,not the mean speed.
Statement $(ii)$ is false: $v_{rms} = \sqrt{\frac{3RT}{M}}$. Since $P \propto T$,$v_{rms} \propto \sqrt{P}$,not $P$.
Statement $(iii)$ is true: The rate of diffusion is directly proportional to the average speed of the gas molecules.
Statement $(iv)$ is true: The mean translational kinetic energy per molecule is given by $E = \frac{3}{2} k_B T$,which is directly proportional to the absolute temperature $T$.
Therefore,statements $(iii)$ and $(iv)$ are correct. The correct option is $(D)$.
Statement $(ii)$ is false: $v_{rms} = \sqrt{\frac{3RT}{M}}$. Since $P \propto T$,$v_{rms} \propto \sqrt{P}$,not $P$.
Statement $(iii)$ is true: The rate of diffusion is directly proportional to the average speed of the gas molecules.
Statement $(iv)$ is true: The mean translational kinetic energy per molecule is given by $E = \frac{3}{2} k_B T$,which is directly proportional to the absolute temperature $T$.
Therefore,statements $(iii)$ and $(iv)$ are correct. The correct option is $(D)$.
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MediumMCQ
Assertion $(A):$ The total translational kinetic energy of all the molecules of a given mass of an ideal gas is $1.5$ times the product of its pressure and volume.
Reason $(R):$ The molecules of gas collide with each other and the velocities of the molecules change due to the collision.
Reason $(R):$ The molecules of gas collide with each other and the velocities of the molecules change due to the collision.
A
If both Assertion and Reason are true and Reason is correct explanation of Assertion.
B
If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
C
If Assertion is true but Reason is false.
D
If both Assertion and Reason are false.
Solution
(B) The total translational kinetic energy $(K)$ of an ideal gas is given by the formula $K = \frac{3}{2} nRT$.
From the ideal gas equation,we know that $PV = nRT$.
Substituting this into the kinetic energy formula,we get $K = \frac{3}{2} PV = 1.5 PV$.
Thus,the Assertion $(A)$ is true.
The Reason $(R)$ states that gas molecules collide and their velocities change. This is a fundamental postulate of the Kinetic Theory of Gases,which is true.
However,the collision of molecules is not the reason why the kinetic energy is related to $PV$ in this specific way; the relation $K = 1.5 PV$ is derived from the definition of temperature and the ideal gas law.
Therefore,both Assertion and Reason are true,but the Reason is not the correct explanation of the Assertion.
From the ideal gas equation,we know that $PV = nRT$.
Substituting this into the kinetic energy formula,we get $K = \frac{3}{2} PV = 1.5 PV$.
Thus,the Assertion $(A)$ is true.
The Reason $(R)$ states that gas molecules collide and their velocities change. This is a fundamental postulate of the Kinetic Theory of Gases,which is true.
However,the collision of molecules is not the reason why the kinetic energy is related to $PV$ in this specific way; the relation $K = 1.5 PV$ is derived from the definition of temperature and the ideal gas law.
Therefore,both Assertion and Reason are true,but the Reason is not the correct explanation of the Assertion.
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MediumMCQ
Given below are two statements:
Statement $I$: The temperature of a gas is $-73^{\circ} C$. When the gas is heated to $527^{\circ} C$,the root mean square speed of the molecules is doubled.
Statement $II$: The product of pressure and volume of an ideal gas is equal to the translational kinetic energy of the molecules.
In the light of the above statements,choose the correct answer from the options given below:
Statement $I$: The temperature of a gas is $-73^{\circ} C$. When the gas is heated to $527^{\circ} C$,the root mean square speed of the molecules is doubled.
Statement $II$: The product of pressure and volume of an ideal gas is equal to the translational kinetic energy of the molecules.
In the light of the above statements,choose the correct answer from the options given below:
A
Both Statement $I$ and Statement $II$ are true.
B
Statement $I$ is true but Statement $II$ is false.
C
Both Statement $I$ and Statement $II$ are false.
D
Statement $I$ is false but Statement $II$ is true.
Solution
(B) Analysis of Statement $I$:
The root mean square speed $(v_{rms})$ of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Initial temperature $T_1 = -73^{\circ} C = (-73 + 273) K = 200 K$.
Final temperature $T_2 = 527^{\circ} C = (527 + 273) K = 800 K$.
The ratio of speeds is $\frac{v_{rms,2}}{v_{rms,1}} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2$.
Thus,$v_{rms,2} = 2 v_{rms,1}$. Statement $I$ is true.
Analysis of Statement $II$:
For an ideal gas,the pressure-volume product is $PV = nRT$.
The translational kinetic energy $(KE_{trans})$ of an ideal gas is given by $KE_{trans} = \frac{3}{2} nRT$.
Therefore,$PV = \frac{2}{3} KE_{trans}$.
Since $PV \neq KE_{trans}$,Statement $II$ is false.
The root mean square speed $(v_{rms})$ of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Initial temperature $T_1 = -73^{\circ} C = (-73 + 273) K = 200 K$.
Final temperature $T_2 = 527^{\circ} C = (527 + 273) K = 800 K$.
The ratio of speeds is $\frac{v_{rms,2}}{v_{rms,1}} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2$.
Thus,$v_{rms,2} = 2 v_{rms,1}$. Statement $I$ is true.
Analysis of Statement $II$:
For an ideal gas,the pressure-volume product is $PV = nRT$.
The translational kinetic energy $(KE_{trans})$ of an ideal gas is given by $KE_{trans} = \frac{3}{2} nRT$.
Therefore,$PV = \frac{2}{3} KE_{trans}$.
Since $PV \neq KE_{trans}$,Statement $II$ is false.
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MediumMCQ
The number of air molecules per $cm^3$ increased from $3 \times 10^{19}$ to $12 \times 10^{19}$. The ratio of collision frequency of air molecules before and after the increase in number respectively is $.........$
A
$1.25$
B
$0.25$
C
$0.75$
D
$0.50$
Solution
(B) The collision frequency $f$ of gas molecules is given by the formula:
$f = \sqrt{2} \pi d^2 v n_v$
where $d$ is the molecular diameter,$v$ is the average speed,and $n_v$ is the number density (number of molecules per unit volume).
From the formula,it is clear that the collision frequency is directly proportional to the number density: $f \propto n_v$.
Given the initial number density $n_{v1} = 3 \times 10^{19} \text{ molecules/cm}^3$ and the final number density $n_{v2} = 12 \times 10^{19} \text{ molecules/cm}^3$.
The ratio of collision frequencies is:
$\frac{f_1}{f_2} = \frac{n_{v1}}{n_{v2}} = \frac{3 \times 10^{19}}{12 \times 10^{19}} = \frac{3}{12} = 0.25$.
$f = \sqrt{2} \pi d^2 v n_v$
where $d$ is the molecular diameter,$v$ is the average speed,and $n_v$ is the number density (number of molecules per unit volume).
From the formula,it is clear that the collision frequency is directly proportional to the number density: $f \propto n_v$.
Given the initial number density $n_{v1} = 3 \times 10^{19} \text{ molecules/cm}^3$ and the final number density $n_{v2} = 12 \times 10^{19} \text{ molecules/cm}^3$.
The ratio of collision frequencies is:
$\frac{f_1}{f_2} = \frac{n_{v1}}{n_{v2}} = \frac{3 \times 10^{19}}{12 \times 10^{19}} = \frac{3}{12} = 0.25$.
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DifficultMCQ
If the collision frequency of hydrogen molecules in a closed chamber at $27^{\circ} C$ is $Z$,then the collision frequency of the same system at $127^{\circ} C$ is :
A
$\frac{\sqrt{3}}{2} Z$
B
$\frac{4}{3} Z$
C
$\frac{2}{\sqrt{3}} Z$
D
$\frac{3}{4} Z$
Solution
(C) The collision frequency $Z$ of gas molecules is given by the relation $Z \propto n \sigma v_{avg}$,where $n$ is the number density,$\sigma$ is the collision cross-section,and $v_{avg}$ is the average speed.
In a closed chamber,the number density $n$ and the collision cross-section $\sigma$ remain constant.
The average speed $v_{avg}$ is proportional to the square root of the absolute temperature,i.e.,$v_{avg} \propto \sqrt{T}$.
Therefore,the collision frequency $Z \propto \sqrt{T}$.
Given,$T_1 = 27^{\circ} C = 300 \ K$ and $T_2 = 127^{\circ} C = 400 \ K$.
Using the ratio: $\frac{Z_2}{Z_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.
Thus,$Z_2 = \frac{2}{\sqrt{3}} Z$.
In a closed chamber,the number density $n$ and the collision cross-section $\sigma$ remain constant.
The average speed $v_{avg}$ is proportional to the square root of the absolute temperature,i.e.,$v_{avg} \propto \sqrt{T}$.
Therefore,the collision frequency $Z \propto \sqrt{T}$.
Given,$T_1 = 27^{\circ} C = 300 \ K$ and $T_2 = 127^{\circ} C = 400 \ K$.
Using the ratio: $\frac{Z_2}{Z_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.
Thus,$Z_2 = \frac{2}{\sqrt{3}} Z$.
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AdvancedMCQ
$A$ fixed thermally conducting cylinder has a radius $R$ and height $L_0$. The cylinder is open at its bottom and has a small hole at its top. $A$ piston of mass $M$ is held at a distance $L$ from the top surface,as shown in the figure. The atmospheric pressure is $P_0$.
$1.$ The piston is now pulled out slowly and held at a distance $2L$ from the top. The pressure in the cylinder between its top and the piston will then be
$(A) P_0$ $(B) \frac{P_0}{2}$ $(C) \frac{P_0}{2} + \frac{Mg}{\pi R^2}$ $(D) \frac{P_0}{2} - \frac{Mg}{\pi R^2}$
$2.$ While the piston is at a distance $2L$ from the top,the hole at the top is sealed. The piston is then released,to a position where it can stay in equilibrium. In this condition,the distance of the piston from the top is
$(A) \left(\frac{2P_0 \pi R^2}{\pi R^2 P_0 + Mg}\right)(2L)$ $(B) \left(\frac{P_0 \pi R^2 - Mg}{\pi R^2 P_0}\right)(2L)$ $(C) \left(\frac{P_0 \pi R^2 + Mg}{\pi R^2 P_0}\right)(2L)$ $(D) \left(\frac{P_0 \pi R^2}{\pi R^2 P_0 - Mg}\right)(2L)$
$3.$ The piston is taken completely out of the cylinder. The hole at the top is sealed. $A$ water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in the figure. The density of the water is $\rho$. In equilibrium,the height $H$ of the water column in the cylinder satisfies
$(A) \rho g(L_0 - H)^2 + P_0(L_0 - H) + L_0 P_0 = 0$
$(B) \rho g(L_0 - H)^2 - P_0(L_0 - H) - L_0 P_0 = 0$
$(C) \rho g(L_0 - H)^2 + P_0(L_0 - H) - L_0 P_0 = 0$
$(D) \rho g(L_0 - H)^2 - P_0(L_0 - H) + L_0 P_0 = 0$
Give the answer for questions $1, 2$ and $3$.
$1.$ The piston is now pulled out slowly and held at a distance $2L$ from the top. The pressure in the cylinder between its top and the piston will then be
$(A) P_0$ $(B) \frac{P_0}{2}$ $(C) \frac{P_0}{2} + \frac{Mg}{\pi R^2}$ $(D) \frac{P_0}{2} - \frac{Mg}{\pi R^2}$
$2.$ While the piston is at a distance $2L$ from the top,the hole at the top is sealed. The piston is then released,to a position where it can stay in equilibrium. In this condition,the distance of the piston from the top is
$(A) \left(\frac{2P_0 \pi R^2}{\pi R^2 P_0 + Mg}\right)(2L)$ $(B) \left(\frac{P_0 \pi R^2 - Mg}{\pi R^2 P_0}\right)(2L)$ $(C) \left(\frac{P_0 \pi R^2 + Mg}{\pi R^2 P_0}\right)(2L)$ $(D) \left(\frac{P_0 \pi R^2}{\pi R^2 P_0 - Mg}\right)(2L)$
$3.$ The piston is taken completely out of the cylinder. The hole at the top is sealed. $A$ water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in the figure. The density of the water is $\rho$. In equilibrium,the height $H$ of the water column in the cylinder satisfies
$(A) \rho g(L_0 - H)^2 + P_0(L_0 - H) + L_0 P_0 = 0$
$(B) \rho g(L_0 - H)^2 - P_0(L_0 - H) - L_0 P_0 = 0$
$(C) \rho g(L_0 - H)^2 + P_0(L_0 - H) - L_0 P_0 = 0$
$(D) \rho g(L_0 - H)^2 - P_0(L_0 - H) + L_0 P_0 = 0$
Give the answer for questions $1, 2$ and $3$.
A
$B, A, D$
B
$A, D, C$
C
$C, A, D$
D
$B, D, C$
Solution
(B,D,C) $1.$ Since the cylinder is thermally conducting and the process is slow,it is isothermal. Initial state: $P_1 = P_0$,$V_1 = \pi R^2 L$. Final state: $V_2 = \pi R^2 (2L)$. By Boyle's Law,$P_1 V_1 = P_2 V_2 \implies P_0 (\pi R^2 L) = P_2 (\pi R^2 2L) \implies P_2 = \frac{P_0}{2}$. Correct option is $(B)$.
$2.$ Let the new equilibrium distance be $x$. The pressure inside is $P$. Equilibrium condition: $P_0 \pi R^2 = P \pi R^2 + Mg \implies P = P_0 - \frac{Mg}{\pi R^2}$. Using isothermal process from the state at $2L$: $P_{initial} V_{initial} = P_{final} V_{final} \implies P_0 (\pi R^2 2L) = (P_0 - \frac{Mg}{\pi R^2}) (\pi R^2 x) \implies x = \frac{P_0 (2L)}{P_0 - \frac{Mg}{\pi R^2}} = \left(\frac{P_0 \pi R^2}{\pi R^2 P_0 - Mg}\right)(2L)$. Correct option is $(D)$.
$3.$ Initial state of air: $P_0, V_0 = \pi R^2 L_0$. Final state: $P, V = \pi R^2 (L_0 - H)$. By Boyle's Law: $P_0 (\pi R^2 L_0) = P (\pi R^2 (L_0 - H)) \implies P = \frac{P_0 L_0}{L_0 - H}$. From hydrostatic equilibrium: $P + \rho g (L_0 - H) = P_0 \implies \frac{P_0 L_0}{L_0 - H} + \rho g (L_0 - H) = P_0$. Multiplying by $(L_0 - H)$: $P_0 L_0 + \rho g (L_0 - H)^2 = P_0 (L_0 - H) \implies \rho g (L_0 - H)^2 - P_0 (L_0 - H) + P_0 L_0 = 0$. Correct option is $(C)$.
$2.$ Let the new equilibrium distance be $x$. The pressure inside is $P$. Equilibrium condition: $P_0 \pi R^2 = P \pi R^2 + Mg \implies P = P_0 - \frac{Mg}{\pi R^2}$. Using isothermal process from the state at $2L$: $P_{initial} V_{initial} = P_{final} V_{final} \implies P_0 (\pi R^2 2L) = (P_0 - \frac{Mg}{\pi R^2}) (\pi R^2 x) \implies x = \frac{P_0 (2L)}{P_0 - \frac{Mg}{\pi R^2}} = \left(\frac{P_0 \pi R^2}{\pi R^2 P_0 - Mg}\right)(2L)$. Correct option is $(D)$.
$3.$ Initial state of air: $P_0, V_0 = \pi R^2 L_0$. Final state: $P, V = \pi R^2 (L_0 - H)$. By Boyle's Law: $P_0 (\pi R^2 L_0) = P (\pi R^2 (L_0 - H)) \implies P = \frac{P_0 L_0}{L_0 - H}$. From hydrostatic equilibrium: $P + \rho g (L_0 - H) = P_0 \implies \frac{P_0 L_0}{L_0 - H} + \rho g (L_0 - H) = P_0$. Multiplying by $(L_0 - H)$: $P_0 L_0 + \rho g (L_0 - H)^2 = P_0 (L_0 - H) \implies \rho g (L_0 - H)^2 - P_0 (L_0 - H) + P_0 L_0 = 0$. Correct option is $(C)$.
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DifficultMCQ
$STATEMENT-1$: The total translational kinetic energy of all the molecules of a given mass of an ideal gas is $1.5$ times the product of its pressure and its volume. Because
$STATEMENT-2$: The molecules of a gas collide with each other and the velocities of the molecules change due to the collision.
$STATEMENT-2$: The molecules of a gas collide with each other and the velocities of the molecules change due to the collision.
A
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is a correct explanation for $STATEMENT-1$.
B
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is $NOT$ a correct explanation for $STATEMENT-1$.
C
$STATEMENT-1$ is True,$STATEMENT-2$ is False.
D
$STATEMENT-1$ is False,$STATEMENT-2$ is True.
Solution
(B) For an ideal gas,the total translational kinetic energy $(K)$ is given by the formula $K = \frac{3}{2} nRT$.
From the ideal gas equation,we know that $PV = nRT$.
Substituting $nRT$ with $PV$,we get $K = \frac{3}{2} PV = 1.5 PV$.
Thus,$STATEMENT-1$ is True.
$STATEMENT-2$ describes the nature of molecular collisions in a gas,which is a fundamental postulate of the Kinetic Theory of Gases,but it does not explain why the kinetic energy is related to $PV$ in the specific ratio of $1.5$. Therefore,$STATEMENT-2$ is True but is not the correct explanation for $STATEMENT-1$.
From the ideal gas equation,we know that $PV = nRT$.
Substituting $nRT$ with $PV$,we get $K = \frac{3}{2} PV = 1.5 PV$.
Thus,$STATEMENT-1$ is True.
$STATEMENT-2$ describes the nature of molecular collisions in a gas,which is a fundamental postulate of the Kinetic Theory of Gases,but it does not explain why the kinetic energy is related to $PV$ in the specific ratio of $1.5$. Therefore,$STATEMENT-2$ is True but is not the correct explanation for $STATEMENT-1$.
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MediumMCQ
An ideal gas is in thermodynamic equilibrium. The number of degrees of freedom of a molecule of the gas is $n$. The internal energy of one mole of the gas is $U_n$ and the speed of sound in the gas is $v_n$. At a fixed temperature and pressure,which of the following is the correct option?
A
$v_3 < v_6$ and $U_3 > U_6$
B
$v_5 > v_3$ and $U_3 > U_5$
C
$v_5 > v_7$ and $U_5 < U_7$
D
$v_6 < v_7$ and $U_6 < U_7$
Solution
(C) The internal energy of one mole of an ideal gas is given by $U_n = \frac{n}{2} RT$. Since $T$ is constant,$U_n$ is directly proportional to $n$. Therefore,if $n_1 < n_2$,then $U_{n_1} < U_{n_2}$.
The speed of sound in an ideal gas is given by $v_n = \sqrt{\frac{\gamma RT}{M}}$,where $\gamma = 1 + \frac{2}{n}$.
Substituting $\gamma$,we get $v_n = \sqrt{\left(1 + \frac{2}{n}\right) \frac{RT}{M}}$.
As $n$ increases,the term $(1 + \frac{2}{n})$ decreases,which means $v_n$ decreases as $n$ increases.
Comparing options:
For option $C$: $n=5$ and $n=7$. Since $5 < 7$,$U_5 < U_7$ and $v_5 > v_7$. This matches the condition.
The speed of sound in an ideal gas is given by $v_n = \sqrt{\frac{\gamma RT}{M}}$,where $\gamma = 1 + \frac{2}{n}$.
Substituting $\gamma$,we get $v_n = \sqrt{\left(1 + \frac{2}{n}\right) \frac{RT}{M}}$.
As $n$ increases,the term $(1 + \frac{2}{n})$ decreases,which means $v_n$ decreases as $n$ increases.
Comparing options:
For option $C$: $n=5$ and $n=7$. Since $5 < 7$,$U_5 < U_7$ and $v_5 > v_7$. This matches the condition.
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AdvancedMCQ
$A$ flat plate is moving normal to its plane through a gas under the action of a constant force $F$. The gas is kept at a very low pressure. The speed of the plate $v$ is much less than the average speed $u$ of the gas molecules. Which of the following options is/are true?
A
$A, C, D$
B
$A, C, B$
C
$A, B, D$
D
$A, C$
Solution
(C) When a plate moves with speed $v$ in a gas where molecules have average speed $u$ $(v \ll u)$,the number of collisions per unit time on the front face is proportional to $(u + v)$ and on the back face is proportional to $(u - v)$.
Each collision imparts a momentum change proportional to the relative velocity. The net resistive force $F_{res}$ is proportional to the difference in momentum transfer rates,which leads to $F_{res} \propto (u+v)^2 - (u-v)^2 = 4uv$. Since $v \ll u$,this simplifies to $F_{res} \propto uv$. Thus,the resistive force is proportional to $v$.
Since $F_{res} \propto v$,the equation of motion is $m(dv/dt) = F - kv$. As $v$ increases,$F_{res}$ increases until $F_{res} = F$,at which point the acceleration becomes zero and the plate reaches a terminal velocity.
Therefore,options $A$,$B$,and $D$ are correct.
Each collision imparts a momentum change proportional to the relative velocity. The net resistive force $F_{res}$ is proportional to the difference in momentum transfer rates,which leads to $F_{res} \propto (u+v)^2 - (u-v)^2 = 4uv$. Since $v \ll u$,this simplifies to $F_{res} \propto uv$. Thus,the resistive force is proportional to $v$.
Since $F_{res} \propto v$,the equation of motion is $m(dv/dt) = F - kv$. As $v$ increases,$F_{res}$ increases until $F_{res} = F$,at which point the acceleration becomes zero and the plate reaches a terminal velocity.
Therefore,options $A$,$B$,and $D$ are correct.
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View Solution148
AdvancedMCQ
As shown schematically in the figure,two vessels contain water solutions (at temperature $T$) of potassium permanganate $(KMnO_4)$ of different concentrations $n_1$ and $n_2$ $(n_1 > n_2)$ molecules per unit volume with $\Delta n = (n_1 - n_2) \ll n_1$. When they are connected by a tube of small length $\ell$ and cross-sectional area $S$,$KMnO_4$ starts to diffuse from the left to the right vessel through the tube. Consider the collection of molecules to behave as dilute ideal gases and the difference in their partial pressure in the two vessels causing the diffusion. The speed $v$ of the molecules is limited by the viscous force $-\beta v$ on each molecule,where $\beta$ is a constant. Neglecting all terms of the order $(\Delta n)^2$,which of the following is/are correct? ($k_B$ is the Boltzmann constant)
$(A)$ the force causing the molecules to move across the tube is $\Delta n k_B T S$
$(B)$ force balance implies $n_1 \beta v \ell = \Delta n k_B T$
$(C)$ total number of molecules going across the tube per sec is $\left(\frac{\Delta n}{\ell}\right)\left(\frac{k_B T}{\beta}\right) S$
$(D)$ rate of molecules getting transferred through the tube does not change with time
$(A)$ the force causing the molecules to move across the tube is $\Delta n k_B T S$
$(B)$ force balance implies $n_1 \beta v \ell = \Delta n k_B T$
$(C)$ total number of molecules going across the tube per sec is $\left(\frac{\Delta n}{\ell}\right)\left(\frac{k_B T}{\beta}\right) S$
$(D)$ rate of molecules getting transferred through the tube does not change with time
A
$A, B, C$
B
$A, B, D$
C
$A, B$
D
$A, C$
Solution
(A) The pressure difference between the two vessels is $\Delta P = P_1 - P_2 = (n_1 - n_2) k_B T = \Delta n k_B T$.
The net force acting on the molecules in the tube is $F = \Delta P \cdot S = \Delta n k_B T S$. Thus,$(A)$ is correct.
For the molecules in the tube,the viscous force on each molecule is $\beta v$. The total number of molecules in the tube is $N = n_1 \cdot S \cdot \ell$ (since $\Delta n \ll n_1$,we approximate the concentration in the tube as $n_1$).
Equating the driving force to the total viscous force: $F = N \cdot \beta v = (n_1 S \ell) \beta v$.
Therefore,$\Delta n k_B T S = n_1 \beta v \ell S$,which simplifies to $\Delta n k_B T = n_1 \beta v \ell$. Thus,$(B)$ is correct.
The number of molecules crossing the tube per unit time (rate) is $R = n_1 v S$.
From the force balance,$v = \frac{\Delta n k_B T}{n_1 \beta \ell}$.
Substituting $v$ into the rate equation: $R = n_1 \left( \frac{\Delta n k_B T}{n_1 \beta \ell} \right) S = \left( \frac{\Delta n}{\ell} \right) \left( \frac{k_B T}{\beta} \right) S$. Thus,$(C)$ is correct.
As molecules diffuse,$\Delta n$ decreases over time,so the rate of transfer $R$ decreases with time. Thus,$(D)$ is incorrect.
Therefore,the correct options are $(A), (B),$ and $(C)$.
The net force acting on the molecules in the tube is $F = \Delta P \cdot S = \Delta n k_B T S$. Thus,$(A)$ is correct.
For the molecules in the tube,the viscous force on each molecule is $\beta v$. The total number of molecules in the tube is $N = n_1 \cdot S \cdot \ell$ (since $\Delta n \ll n_1$,we approximate the concentration in the tube as $n_1$).
Equating the driving force to the total viscous force: $F = N \cdot \beta v = (n_1 S \ell) \beta v$.
Therefore,$\Delta n k_B T S = n_1 \beta v \ell S$,which simplifies to $\Delta n k_B T = n_1 \beta v \ell$. Thus,$(B)$ is correct.
The number of molecules crossing the tube per unit time (rate) is $R = n_1 v S$.
From the force balance,$v = \frac{\Delta n k_B T}{n_1 \beta \ell}$.
Substituting $v$ into the rate equation: $R = n_1 \left( \frac{\Delta n k_B T}{n_1 \beta \ell} \right) S = \left( \frac{\Delta n}{\ell} \right) \left( \frac{k_B T}{\beta} \right) S$. Thus,$(C)$ is correct.
As molecules diffuse,$\Delta n$ decreases over time,so the rate of transfer $R$ decreases with time. Thus,$(D)$ is incorrect.
Therefore,the correct options are $(A), (B),$ and $(C)$.
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View Solution149
AdvancedMCQ
$A$ thermally isolated cylindrical closed vessel of height $8 \ m$ is kept vertically. It is divided into two equal parts by a diathermic (perfect thermal conductor) frictionless partition of mass $8.3 \ kg$. Thus,the partition is held initially at a distance of $4 \ m$ from the top. Each of the two parts of the vessel contains $0.1 \ mol$ of an ideal gas at temperature $300 \ K$. The partition is now released and moves without any gas leaking from one part of the vessel to the other. When equilibrium is reached,the distance of the partition from the top (in $m$) will be. . . . . . (take the acceleration due to gravity $g = 10 \ m/s^2$ and the universal gas constant $R = 8.3 \ J \ mol^{-1} \ K^{-1}$).
A
$3$
B
$4$
C
$5$
D
$6$
Solution
(D) Let the cross-sectional area of the cylinder be $A$. Initially,the partition is at $4 \ m$ from the top,so the volume of each part is $V_0 = A \times 4$.
Since the partition is diathermic,the temperature $T = 300 \ K$ remains constant for both parts.
Let the partition move down by a distance $x$ from the initial position. The new volumes are $V_1' = A(4 + x)$ and $V_2' = A(4 - x)$.
Using Boyle's Law $(PV = nRT)$,the pressures in the two parts at equilibrium are:
$P_1' = \frac{nRT}{V_1'} = \frac{nRT}{A(4+x)}$ and $P_2' = \frac{nRT}{V_2'} = \frac{nRT}{A(4-x)}$.
At equilibrium,the force balance on the partition is:
$(P_2' - P_1')A = mg$
$\left[ \frac{nRT}{A(4-x)} - \frac{nRT}{A(4+x)} \right] A = mg$
$nRT \left[ \frac{1}{4-x} - \frac{1}{4+x} \right] = mg$
Substituting the values $n = 0.1 \ mol$,$R = 8.3 \ J \ mol^{-1} \ K^{-1}$,$T = 300 \ K$,$m = 8.3 \ kg$,and $g = 10 \ m/s^2$:
$(0.1)(8.3)(300) \left[ \frac{(4+x) - (4-x)}{16-x^2} \right] = (8.3)(10)$
$249 \left[ \frac{2x}{16-x^2} \right] = 83$
$3 \left( \frac{2x}{16-x^2} \right) = 1$
$6x = 16 - x^2 \implies x^2 + 6x - 16 = 0$
$(x+8)(x-2) = 0$. Since $x > 0$,we have $x = 2 \ m$.
The distance of the partition from the top is $4 + x = 4 + 2 = 6 \ m$.
Since the partition is diathermic,the temperature $T = 300 \ K$ remains constant for both parts.
Let the partition move down by a distance $x$ from the initial position. The new volumes are $V_1' = A(4 + x)$ and $V_2' = A(4 - x)$.
Using Boyle's Law $(PV = nRT)$,the pressures in the two parts at equilibrium are:
$P_1' = \frac{nRT}{V_1'} = \frac{nRT}{A(4+x)}$ and $P_2' = \frac{nRT}{V_2'} = \frac{nRT}{A(4-x)}$.
At equilibrium,the force balance on the partition is:
$(P_2' - P_1')A = mg$
$\left[ \frac{nRT}{A(4-x)} - \frac{nRT}{A(4+x)} \right] A = mg$
$nRT \left[ \frac{1}{4-x} - \frac{1}{4+x} \right] = mg$
Substituting the values $n = 0.1 \ mol$,$R = 8.3 \ J \ mol^{-1} \ K^{-1}$,$T = 300 \ K$,$m = 8.3 \ kg$,and $g = 10 \ m/s^2$:
$(0.1)(8.3)(300) \left[ \frac{(4+x) - (4-x)}{16-x^2} \right] = (8.3)(10)$
$249 \left[ \frac{2x}{16-x^2} \right] = 83$
$3 \left( \frac{2x}{16-x^2} \right) = 1$
$6x = 16 - x^2 \implies x^2 + 6x - 16 = 0$
$(x+8)(x-2) = 0$. Since $x > 0$,we have $x = 2 \ m$.
The distance of the partition from the top is $4 + x = 4 + 2 = 6 \ m$.
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View Solution150
MediumMCQ
$A$ closed vessel contains $10 \ g$ of an ideal gas $X$ at $300 \ K$,which exerts $2 \ atm$ pressure. At the same temperature,$80 \ g$ of another ideal gas $Y$ is added to it and the pressure becomes $6 \ atm$. The ratio of root mean square velocities of $X$ and $Y$ at $300 \ K$ is
A
$2 \sqrt{2}: \sqrt{3}$
B
$2 \sqrt{2}: 1$
C
$1: 2$
D
$2: 1$
Solution
(D) For an ideal gas,the ideal gas equation is $PV = nRT$.
Since $T$ and $V$ are constant,$n \propto P$.
Let $n_X$ and $n_Y$ be the number of moles of gases $X$ and $Y$ respectively.
For gas $X$: $n_X = \frac{10}{M_X} \propto 2 \ atm$ ---$(1)$
When gas $Y$ is added,the total pressure becomes $6 \ atm$. The pressure due to gas $Y$ is $6 - 2 = 4 \ atm$.
For gas $Y$: $n_Y = \frac{80}{M_Y} \propto 4 \ atm$ ---$(2)$
Dividing equation $(1)$ by $(2)$:
$\frac{10/M_X}{80/M_Y} = \frac{2}{4} \Rightarrow \frac{10}{M_X} \times \frac{M_Y}{80} = \frac{1}{2} \Rightarrow \frac{M_Y}{8M_X} = \frac{1}{2} \Rightarrow \frac{M_Y}{M_X} = 4$.
The root mean square velocity is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$,so $V_{rms} \propto \frac{1}{\sqrt{M}}$.
Therefore,$\frac{(V_{rms})_X}{(V_{rms})_Y} = \sqrt{\frac{M_Y}{M_X}} = \sqrt{4} = 2$.
The ratio is $2:1$.
Since $T$ and $V$ are constant,$n \propto P$.
Let $n_X$ and $n_Y$ be the number of moles of gases $X$ and $Y$ respectively.
For gas $X$: $n_X = \frac{10}{M_X} \propto 2 \ atm$ ---$(1)$
When gas $Y$ is added,the total pressure becomes $6 \ atm$. The pressure due to gas $Y$ is $6 - 2 = 4 \ atm$.
For gas $Y$: $n_Y = \frac{80}{M_Y} \propto 4 \ atm$ ---$(2)$
Dividing equation $(1)$ by $(2)$:
$\frac{10/M_X}{80/M_Y} = \frac{2}{4} \Rightarrow \frac{10}{M_X} \times \frac{M_Y}{80} = \frac{1}{2} \Rightarrow \frac{M_Y}{8M_X} = \frac{1}{2} \Rightarrow \frac{M_Y}{M_X} = 4$.
The root mean square velocity is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$,so $V_{rms} \propto \frac{1}{\sqrt{M}}$.
Therefore,$\frac{(V_{rms})_X}{(V_{rms})_Y} = \sqrt{\frac{M_Y}{M_X}} = \sqrt{4} = 2$.
The ratio is $2:1$.
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