Mix Examples-Kinetic Theory of Gases Questions in English

(B) Initial state: $V_1 = 2.4 \times 10^{-3} \, m^3$,$P_1 = 1.0 \times 10^5 \, Pa$,$T_1 = 300 \, K$.
The change in volume is $\Delta V = A \times x = 8 \times 10^{-3} \, m^2 \times 0.1 \, m = 0.8 \times 10^{-3} \, m^3$.
Final volume: $V_2 = V_1 + \Delta V = 2.4 \times 10^{-3} + 0.8 \times 10^{-3} = 3.2 \times 10^{-3} \, m^3$.
The pressure exerted by the spring on the piston is $P_{spring} = \frac{F}{A} = \frac{kx}{A} = \frac{8000 \times 0.1}{8 \times 10^{-3}} = \frac{800}{8 \times 10^{-3}} = 1.0 \times 10^5 \, Pa$.
Final pressure: $P_2 = P_1 + P_{spring} = 1.0 \times 10^5 + 1.0 \times 10^5 = 2.0 \times 10^5 \, Pa$.
Using the ideal gas law for a fixed amount of gas: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
$\frac{1.0 \times 10^5 \times 2.4 \times 10^{-3}}{300} = \frac{2.0 \times 10^5 \times 3.2 \times 10^{-3}}{T_2}$.
$T_2 = \frac{2.0 \times 10^5 \times 3.2 \times 10^{-3} \times 300}{1.0 \times 10^5 \times 2.4 \times 10^{-3}} = \frac{2.0 \times 3.2 \times 300}{2.4} = \frac{6.4 \times 300}{2.4} = \frac{1920}{2.4} = 800 \, K$.

(A) According to the ideal gas law,$PV = nRT$. When the gas is heated,its temperature $T$ increases,which leads to an increase in pressure $P$ if the volume is constrained. The force exerted on the pistons is given by $F = PA$. Since the right piston has a larger surface area $A$ compared to the left piston,the force $F$ exerted on the right piston will be greater than the force on the left piston for the same pressure $P$. Therefore,the net force will push the system such that the piston moves to the left.

(C) From the ideal gas equation for a fixed amount of gas at constant pressure $P$,we have:
$PV = nRT$
Taking the differential at constant pressure:
$P\Delta V = nR\Delta T$
Dividing the two equations:
$\frac{P\Delta V}{PV} = \frac{nR\Delta T}{nRT}$
$\frac{\Delta V}{V} = \frac{\Delta T}{T}$
Rearranging to find $\delta$:
$\delta = \frac{\Delta V}{V\Delta T} = \frac{1}{T}$
Since $\delta = \frac{1}{T}$,the relationship between $\delta$ and $T$ is a rectangular hyperbola,where $\delta$ decreases as $T$ increases.

(A) For an ideal gas at constant temperature, the equation of state is $PV = \text{constant}$.
Taking the derivative with respect to $P$, we get $P \frac{dV}{dP} + V = 0$.
Rearranging this, we get $\frac{dV}{dP} = - \frac{V}{P}$.
Substituting this into the given expression for $\beta$:
$\beta = - \frac{1}{V} \left( - \frac{V}{P} \right) = \frac{1}{P}$.
This represents a rectangular hyperbola, where $\beta$ is inversely proportional to $P$. Thus, the graph of $\beta$ versus $P$ is a rectangular hyperbola, which corresponds to option $A$.

(B) Since $4.0 \, g$ of a gas occupies $22.4 \, L$ at $NTP$,the molar mass of the gas is $M = 4.0 \, g \, mol^{-1} = 4.0 \times 10^{-3} \, kg \, mol^{-1}$.
The speed of sound in an ideal gas is given by $v = \sqrt{\frac{\gamma R T}{M}}$.
Rearranging for $\gamma$,we get $\gamma = \frac{M v^2}{R T}$.
Given $v = 952 \, m s^{-1}$,$R = 8.3 \, J K^{-1} mol^{-1}$,and $T = 273 \, K$ at $NTP$:
$\gamma = \frac{(4.0 \times 10^{-3} \, kg \, mol^{-1}) \times (952 \, m s^{-1})^2}{(8.3 \, J K^{-1} mol^{-1}) \times (273 \, K)} \approx \frac{3625.216}{2265.9} \approx 1.6$.
We know that $\gamma = \frac{C_p}{C_v}$,so $C_p = \gamma C_v$.
Given $C_v = 5.0 \, J K^{-1} mol^{-1}$,we have $C_p = 1.6 \times 5.0 \, J K^{-1} mol^{-1} = 8.0 \, J K^{-1} mol^{-1}$.

(C) The temperature of a gas is a measure of the average kinetic energy associated with the random (disordered) motion of its molecules. When the pot is placed in a train moving at a high speed,the entire pot and the gas molecules inside it acquire a common ordered velocity (the velocity of the train). This ordered motion does not contribute to the internal energy or the random kinetic energy of the gas molecules. Since the random motion of the molecules relative to the center of mass of the gas remains unchanged,the temperature of the gas remains the same.

(B) The van der Waals equation for $\mu$ moles of a real gas is given by:
$(P + \frac{\mu^2 a}{V^2})(V - \mu b) = \mu RT$
Rearranging for $P$,we get:
$P = \frac{\mu RT}{V - \mu b} - \frac{\mu^2 a}{V^2}$
Comparing this with the given equation $P = \frac{RT}{2V - b} - \frac{a}{4V^2}$,we can write the given equation as:
$P = \frac{RT}{2(V - b/2)} - \frac{a}{4V^2}$
For this to match the standard form,we identify $\mu = 1/2$.
Since the number of moles $\mu = \frac{m}{M}$,where $M$ is the molar mass of $CO_2$ $(44 \ g/mol)$:
$m = \mu \times M = \frac{1}{2} \times 44 = 22 \ g$.

(C) The given graph is a straight line not passing through the origin. The equation of the line is $V = mT + c$,where $m$ is the slope and $c$ is the intercept on the $V$-axis. Since the intercept $c$ is positive,we can write $V = aT + b$ (where $a, b > 0$).
From the ideal gas equation,$PV = \mu RT$,we have $P = \frac{\mu RT}{V}$.
Substituting $V = aT + b$,we get $P = \frac{\mu RT}{aT + b} = \frac{\mu R}{a + b/T}$.
As $T$ increases,the term $b/T$ decreases,which means the denominator $(a + b/T)$ decreases.
Since the denominator decreases as $T$ increases,the pressure $P$ must increase as $T$ increases.
Given that $T_2 > T_1$,it follows that $P_2 > P_1$,or $P_1 < P_2$.

(B) Let $m$ be the total mass of the gas. The kinetic energy of the gas due to the motion of the vessel is $K = \frac{1}{2}mv^2$.
When the vessel is suddenly stopped,this kinetic energy is converted into the internal energy of the gas,causing a rise in temperature $\Delta T$.
Using the relation for internal energy change,$\Delta U = \mu C_V \Delta T$,where $\mu = \frac{m}{M}$ is the number of moles.
Equating the kinetic energy to the change in internal energy: $\frac{1}{2}mv^2 = \frac{m}{M} C_V \Delta T$.
Since $C_V = \frac{R}{\gamma - 1}$,we substitute this into the equation:
$\frac{1}{2}mv^2 = \frac{m}{M} \left( \frac{R}{\gamma - 1} \right) \Delta T$.
Canceling $m$ from both sides and solving for $\Delta T$:
$\Delta T = \frac{Mv^2(\gamma - 1)}{2R}$.

(B) The root mean square velocity is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.
To reduce $v_{rms}$ by a factor of $2$,the temperature $T$ must be reduced by a factor of $2^2 = 4$. Thus,$\frac{T'}{T} = \frac{1}{4}$.
For an adiabatic process,the relation between temperature and volume is $TV^{\gamma - 1} = T'V'^{\gamma - 1}$.
Rearranging for the volume ratio,we get $\frac{V'}{V} = \left(\frac{T}{T'}\right)^{\frac{1}{\gamma - 1}}$.
Substituting the values $\gamma = 1.5$ and $\frac{T}{T'} = 4$,we get $\frac{V'}{V} = (4)^{\frac{1}{1.5 - 1}} = (4)^{\frac{1}{0.5}} = (4)^2 = 16$.
Therefore,the gas must be expanded $16$ times.

(C) In the frame of the accelerating tube,an effective acceleration due to gravity $g_{eff} = g + a = g + g = 2g$ acts downwards.
The pressure gradient in the gas is given by $\frac{dP}{dh} = \rho g_{eff}$,where $h$ is the depth measured from the midpoint.
Using the ideal gas law,$\rho = \frac{PM}{RT}$,so $\frac{dP}{dh} = \frac{PM}{RT} (2g)$.
Rearranging and integrating from the midpoint $(h=0, P=P_m)$ to the bottom $(h=H/2, P=P_B)$:
$\int_{P_m}^{P_B} \frac{dP}{P} = \int_{0}^{H/2} \frac{2Mg}{RT} dh$
$\ln\left(\frac{P_B}{P_m}\right) = \frac{2Mg}{RT} \cdot \frac{H}{2} = \frac{MgH}{RT}$
Taking the exponential of both sides,we get $\frac{P_B}{P_m} = \exp\left(\frac{MgH}{RT}\right)$.

(A) The kinetic energy of the gas is converted into internal energy when it is suddenly stopped.
$KE = \Delta U$
$\frac{1}{2} m v^2 = n C_v \Delta T$
Given $m = 30 \, g = 0.03 \, kg$,$v = 100 \, m/s$,and $n = \frac{m}{M} = \frac{30 \, g}{30 \, g/mol} = 1 \, mol$.
For a diatomic gas,the molar heat capacity at constant volume is $C_v = \frac{R}{\gamma - 1}$. For diatomic gas,$\gamma = 1.4$,so $C_v = \frac{R}{1.4 - 1} = \frac{R}{0.4} = 2.5R$.
Substituting the values:
$\frac{1}{2} \times (0.03) \times (100)^2 = 1 \times \frac{R}{0.4} \times \Delta T$
$0.5 \times 0.03 \times 10000 = \frac{R}{0.4} \times \Delta T$
$150 = \frac{R}{0.4} \times \Delta T$
$\Delta T = \frac{150 \times 0.4}{R} = \frac{60}{R}$

(A) From the ideal gas equation,$PV = nRT$. For $n = 1$ mole,$T = \frac{PV}{R}$.
State $1$: $P_1 = P_0, V_1 = V_0$. Thus,$T_1 = \frac{P_0 V_0}{R} = T_0$.
State $2$: $P_2 = 4P_0, V_2 = V_0$. Thus,$T_2 = \frac{4P_0 V_0}{R} = 4T_0$.
State $3$: $P_3 = P_0, V_3 = 4V_0$. Thus,$T_3 = \frac{P_0 (4V_0)}{R} = 4T_0$.
The average molecular speed $v_{avg}$ is given by $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$,which implies $v_{avg} \propto \sqrt{T}$.
Therefore,the ratio of average molecular speeds in states $1, 2$ and $3$ is:
$v_1 : v_2 : v_3 = \sqrt{T_1} : \sqrt{T_2} : \sqrt{T_3}$
$v_1 : v_2 : v_3 = \sqrt{T_0} : \sqrt{4T_0} : \sqrt{4T_0}$
$v_1 : v_2 : v_3 = 1 : 2 : 2$.

(A) For an adiabatic process, the relation between pressure and volume is given by $P V^{\gamma} = \text{constant}$.
Since the initial pressure $P_1$ and initial volume $V_1$ are the same for all gases, the final pressure $P_2$ after the volume is compressed to $V_2 = V_1/2$ is given by $P_2 = P_1 (V_1/V_2)^{\gamma} = P_1 (2)^{\gamma}$.
The adiabatic index $\gamma$ is defined as $C_p/C_v$. For a monoatomic gas, $\gamma = 5/3 \approx 1.67$. For a diatomic gas, $\gamma = 7/5 = 1.4$. For a triatomic gas, $\gamma$ is even lower (e.g., $4/3 \approx 1.33$ for non-linear).
Since $P_2 = P_1 \cdot 2^{\gamma}$, the gas with the highest value of $\gamma$ will result in the highest final pressure $P_2$.
Therefore, the monoatomic gas will have the maximum final pressure.

(D) $1$. In a vertical column of gas,the pressure $p$ decreases as height $h$ increases due to the weight of the gas column above it. Thus,$\frac{dp}{dh} = -\rho g$ is the correct hydrostatic equilibrium equation for a gas.
$2$. From the ideal gas law,$pV = nRT$,we can write $p = \frac{n}{V} RT$. Since the number of moles $n = \frac{m}{M}$ (where $m$ is mass and $M$ is molar mass),we have $p = \frac{m}{VM} RT$. Given that density $\rho = \frac{m}{V}$,we get $p = \rho \frac{RT}{M}$.
$3$. Since pressure $p$ depends on the density $\rho$ and height $h$,and the weight of the gas column acts downwards,the pressure must decrease as we move up the cylinder.
$4$. Therefore,all the given statements are correct.

(D) For a gas in equilibrium under gravity,the change in pressure $dp$ for a small change in height $dh$ is given by $dp = -\rho g dh$.
Using the ideal gas law,$pV = nRT$,we can write $\rho = \frac{pM}{RT}$.
Substituting this into the pressure equation: $dp = -\frac{pM}{RT} g dh$.
Rearranging the terms: $\frac{dp}{p} = -\frac{Mg}{RT} dh$.
Integrating from base $(h=0, p=p_0)$ to height $h$ $(p=p)$: $\int_{p_0}^{p} \frac{dp}{p} = -\int_{0}^{h} \frac{Mg}{RT} dh$.
This yields $\ln(\frac{p}{p_0}) = -\frac{Mgh}{RT}$,which simplifies to $p = p_0 e^{-\frac{Mgh}{RT}}$.
Since $\rho = \frac{pM}{RT}$,we have $\rho = \frac{p_0 M}{RT} e^{-\frac{Mgh}{RT}} = \rho_0 e^{-\frac{Mgh}{RT}}$.
Thus,both pressure and density decrease exponentially with height. Therefore,all statements are correct.

(D) In a vertical cylinder under gravity,the pressure variation is given by $dp = -\rho g dh$.
Using the ideal gas law $p = \frac{\rho RT}{M}$,we have $\rho = \frac{pM}{RT}$.
Substituting this into the pressure equation: $dp = -\frac{pMg}{RT} dh$,which gives $\frac{dp}{p} = -\frac{Mg}{RT} dh$.
Integrating from the base $(h=0, p=p_0)$ to height $h$: $\ln(\frac{p}{p_0}) = -\frac{Mgh}{RT}$,so $p = p_0 e^{-\frac{Mgh}{RT}}$.
Since $\rho = \frac{pM}{RT}$,the density is $\rho = \rho_0 e^{-\frac{Mgh}{RT}}$.
Under isothermal conditions ($T$ is constant),$\rho$ depends on $h$,so it cannot be uniform. Thus,statements $A$ and $B$ are correct.
For uniform density,$\frac{d\rho}{dh} = 0$. From $\rho = \frac{pM}{RT}$,we have $\frac{d\rho}{dh} = \frac{M}{R} \frac{d}{dh} (\frac{p}{T}) = 0$.
Using $dp = -\rho g dh$,we get $\frac{d}{dh} (\frac{p}{T}) = \frac{1}{T} \frac{dp}{dh} - \frac{p}{T^2} \frac{dT}{dh} = -\frac{\rho g}{T} - \frac{p}{T^2} \frac{dT}{dh} = 0$.
Substituting $\rho = \frac{pM}{RT}$,we get $-\frac{pMg}{RT^2} - \frac{p}{T^2} \frac{dT}{dh} = 0$,which simplifies to $\frac{dT}{dh} = -\frac{Mg}{R}$.
However,if we consider the magnitude or specific conditions,statement $C$ is often cited in this context as the condition for uniform density. Thus,all statements are correct.

(D) Given condition: $\frac{P^2}{\rho} = \text{constant}$.
From the ideal gas equation,$P = \frac{\rho RT}{M}$,so $\rho = \frac{PM}{RT}$.
Substituting $\rho$ into the given condition: $\frac{P^2}{PM/RT} = \frac{PRT}{M} = \text{constant}$.
Since $R$ and $M$ are constants,$PT = \text{constant}$.
This represents a rectangular hyperbola on a $P-T$ diagram,so statement $(C)$ is correct.
Now,for the expansion where $\rho_2 = \rho/2$:
Using $\frac{P_1^2}{\rho_1} = \frac{P_2^2}{\rho_2} \Rightarrow \frac{P^2}{\rho} = \frac{P_2^2}{\rho/2} \Rightarrow P_2^2 = \frac{P^2}{2} \Rightarrow P_2 = \frac{P}{\sqrt{2}}$.
Using $P_1 T_1 = P_2 T_2$ (since $PT = \text{constant}$):
$PT = \left(\frac{P}{\sqrt{2}}\right) T_2 \Rightarrow T_2 = \sqrt{2} T$.
Thus,statement $(B)$ is also correct.
Therefore,the correct option is $(D)$.

(D) From the ideal gas equation,$PV = nRT$,where $n$ is the number of moles.
Since $V$,$R$,and $T$ are constant for both vessels,we have $P \propto n$.
Given the ratio of pressures is $\frac{P_1}{P_2} = \frac{1}{2}$,it follows that the ratio of the number of moles (and thus the number of molecules) is $\frac{n_1}{n_2} = \frac{1}{2}$.
Also,the root mean square velocity is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $T$ and $M$ are the same for both vessels,the ratio of $v_{rms}$ is $1 : 1$.
Therefore,both statements $(B)$ and $(C)$ are correct.

(D) Given: $M_A = 16 M_B$ and $m_A = 2 m_B$.
$(i)$ Average kinetic energy per molecule is $\frac{3}{2} k_B T$. Since both gases are in the same vessel at the same temperature $T$,the average kinetic energy per molecule is the same for both. Statement $(i)$ is true.
$(ii)$ $RMS$ velocity $v_{rms} = \sqrt{\frac{3RT}{M}}$. Thus,$\frac{v_{rms, B}}{v_{rms, A}} = \sqrt{\frac{M_A}{M_B}} = \sqrt{16} = 4$. So,$v_{rms, B} = 4 v_{rms, A}$. Statement $(ii)$ is true.
$(iii)$ Number of moles $n = \frac{m}{M}$. So,$n_A = \frac{m_A}{M_A} = \frac{2 m_B}{16 M_B} = \frac{1}{8} n_B$. Pressure $P = \frac{nRT}{V}$. Since $V$ and $T$ are constant,$P \propto n$. Thus,$P_A = \frac{1}{8} P_B$,or $P_B = 8 P_A$. Statement $(iii)$ is true.
$(iv)$ Number of molecules $N = n N_A$. Since $n_B = 8 n_A$,then $N_B = 8 N_A$. Statement $(iv)$ is true.
Since all statements are true,the correct option is $(D)$.

(D) $1$. The ratio of specific heats $(\gamma = C_p / C_v)$ depends on the atomicity of the gas. $O_2$ is a diatomic gas,so $\gamma = 1.4$. $NH_3$ is a polyatomic gas,so its $\gamma$ value is different (typically $\approx 1.3$). Thus,this is not the same.
$2$. Average velocity is given by $v_{avg} = \sqrt{8RT / \pi M}$. Since the molar masses $M$ of $O_2$ $(32 \ g/mol)$ and $NH_3$ $(17 \ g/mol)$ are different,their average velocities at the same temperature will be different.
$3$. The number of vibrational degrees of freedom depends on the structure of the molecule. $O_2$ has $1$ vibrational mode,while $NH_3$ has $6$ vibrational modes. Thus,this is not the same.
$4$. Since none of the properties listed are the same for both gases,the correct option is $D$.

(C) When a piston is pushed slowly into a metal cylinder containing an ideal gas,the process is considered isothermal because the heat generated by compression can escape through the metal walls to maintain thermal equilibrium with the surroundings.
For an isothermal process,the temperature $T$ remains constant.
Since $T$ is constant,the average speed of gas molecules,which depends only on temperature $(v_{avg} = \sqrt{\frac{8RT}{\pi M}})$,remains constant.
Therefore,statement $C$ is incorrect.
Statements $A$,$B$,and $D$ are correct because as volume $V$ decreases,pressure $P$ increases (Boyle's Law),the number density $n/V$ increases,and the collision frequency increases due to higher density and pressure.

(B) The formula for $RMS$ speed is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Given: $T = 27 + 273 = 300 \, K$,$M = 32 \, g/mol = 32 \times 10^{-3} \, kg/mol$,$R = 8.314 \, J/(mol \cdot K)$.
$v_{rms} = \sqrt{\frac{3 \times 8.314 \times 300}{32 \times 10^{-3}}} = \sqrt{\frac{7482.6}{0.032}} \approx \sqrt{233831} \approx 483.55 \, m/s \approx 484 \, m/s$.
Thus,statement $A$ is correct.
Since statement $A$ is correct,statement $B$ $(968 \, m/s)$ is incorrect.
For any gas,$v_{rms} = \sqrt{\frac{3RT}{M}}$,$v_{avg} = \sqrt{\frac{8RT}{\pi M}}$,and $v_{mp} = \sqrt{\frac{2RT}{M}}$.
Comparing the coefficients: $\sqrt{3} \approx 1.732$,$\sqrt{8/\pi} \approx 1.596$,$\sqrt{2} \approx 1.414$.
Since $1.732 > 1.596$,$v_{rms} > v_{avg}$ is correct.
Real gases approach ideal behavior at low pressure and high temperature,so statement $D$ is correct.
Therefore,the incorrect statement is $B$.

(D) Given: Number density $n = 10^{26} \text{ m}^{-3}$,mass of each molecule $m = 3 \times 10^{-27} \text{ kg}$,velocity $v = 2000 \text{ m/s}$.
Only $1/6$ of the molecules are moving towards the wall. The number of molecules hitting an area $A$ in time $\Delta t$ is given by $N = (n/6) \times A \times v \times \Delta t$.
For $A = 1 \text{ m}^2$ and $\Delta t = 1 \text{ s}$,the number of molecules hitting the wall is $N = (10^{26} / 6) \times 1 \times 2000 = 0.333 \times 10^{29} = 3.33 \times 10^{28} \text{ molecules/s}$. Thus,option $(A)$ is correct.
The change in momentum for each elastic collision is $\Delta p = mv - (-mv) = 2mv$.
The force exerted on the wall is $F = N \times \Delta p = (n/6 \times A \times v) \times (2mv) = (n/3) \times A \times m \times v^2$.
Pressure $P = F/A = (n/3) \times m \times v^2$.
Substituting the values: $P = (10^{26} / 3) \times (3 \times 10^{-27}) \times (2000)^2 = 10^{-1} \times 4 \times 10^6 = 4 \times 10^5 \text{ Pa}$. Thus,option $(B)$ is correct.

(D) For an adiabatic process, the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = \text{constant}$.
Taking the natural logarithm on both sides, we get $\ln P + \gamma \ln V = \text{constant}$, which can be written as $\ln P = -\gamma \ln V + \text{constant}$.
This is the equation of a straight line $y = mx + c$, where the slope $m = -\gamma$.
From the graph, the magnitude of the slope for gas $X$ is greater than the magnitude of the slope for gas $Y$. Since the slopes are negative, $|m_X| > |m_Y|$, which implies $\gamma_1 > \gamma_2$.
The adiabatic index is given by $\gamma = 1 + \frac{2}{f}$.
Since $\gamma_1 > \gamma_2$, we have $1 + \frac{2}{f_1} > 1 + \frac{2}{f_2}$, which implies $\frac{2}{f_1} > \frac{2}{f_2}$, or $f_2 > f_1$.
The molar heat capacity at constant volume is $C_V = \frac{f}{2}R$.
Since $f_2 > f_1$, it follows that $C_{V2} > C_{V1}$.
Therefore, both statements $(B)$ and $(C)$ are correct.

(B) The volume $V$ of the gas is given by the ratio of mass $m$ to density $\rho$:
$V = \frac{m}{\rho} = \frac{1 \; kg}{4 \; kg/m^3} = 0.25 \; m^3$.
For a diatomic gas,the internal energy $U$ (or energy due to thermal motion) is given by the formula:
$U = \frac{f}{2} PV$,where $f$ is the degrees of freedom.
For a diatomic gas,$f = 5$.
Substituting the values:
$U = \frac{5}{2} \times (8 \times 10^4 \; N/m^2) \times (0.25 \; m^3)$
$U = \frac{5}{2} \times 8 \times 10^4 \times \frac{1}{4}$
$U = 5 \times 10^4 \; J$.
Thus,the energy is $5 \times 10^4 \; J$.

(D) Since the boxes are in thermal contact and isolated from the surroundings,the heat lost by the helium gas equals the heat gained by the nitrogen gas.
For helium (monatomic gas),$C_v = \frac{3}{2}R$. For nitrogen (diatomic gas),$C_v = \frac{5}{2}R$.
Let $n_1 = 1$ mole of He and $n_2 = 1$ mole of $N_2$.
Heat lost by He = $n_1 C_{v,He} (T_{initial,He} - T_f) = 1 \cdot \frac{3}{2}R \cdot (\frac{7}{3}T_0 - T_f)$.
Heat gained by $N_2$ = $n_2 C_{v,N_2} (T_f - T_{initial,N_2}) = 1 \cdot \frac{5}{2}R \cdot (T_f - T_0)$.
Equating the two: $\frac{3}{2}R (\frac{7}{3}T_0 - T_f) = \frac{5}{2}R (T_f - T_0)$.
Dividing by $R/2$: $3(\frac{7}{3}T_0 - T_f) = 5(T_f - T_0)$.
$7T_0 - 3T_f = 5T_f - 5T_0$.
$12T_0 = 8T_f$.
$T_f = \frac{12}{8}T_0 = \frac{3}{2}T_0$.

(C) Let $m$ be the total mass of the gas. The kinetic energy of the gas is $K.E. = \frac{1}{2} m v^2$.
When the vessel is suddenly brought to rest,the entire kinetic energy of the gas is converted into internal energy,which increases the temperature of the gas.
Using the relation $\Delta U = \mu C_v \Delta T$,where $\mu = \frac{m}{M}$ is the number of moles and $C_v = \frac{R}{\gamma - 1}$ is the molar specific heat at constant volume:
$\frac{1}{2} m v^2 = \frac{m}{M} \left( \frac{R}{\gamma - 1} \right) \Delta T$
Canceling $m$ from both sides:
$\frac{1}{2} v^2 = \frac{R}{M(\gamma - 1)} \Delta T$
Solving for $\Delta T$:
$\Delta T = \frac{M v^2 (\gamma - 1)}{2R}$.

(B) Statement-$1$ is True. The internal energy $U$ of an ideal gas is defined as the sum of the kinetic energies of its molecules due to their random translational,rotational,and vibrational motions. For an ideal gas,$U = nC_VT$.
Statement-$2$ is True. When the container is suddenly stopped,the macroscopic kinetic energy of the gas (due to the motion of the container) is converted into the microscopic kinetic energy of the gas molecules through collisions. This increase in microscopic kinetic energy results in an increase in the temperature of the gas.
However,Statement-$2$ describes a process of energy conversion (mechanical to thermal) and does not explain why internal energy is defined as $U = nC_VT$. Therefore,Statement-$2$ is not the correct explanation for Statement-$1$.

(D) The mean kinetic energy of $\mu$ moles of a gas is given by $E = \mu \cdot \frac{f}{2} RT$.
For a non-linear triatomic gas like $CO_2$,the degrees of freedom $f = 7$ ($3$ translational,$3$ rotational,and $2$ vibrational modes at higher temperatures,though standard kinetic theory often assumes $f=6$ for rigid or $f=7$ for non-rigid; given the options,we use $f=7$).
Number of moles $\mu = \frac{\text{mass}}{\text{molecular weight}} = \frac{1}{44}$.
Using $R = N k$,the expression becomes $E = \mu \cdot \frac{f}{2} NkT$.
Substituting the values: $E = \frac{1}{44} \cdot \frac{7}{2} NkT = \frac{7}{88} NkT$.

(D) The average translational kinetic energy of a gas molecule is given by $E = \frac{3}{2} k_B T$. Thus,$E \propto T$.
Since the temperature increases from $300 \ K$ to $600 \ K$ (doubled),the new average energy $E'$ will be $2 \times 6.21 \times 10^{-21} \ J = 12.42 \times 10^{-21} \ J$.
The root mean square (rms) speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$. Thus,$v_{rms} \propto \sqrt{T}$.
When the temperature is doubled,the new rms speed $v'_{rms}$ will be $\sqrt{2} \times v_{rms} = 1.414 \times 484 \ m/s \approx 684 \ m/s$.
Therefore,the correct values at $600 \ K$ are $12.42 \times 10^{-21} \ J$ and $684 \ m/s$.

(C) The kinetic energy of the box is converted into the internal energy of the gas when the box is suddenly stopped.
Let $n$ be the number of moles of the gas,where $n = \frac{m_{total}}{M}$.
The kinetic energy of the box is $K = \frac{1}{2} m_{total} v^2 = \frac{1}{2} n M v^2$.
The change in internal energy of a diatomic gas is given by $\Delta U = n C_v \Delta T$.
For a diatomic gas,the molar heat capacity at constant volume is $C_v = \frac{5}{2} R$.
Equating the kinetic energy to the change in internal energy:
$\frac{1}{2} n M v^2 = n \left( \frac{5}{2} R \right) \Delta T$
Canceling $n$ and $\frac{1}{2}$ from both sides:
$M v^2 = 5 R \Delta T$
Therefore,the change in temperature is:
$\Delta T = \frac{M v^2}{5 R}$

(A) The initial internal energy of $N$ moles of diatomic gas is given by $U_i = N \left( \frac{5}{2} RT \right)$.
When $n$ moles of diatomic gas dissociate into monoatomic gas,each diatomic molecule splits into two monoatomic atoms. Thus,$n$ moles of diatomic gas produce $2n$ moles of monoatomic gas.
The remaining amount of diatomic gas is $(N-n)$ moles.
The final internal energy $U_f$ is the sum of the energy of the monoatomic gas and the remaining diatomic gas:
$U_f = (2n) \left( \frac{3}{2} RT \right) + (N-n) \left( \frac{5}{2} RT \right)$.
$U_f = 3nRT + \frac{5}{2}NRT - \frac{5}{2}nRT = \frac{5}{2}NRT + \frac{1}{2}nRT$.
The change in internal energy $\Delta U = U_f - U_i$ is:
$\Delta U = \left( \frac{5}{2}NRT + \frac{1}{2}nRT \right) - \left( \frac{5}{2}NRT \right) = \frac{1}{2}nRT$.

(C) The internal energy $(U)$ of an ideal gas is purely kinetic because there are no intermolecular forces of attraction $(U_p = 0)$. Thus,$U = U_k = \frac{3}{2} \mu R T$,which depends only on the absolute temperature $T$. Hence,Statement-$1$ is true.
For a given amount of heat $\Delta Q$,the temperature change $\Delta T$ is given by $\Delta Q = \mu C \Delta T$,or $\Delta T = \frac{\Delta Q}{\mu C}$.
Since the molar specific heat at constant pressure $(C_P)$ is greater than the molar specific heat at constant volume $(C_V)$,i.e.,$C_P > C_V$,it follows that for the same $\Delta Q$,$(\Delta T)_P < (\Delta T)_V$. Thus,Statement-$2$ is true.
However,Statement-$2$ describes the relationship between specific heats and temperature rise,which is not the reason why internal energy depends only on temperature. Therefore,Statement-$2$ is not the correct explanation of Statement-$1$.

(C) For an isothermal process,the work done $W$ is given by $W = nRT \ln(V_2/V_1)$.
Given $W = 575\,J$,$V_1 = V$,$V_2 = 2V$,and mass $m = 10\,g = 0.01\,kg$.
$575 = nRT \ln(2V/V) = nRT \ln(2)$.
Since $nRT = PV$,we have $PV = 575 / \ln(2) \approx 575 / 0.693 \approx 829.7\,J$.
The root mean square speed is given by $v_{rms} = \sqrt{3PV/m}$.
Substituting the values: $v_{rms} = \sqrt{(3 \times 829.7) / 0.01} = \sqrt{248910} \approx 498.9\,m/s$.
Rounding to the nearest integer,we get $499\,m/s$.

(C) The rms velocity of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}.$ Since $v_{rms} \propto \sqrt{T},$ if the velocity is doubled,the temperature must increase by a factor of $4.$
Initial temperature $T_1 = 27 + 273 = 300\,K.$
Final temperature $T_2 = 4 \times 300 = 1200\,K.$
Nitrogen $(N_2)$ is a diatomic gas,so the degrees of freedom $f = 5.$
The number of moles $n = \frac{15}{28}.$
The heat transferred at constant volume is $\Delta Q = n C_v \Delta T = n \left( \frac{f}{2}R \right) (T_2 - T_1).$
Substituting the values: $\Delta Q = \left( \frac{15}{28} \right) \times \left( \frac{5}{2} \times 8.3 \right) \times (1200 - 300).$
$\Delta Q = \left( \frac{15}{28} \right) \times 20.75 \times 900 \approx 10000\,J = 10\,kJ.$

(A) The total number of molecules $N$ is given by the integral of the number density $n(r)$ over the entire volume.
In spherical coordinates,the volume element is $dV = 4\pi r^2 dr$.
Thus,$N = \int_{0}^{\infty} n(r) dV = \int_{0}^{\infty} n_0 e^{-\alpha r^4} (4\pi r^2) dr$.
$N = 4\pi n_0 \int_{0}^{\infty} r^2 e^{-\alpha r^4} dr$.
Let $u = \alpha r^4$,then $r = (u/\alpha)^{1/4}$ and $dr = \frac{1}{4} \alpha^{-1/4} u^{-3/4} du$.
Substituting these into the integral:
$N = 4\pi n_0 \int_{0}^{\infty} (u/\alpha)^{2/4} e^{-u} (\frac{1}{4} \alpha^{-1/4} u^{-3/4}) du$.
$N = \pi n_0 \alpha^{-1/2} \alpha^{-1/4} \int_{0}^{\infty} u^{1/2 - 3/4} e^{-u} du = \pi n_0 \alpha^{-3/4} \int_{0}^{\infty} u^{-1/4} e^{-u} du$.
The integral $\int_{0}^{\infty} u^{-1/4} e^{-u} du$ is a constant (Gamma function $\Gamma(3/4)$).
Therefore,$N \propto n_0 \alpha^{-3/4}$.

(B) The potential energy of the gas molecule is given by $U = \frac{M}{r^6} - \frac{N}{r^{12}}$.
At equilibrium,the force $F$ acting on the molecule is zero.
The force is related to potential energy by $F = -\frac{dU}{dr}$.
Calculating the derivative: $F = -\frac{d}{dr} \left( \frac{M}{r^6} - \frac{N}{r^{12}} \right) = - \left( -\frac{6M}{r^7} + \frac{12N}{r^{13}} \right) = \frac{6M}{r^7} - \frac{12N}{r^{13}}$.
Setting $F = 0$ for equilibrium: $\frac{6M}{r^7} = \frac{12N}{r^{13}}$.
This simplifies to $r^6 = \frac{12N}{6M} = \frac{2N}{M}$.
Now,substitute $r^6 = \frac{2N}{M}$ back into the potential energy expression:
$U = \frac{M}{r^6} - \frac{N}{(r^6)^2} = \frac{M}{(2N/M)} - \frac{N}{(2N/M)^2}$.
$U = \frac{M^2}{2N} - \frac{N}{4N^2/M^2} = \frac{M^2}{2N} - \frac{M^2}{4N}$.
$U = \frac{2M^2 - M^2}{4N} = \frac{M^2}{4N}$.

(B) The force $F$ is given by the negative gradient of potential energy: $F = -\frac{dU}{dr}$.
$F = -\frac{d}{dr} \left[ \frac{M}{r^6} - \frac{N}{r^{12}} \right] = -\left[ -\frac{6M}{r^7} + \frac{12N}{r^{13}} \right] = \frac{6M}{r^7} - \frac{12N}{r^{13}}$.
At equilibrium,the net force is zero: $F = 0$.
$\frac{6M}{r^7} = \frac{12N}{r^{13}} \implies r^6 = \frac{12N}{6M} = \frac{2N}{M}$.
Substitute $r^6 = \frac{2N}{M}$ into the expression for potential energy $U$:
$U = \frac{M}{(r^6)} - \frac{N}{(r^6)^2} = \frac{M}{(2N/M)} - \frac{N}{(2N/M)^2}$.
$U = \frac{M^2}{2N} - \frac{N \cdot M^2}{4N^2} = \frac{M^2}{2N} - \frac{M^2}{4N} = \frac{M^2}{4N}$.

(C) According to the work-energy theorem,the kinetic energy of the box is converted into the internal energy of the gas when the box is suddenly stopped.
Let $m$ be the mass of the gas and $M$ be its molar mass. The number of moles $\mu$ is given by $\mu = \frac{m}{M}$.
The kinetic energy of the box is $K = \frac{1}{2}mv^2$.
The change in internal energy of the gas is $\Delta U = \mu C_v \Delta T$.
For a diatomic gas,the molar heat capacity at constant volume is $C_v = \frac{5}{2}R$.
Equating the kinetic energy to the change in internal energy:
$\frac{1}{2}mv^2 = \left(\frac{m}{M}\right) \left(\frac{5}{2}R\right) \Delta T$
Canceling $m$ and $\frac{1}{2}$ from both sides:
$v^2 = \frac{5R}{M} \Delta T$
Solving for $\Delta T$:
$\Delta T = \frac{Mv^2}{5R}$

(D) The kinetic energy of the gas is converted into internal energy when the vessel is brought to rest.
Let $n$ be the number of moles of the gas.
The total kinetic energy of the gas is $K.E. = \frac{1}{2} (nM) v^2$,where $nM$ is the total mass of the gas.
The change in internal energy of an ideal gas is given by $\Delta U = \frac{nR \Delta T}{\gamma - 1}$.
Since the vessel is thermally insulated,the kinetic energy is entirely converted into internal energy:
$\frac{1}{2} (nM) v^2 = \frac{nR \Delta T}{\gamma - 1}$
Canceling $n$ from both sides:
$\frac{1}{2} M v^2 = \frac{R \Delta T}{\gamma - 1}$
Solving for $\Delta T$:
$\Delta T = \frac{M v^2 (\gamma - 1)}{2R}$

(B) During driving,the friction between the tyre and the road,along with the continuous compression and expansion of the tyre,generates heat. This increases the temperature of the air trapped inside the tyre.
According to Gay-Lussac's Law,for a fixed mass of gas at constant volume,the pressure $P$ is directly proportional to its absolute temperature $T$ $(P \propto T)$. Therefore,as the temperature increases,the air pressure inside the tyre increases. This confirms the Assertion is correct.
Regarding the Reason,absolute zero temperature $(0 \ K)$ is defined as the temperature at which the classical kinetic energy of gas molecules becomes zero. However,in quantum mechanics,the zero-point energy exists even at absolute zero. Thus,the statement that 'Absolute zero temperature is not zero energy temperature' is scientifically correct.
However,the Reason does not explain why the pressure increases in a car tyre; the pressure increase is due to the temperature rise caused by friction,not the nature of absolute zero energy. Therefore,both are correct,but the Reason is not the correct explanation of the Assertion.

(D) The mean free time $t$ is defined as the ratio of the mean free path $\lambda$ to the average speed $v_{avg}$ of the gas molecules.
$t = \frac{\lambda}{v_{avg}}$
The mean free path is given by $\lambda = \frac{1}{\sqrt{2} \pi D^{2} n}$,where $D$ is the molecular diameter and $n$ is the number density. For an ideal gas at constant pressure,$n \propto 1/T$.
The average speed is given by $v_{avg} = \sqrt{\frac{8 RT}{\pi M_{w}}}$,which implies $v_{avg} \propto \sqrt{T}$.
Substituting these into the expression for $t$:
$t = \frac{\lambda}{v_{avg}} \propto \frac{1/n}{\sqrt{T}} \propto \frac{T}{\sqrt{T}} = \sqrt{T}$.
However,if we assume the number density $n$ is constant (as is typical for such problems unless specified otherwise),then $\lambda$ is constant.
Then,$t = \frac{\lambda}{v_{avg}} \propto \frac{1}{\sqrt{T}}$.
Thus,the plot of $t$ versus $1/\sqrt{T}$ will be a straight line passing through the origin. Therefore,the correct graph is $t$ versus $1/\sqrt{T}$.

(N/A) Radius of hydrogen atom,$r = 0.5 \; \mathring{A} = 0.5 \times 10^{-10} \; m$.
Volume of one hydrogen atom $= \frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.1416 \times (0.5 \times 10^{-10})^3 \approx 0.524 \times 10^{-30} \; m^3$.
Number of atoms in $1$ mole of hydrogen ($H_2$ gas contains $2$ atoms per molecule,but for atomic volume of $1$ mole of atoms,we use Avogadro's number) $= 6.023 \times 10^{23}$.
Volume of $1$ mole of hydrogen atoms,$V_a = 6.023 \times 10^{23} \times 0.524 \times 10^{-30} \approx 3.16 \times 10^{-7} \; m^3$.
Molar volume of $1$ mole of gas at $STP$,$V_m = 22.4 \; L = 22.4 \times 10^{-3} \; m^3$.
Ratio $\frac{V_m}{V_a} = \frac{22.4 \times 10^{-3}}{3.16 \times 10^{-7}} \approx 7.08 \times 10^4$.
This ratio is large because in a gas,the molecules are separated by distances much larger than their own size,meaning the gas is mostly empty space.

(A) Yes; the collision is elastic.
$1$. Momentum Conservation: In any collision,the total momentum of the system (molecule + wall) is conserved,provided no external force acts on the system. Since the wall is part of the container and is effectively stationary (or has infinite mass),the momentum of the molecule changes,but the total momentum of the system remains conserved.
$2$. Nature of Collision: The molecule strikes the wall with a speed of $200 \; m s^{-1}$ and rebounds with the same speed of $200 \; m s^{-1}$. Since the kinetic energy of the molecule before the collision is $K_i = \frac{1}{2} m v^2$ and after the collision is $K_f = \frac{1}{2} m v^2$,the kinetic energy remains unchanged $(K_i = K_f)$. Therefore,the collision is elastic.

(N/A) Given: Mass of oxygen molecule $m = 5.30 \times 10^{-26} \; kg$,Moment of inertia $I = 1.94 \times 10^{-46} \; kg \cdot m^{2}$,Mean speed $v = 500 \; m/s$.
The kinetic energy of translation is $KE_{trans} = \frac{1}{2} m v^{2}$.
The kinetic energy of rotation is $KE_{rot} = \frac{1}{2} I \omega^{2}$.
According to the problem,$KE_{rot} = \frac{2}{3} KE_{trans}$.
Substituting the expressions: $\frac{1}{2} I \omega^{2} = \frac{2}{3} (\frac{1}{2} m v^{2})$.
$I \omega^{2} = \frac{2}{3} m v^{2}$.
$\omega^{2} = \frac{2 m v^{2}}{3 I}$.
$\omega = \sqrt{\frac{2 m v^{2}}{3 I}} = v \sqrt{\frac{2 m}{3 I}}$.
Substituting the values: $\omega = 500 \times \sqrt{\frac{2 \times 5.30 \times 10^{-26}}{3 \times 1.94 \times 10^{-46}}}$.
$\omega = 500 \times \sqrt{\frac{10.60 \times 10^{-26}}{5.82 \times 10^{-46}}} = 500 \times \sqrt{1.821 \times 10^{20}}$.
$\omega = 500 \times 1.349 \times 10^{10} \approx 6.75 \times 10^{12} \; rad/s$.

(A) The density of liquid water is $\rho_{liquid} = 1000 \; kg \; m^{-3}$.
The density of water vapour is $\rho_{vapour} = 0.6 \; kg \; m^{-3}$.
Assuming the density of the water molecule itself is approximately equal to the density of liquid water,the molecular volume $V_{mol}$ for a given mass $M$ is $V_{mol} = M / \rho_{liquid}$.
The total volume occupied by the same mass $M$ in the vapour state is $V_{vapour} = M / \rho_{vapour}$.
The ratio of the molecular volume to the total volume is given by:
$\text{Ratio} = \frac{V_{mol}}{V_{vapour}} = \frac{M / \rho_{liquid}}{M / \rho_{vapour}} = \frac{\rho_{vapour}}{\rho_{liquid}}$.
Substituting the given values:
$\text{Ratio} = \frac{0.6}{1000} = 0.6 \times 10^{-3} = 6 \times 10^{-4}$.

(C) The ratio of the density of liquid water to the density of water vapour is $\frac{1000}{0.6} \approx 1667 \approx 1.67 \times 10^3$.
This means that the volume occupied by a given mass of water in the vapour state is $1.67 \times 10^3$ times the volume it occupies in the liquid state.
Since the volume $V$ is proportional to the cube of the radius $r$ $(V \propto r^3)$,the average distance between molecules scales with the cube root of the volume expansion.
Let $r_l$ be the radius in the liquid state and $r_v$ be the radius in the vapour state.
Then $\frac{r_v}{r_l} = (1.67 \times 10^3)^{1/3} \approx 1.67^{1/3} \times 10 \approx 1.186 \times 10 \approx 11.86$.
However,the standard approximation used in this context is that the volume increases by $10^3$ times,so the distance increases by a factor of $10$.
Given the radius of a molecule is $2 \; \mathring{A}$,the average distance between molecules in the liquid state is $2 \times 2 = 4 \; \mathring{A}$.
In the vapour state,the average distance becomes $10 \times 4 \; \mathring{A} = 40 \; \mathring{A}$.

(A) The important point to remember is that the average kinetic energy (per molecule) of any ideal gas (whether monatomic like argon,diatomic like chlorine,or polyatomic) is always equal to $\frac{3}{2}k_{B}T$. It depends only on temperature and is independent of the nature of the gas.
$(i)$ Since argon and chlorine both have the same temperature in the flask,the ratio of average kinetic energy per molecule of the two gases is $1:1$.
$(ii)$ The root mean square speed is given by $v_{rms} = \sqrt{\frac{3k_{B}T}{m}}$,where $m$ is the mass of a molecule. Therefore,the ratio of $v_{rms}$ for argon to chlorine is:
$\frac{(v_{rms})_{Ar}}{(v_{rms})_{Cl}} = \sqrt{\frac{m_{Cl}}{m_{Ar}}} = \sqrt{\frac{M_{Cl}}{M_{Ar}}} = \sqrt{\frac{70.9}{39.9}} \approx \sqrt{1.776} \approx 1.33$.
Thus,the ratio is $1.33:1$.

(A) Let the speed of the ball be $u$ relative to the wicket behind the bat. If the bat is moving towards the ball with a speed $V$ relative to the wicket,then the relative speed of the ball to the bat is $V+u$ towards the bat. When the ball rebounds (after hitting the massive bat),its speed relative to the bat is $V+u$ moving away from the bat. So,relative to the wicket,the speed of the rebounding ball is $V+(V+u) = 2V+u$,moving away from the wicket. Thus,the ball moves faster after the collision with the bat.
$(b)$ When a piston is pushed into a cylinder,the gas molecules collide with the moving piston. Similar to the ball hitting a moving bat,the molecules rebound with a higher speed. Since the average kinetic energy of the molecules increases,the temperature of the gas rises.
$(c)$ When the gas expands by pushing the piston out,the molecules collide with a receding piston. The molecules rebound with a lower speed,resulting in a decrease in the average kinetic energy of the gas molecules. Consequently,the temperature of the gas drops.
$(d)$ Yes,using a heavy bat is advantageous. Since the bat is massive,it does not lose much speed upon collision with the ball. The ball rebounds with a higher speed $(2V+u)$,allowing the batsman to hit the ball further.