Gas Laws (Charles, Boyle's, Avagadro's, Gay Lussacs and Dalton's law) and Ideal gas Equation Questions in English

(A) The ideal gas equation is given by $PV = \mu RT$,where $\mu$ is the number of moles.
Number of moles $\mu = \frac{\text{mass } (m)}{\text{molar mass } (M)}$.
For oxygen gas $(O_2)$,the molar mass $M = 32 \, g/mol$.
Given mass $m = 5 \, g$.
Therefore,$\mu = \frac{5}{32} \, mol$.
Substituting this into the ideal gas equation,we get $PV = \frac{5}{32}RT$.

(D) From the ideal gas equation,$PV = nRT$.
Since $n = \frac{m}{M}$,where $m$ is the mass and $M$ is the molar mass,we have $PV = \frac{m}{M}RT$.
Rearranging for density $\rho = \frac{m}{V}$,we get $P = \frac{\rho RT}{M}$.
This implies that $\frac{P}{\rho T} = \frac{R}{M} = \text{constant}$.
Therefore,$\frac{P_1}{\rho_1 T_1} = \frac{P_2}{\rho_2 T_2}$.
Given: $P_2 = 2P_1$ and $T_2 = 4T_1$.
Substituting these values: $\frac{P_1}{\rho_1 T_1} = \frac{2P_1}{\rho_2 (4T_1)}$.
Simplifying,$1 = \frac{2}{4} \cdot \frac{\rho_1}{\rho_2}$,which gives $1 = \frac{1}{2} \cdot \frac{\rho_1}{\rho_2}$.
Thus,$\rho_2 = 0.5 \rho_1$.
The density becomes $0.5$ times the original density.

(D) According to Charles's Law,for a fixed mass of gas at constant pressure,the volume $V$ is directly proportional to the absolute temperature $T$ $(V \propto T)$.
Let the initial volume at $T_1 = 0^{\circ}C = 273 \, K$ be $V_1 = V$.
We want the final volume to be $V_2 = 2V$.
Using the relation $\frac{V_1}{T_1} = \frac{V_2}{T_2}$:
$\frac{V}{273} = \frac{2V}{T_2}$
$T_2 = 2 \times 273 = 546 \, K$.
To convert this to Celsius: $T_2(^{\circ}C) = 546 - 273 = 273^{\circ}C$.

(C) According to Dalton's Law of Partial Pressures,the total pressure exerted by a mixture of non-reacting gases in a fixed volume is equal to the sum of the partial pressures of the individual gases.
Since all three gases are mixed and placed into a single container of the same volume $V$,the total pressure $P$ is the sum of the individual pressures.
Therefore,$P = P_1 + P_2 + P_3$.

(C) We know the ideal gas equation is given by $PV = nRT$,where $n$ is the number of moles.
Since $n = \frac{m}{M}$,where $m$ is the mass of the gas and $M$ is the molar mass,the equation becomes $PV = \frac{m}{M} RT$.
Rearranging for pressure $P$,we get $P = \left( \frac{RT}{MV} \right) m$.
Given that the volume $V$ and temperature $T$ are constant,and the molar mass $M$ and gas constant $R$ are also constants,the term $\left( \frac{RT}{MV} \right)$ is a constant.
Therefore,$P \propto m$.
This shows that the pressure of the gas varies linearly with its mass.
Thus,the correct choice is option $C$.

(C) For a gas in a closed vessel,volume $V$ is constant. According to Gay-Lussac's Law,$P \propto T$ (where $T$ is in Kelvin).
Let the initial pressure be $P_1$ and initial temperature be $T_1$ (in Kelvin).
When the temperature increases by $5^{\circ}C$,the new temperature $T_2 = T_1 + 5$.
The pressure increases by $1\%$,so the new pressure $P_2 = P_1 + 0.01 P_1 = 1.01 P_1$.
Using the relation $\frac{P_1}{P_2} = \frac{T_1}{T_2}$,we get:
$\frac{P_1}{1.01 P_1} = \frac{T_1}{T_1 + 5}$
$T_1 + 5 = 1.01 T_1$
$0.01 T_1 = 5$
$T_1 = \frac{5}{0.01} = 500 \ K$.
To convert Kelvin to Celsius: $t(^{\circ}C) = T(K) - 273$.
$t = 500 - 273 = 227^{\circ}C$.

(C) Using the ideal gas equation,$PV = \mu RT = \frac{m}{M}RT$,where $m$ is the mass of the gas.
Since $V$ and $M$ are constant,we have $P \propto mT$.
Let the initial state be $(P_1, m_1, T_1)$ and the final state be $(P_2, m_2, T_2)$.
Given: $P_1 = 20 \ atm$,$T_1 = 27^{\circ}C = 300 \ K$.
After releasing half the gas,$m_2 = \frac{1}{2}m_1$.
The temperature is raised by $50^{\circ}C$,so $T_2 = (27 + 50)^{\circ}C = 77^{\circ}C = 350 \ K$.
Using the ratio: $\frac{P_2}{P_1} = \frac{m_2}{m_1} \times \frac{T_2}{T_1}$.
$\frac{P_2}{20} = \frac{1}{2} \times \frac{350}{300} = \frac{1}{2} \times \frac{7}{6} = \frac{7}{12}$.
$P_2 = 20 \times \frac{7}{12} = \frac{140}{12} \approx 11.67 \ atm$.
Thus,the final pressure is approximately $11.7 \ atm$.

(B) From the ideal gas equation,$PV = \mu RT = \frac{m}{M}RT$,where $m$ is the mass,$M$ is the molar mass,and $V$ is the volume.
Since density $\rho = \frac{m}{V}$,we can write $P = \frac{\rho RT}{M}$.
Therefore,the ratio of density to pressure is $\frac{\rho}{P} = \frac{M}{RT}$.
At $0^{\circ}C$ $(T_1 = 273 \ K)$,the ratio is $\left( \frac{\rho}{P} \right)_1 = \frac{M}{R(273)} = x$ --- $(i)$.
At $100^{\circ}C$ $(T_2 = 373 \ K)$,the ratio is $\left( \frac{\rho}{P} \right)_2 = \frac{M}{R(373)}$ --- $(ii)$.
Dividing equation $(ii)$ by equation $(i)$,we get $\frac{(\rho/P)_2}{x} = \frac{M/R(373)}{M/R(273)} = \frac{273}{373}$.
Thus,the ratio at $100^{\circ}C$ is $\frac{273}{373}x$.

(A) Given: Mass of ${O_2}$ $(m)$ = $2 \, g$,Temperature $(T)$ = $27^{\circ}C = 300 \, K$,Pressure $(P)$ = $76 \, cm$ of $Hg = 1 \, atm = 1.013 \times 10^5 \, Pa$.
Using the Ideal Gas Equation: $PV = nRT = \frac{m}{M}RT$.
Here,$M$ (molar mass of ${O_2}$) = $32 \, g/mol$ and $R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$.
$V = \frac{mRT}{MP} = \frac{2 \times 0.0821 \times 300}{32 \times 1} = \frac{49.26}{32} \approx 1.54 \, L$.
Using $R = 8.314 \, J \cdot K^{-1} \cdot mol^{-1}$ and $P = 1.013 \times 10^5 \, Pa$:
$V = \frac{(2/32) \times 8.314 \times 300}{1.013 \times 10^5} = 0.001539 \, m^3 = 1.539 \, L$.
Rounding to the nearest option,the correct answer is $1.53 \, L$.

(C) The pressure $P$ is given by $P = h \rho g$,where $h = 1.2 \times 10^{-7} \text{ mm} = 1.2 \times 10^{-10} \text{ m}$,$\rho = 13600 \text{ kg/m}^3$,and $g = 9.8 \text{ m/s}^2$.
Using the ideal gas law $PV = NkT$,the number of molecules $N$ is given by $N = \frac{PV}{kT}$.
Here,$V = 100 \text{ cm}^3 = 100 \times 10^{-6} \text{ m}^3 = 10^{-4} \text{ m}^3$,$k = 1.38 \times 10^{-23} \text{ J/K}$,and $T = 27 + 273 = 300 \text{ K}$.
Substituting the values:
$N = \frac{(1.2 \times 10^{-10} \times 13600 \times 9.8) \times 10^{-4}}{1.38 \times 10^{-23} \times 300}$
$N = \frac{1.59936 \times 10^{-8} \times 10^{-4}}{4.14 \times 10^{-21}}$
$N = \frac{1.59936 \times 10^{-12}}{4.14 \times 10^{-21}} \approx 0.386 \times 10^9 \times 10^3 = 3.86 \times 10^{11}$ molecules.

(B) For an ideal gas in a closed vessel, the volume $V$ is constant. According to Gay-Lussac's Law, $P \propto T$ or $\frac{P_1}{P_2} = \frac{T_1}{T_2}$.
Let the initial pressure be $P$ and initial temperature be $T$ (in Kelvin).
Given that the pressure increases by $0.5\%$, the new pressure $P_2 = P + 0.005P = 1.005P$.
The new temperature $T_2 = T + 2$.
Substituting these values into the gas law equation:
$\frac{P}{1.005P} = \frac{T}{T + 2}$
$\frac{1}{1.005} = \frac{T}{T + 2}$
$T + 2 = 1.005T$
$0.005T = 2$
$T = \frac{2}{0.005} = 400 \, K$.
To convert Kelvin to Celsius: $T(^\circ C) = T(K) - 273 = 400 - 273 = 127^\circ C$.

(C) According to the ideal gas law,$PV = nRT$. Since the volume $V$ and the amount of gas $n$ remain constant,we have $P \propto T$,which implies $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.
Given:
Initial pressure $P_1 = 1 \text{ atm}$
Initial temperature $T_1 = 35 + 273 = 308 \text{ K}$
Final pressure $P_2 = 3 \text{ atm}$
Final temperature $T_2 = ?$
Substituting the values into the relation:
$\frac{1}{308} = \frac{3}{T_2}$
Solving for $T_2$:
$T_2 = 3 \times 308 = 924 \text{ K}$
Converting the temperature back to Celsius:
$T_2(^{\circ}C) = 924 - 273 = 651^{\circ}C$.

(D) Given:
Total pressure of moist gas $= 735 \, mm$ of mercury.
Aqueous vapour pressure $= 23.8 \, mm$.
According to Dalton's law of partial pressures,the total pressure of a moist gas is the sum of the pressure of the dry gas and the aqueous vapour pressure.
$P_{\text{total}} = P_{\text{dry gas}} + P_{\text{aqueous vapour}}$
$735 \, mm = P_{\text{dry gas}} + 23.8 \, mm$
$P_{\text{dry gas}} = 735 \, mm - 23.8 \, mm$
$P_{\text{dry gas}} = 711.2 \, mm$.
Therefore,the correct choice is option $D$.

(B) Using the ideal gas law,$\frac{PV}{T} = \text{constant}$,we have:
$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
Given:
$P_1 = 200\, kPa$
$T_1 = 22 + 273 = 295\, K$
$T_2 = 42 + 273 = 315\, K$
$V_1 = V$
$V_2 = V + 0.02V = 1.02V$
Substituting the values:
$\frac{200 \times V}{295} = \frac{P_2 \times 1.02V}{315}$
$P_2 = \frac{200 \times 315}{295 \times 1.02}$
$P_2 = \frac{63000}{300.9} \approx 209.37\, kPa$
Thus,the pressure is approximately $209\, kPa$.

(A) According to Charles's Law,at constant pressure,the volume of a gas is directly proportional to its absolute temperature $(V \propto T)$.
Given:
Initial volume $V_{1} = V$
Initial temperature $T_{1} = 0^{\circ}C = 273 \; K$
Final volume $V_{2} = 3V$
Using the relation $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$:
$T_{2} = \frac{V_{2} \times T_{1}}{V_{1}}$
$T_{2} = \frac{3V \times 273}{V} = 819 \; K$
To convert the temperature to Celsius $(^{\circ}C)$:
$T(^{\circ}C) = T(K) - 273$
$T(^{\circ}C) = 819 - 273 = 546^{\circ}C$.

(B) According to Charles's Law,for a fixed mass of gas at constant pressure,the volume is directly proportional to the absolute temperature: $V \propto T$ or $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: Initial volume $V_1 = V$,Initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$,Final volume $V_2 = 1.5V$.
Using the formula $T_2 = \left( \frac{V_2}{V_1} \right) T_1$,we get $T_2 = \left( \frac{1.5V}{V} \right) \times 300 \ K = 1.5 \times 300 \ K = 450 \ K$.
To convert the final temperature back to Celsius: $T(^{\circ}C) = T(K) - 273 = 450 - 273 = 177^{\circ}C$.

(A) According to the ideal gas law,$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
Given:
$P_1 = 10^5\,N/m^2$ (atmospheric pressure),
$V_1 = 1\,m^3$,
$T_1 = 300\,K$.
New conditions:
$V_2 = 2V_1 = 2\,m^3$,
$T_2 = 2T_1 = 600\,K$.
Substituting these values into the equation:
$\frac{10^5 \times 1}{300} = \frac{P_2 \times 2}{600}$.
$\frac{10^5}{300} = \frac{P_2}{300}$.
$P_2 = 10^5\,N/m^2$.
Therefore,the pressure remains the same.

(C) Using the Ideal Gas Law: $PV = nRT$,where $n = \frac{m}{M}$.
Here,$P = 22.4\,atm$,$V = 2\,L$,$T = 273\,K$,and the molar mass of $N_2$ is $M = 28\,g/mol$.
The gas constant $R = 0.0821\,L \cdot atm \cdot K^{-1} \cdot mol^{-1}$.
Substituting the values: $22.4 \times 2 = n \times 0.0821 \times 273$.
Since $0.0821 \times 273 \approx 22.4$,we get $22.4 \times 2 = n \times 22.4$,which implies $n = 2\,mol$.
The mass $m = n \times M = 2\,mol \times 28\,g/mol = 56\,g$.

(B) For $1$ mole of an ideal gas,the ideal gas equation is given by $PV = nRT$.
Since $n = 1$ mole,the equation simplifies to $PV = RT$.
Rearranging this,we get $PV/T = R$.
The universal gas constant $R$ is a fundamental physical constant.
In $SI$ units,the value of $R$ is approximately $8.314$ $J$ $mol^{-1}K^{-1}$.
Therefore,the value of $PV/T$ is nearly $8.3$ $J$ $mol^{-1}K^{-1}$.

(A) The volume of the gas inside the tyre is essentially constant. According to Gay-Lussac's law,for a given mass of gas at a constant volume,the pressure is directly proportional to its absolute temperature $(P \propto T)$.
When the tyre is kept in sunlight,the temperature of the trapped air increases. Consequently,the pressure of the gas inside the tyre also increases.
This increased pressure exerts a greater force on the inner walls of the tyre,eventually exceeding the structural limit of the material,causing the tyre to burst.
Therefore,the correct choice is option $A$.

(A) Boyle's law states that for a given mass of an ideal gas at a constant temperature, the volume of the gas is inversely proportional to its pressure.
i.e., $V \propto 1/P$ or $PV = k$ (where $k$ is a constant).
Therefore, in Boyle's law, the temperature $(T)$ and the mass of the gas remain constant.
Looking at the options provided, option $A$ represents the mathematical expression of Boyle's law $(PV = \text{constant})$, which is the result of the law. However, the question asks what remains constant in the law's definition. Since temperature is the primary physical quantity held constant, and $PV$ is the product that remains constant, option $A$ is the standard representation.
Correct choice: $A$

(A) The ideal gas equation is given by $PV = \mu RT$,where $\mu = \frac{m}{M}$ is the number of moles.
Substituting $\mu$,we get $PV = \frac{m}{M}RT$.
Since density $d = \frac{m}{V}$,we can write $V = \frac{m}{d}$.
Substituting $V$ in the equation: $P(\frac{m}{d}) = \frac{m}{M}RT$.
Simplifying,we get $P = \frac{d}{M}RT$,which implies $\frac{P}{dT} = \frac{R}{M} = \text{constant}$.
Therefore,for two different states,we have $\frac{P_1}{d_1 T_1} = \frac{P_2}{d_2 T_2}$.

(C) According to the ideal gas law for one mole of gas,$PV = RT$.
Since pressure $P$ is constant,$V = (R/P)T$,which implies $V \propto T$.
Let the initial volume be $V_1$ at temperature $T_1 = T$.
If the temperature increases by $1 \ K$,the new temperature is $T_2 = T + 1$.
The new volume $V_2$ is given by $V_2 = (R/P)(T + 1)$.
The increase in volume is $\Delta V = V_2 - V_1 = (R/P)(T + 1) - (R/P)T = R/P$.
The ratio of the increase in volume to the original volume is $\frac{\Delta V}{V_1} = \frac{R/P}{(R/P)T} = \frac{1}{T}$.

(C) For an ideal gas,the ideal gas equation is given by $PV = \frac{m}{M}RT$,where $P$ is pressure,$V$ is volume,$m$ is mass,$M$ is molar mass,$R$ is the universal gas constant,and $T$ is temperature.
Since the flasks are connected,the pressure $P$ is the same in both flasks. Also,the gas is the same,so $M$ is constant.
Thus,$V \propto \frac{mT}{P} \Rightarrow V \propto mT$.
We can write the ratio as: $\frac{V_1}{V_2} = \frac{m_1 T_1}{m_2 T_2}$.
Given $V_1 = 2V_2$,$T_1 = 100\, K$,$T_2 = 200\, K$,and $m_1 = m$.
Substituting these values: $\frac{2V_2}{V_2} = \frac{m \times 100}{m_2 \times 200}$.
$2 = \frac{m}{2m_2}$.
$4m_2 = m \Rightarrow m_2 = \frac{m}{4}$.

(C) We know that from the ideal gas equation,$PV = n_{moles}RT$.
Here,$n_{moles} = \frac{N}{N_{A}}$,where $N$ is the number of molecules of the gas and $N_{A}$ is the Avogadro's number.
Substituting the value of $n_{moles}$ into the ideal gas equation,we get $PV = \frac{N}{N_{A}}RT$.
This implies that the pressure $P$ of the gas is directly proportional to the number of molecules $N$ of the gas,provided the temperature $T$ and volume $V$ remain constant $(P \propto N)$.
Therefore,if the number of molecules is doubled $(N' = 2N)$,the pressure will also be doubled $(P' = 2P)$.

(A) According to the ideal gas law,$PV = nRT$,where $n = N/N_A$. Since $P$,$V$,and $T$ are the same for both gases,the number of moles $n$ must be the same,which implies the total number of molecules $N$ is constant.
Average kinetic energy is given by $K_{avg} = (3/2)k_BT$,which depends only on temperature $T$. Since $T$ is constant,the average kinetic energy is also constant.
However,in the context of comparing two different gases,the total number of molecules is a fundamental consequence of Avogadro's Law under these conditions. Both $A$ and $B$ are constant,but the total number of molecules is the primary result of the given conditions.

(A) The number of moles of gas remains constant in both states. According to the ideal gas law,$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
Given at sea level:
$P_1 = 1 \, atm$
$V_1 = 30,000 \, cc = 30 \, L$
$T_1 = 27^{\circ}C = 27 + 273 = 300 \, K$
Given at Mount Everest site:
$V_2 = 5.2 \, L$
$T_2 = -73^{\circ}C = -73 + 273 = 200 \, K$
$P_2 = ?$
Substituting the values into the equation:
$\frac{1 \times 30}{300} = \frac{P_2 \times 5.2}{200}$
$0.1 = \frac{P_2 \times 5.2}{200}$
$P_2 = \frac{0.1 \times 200}{5.2} = \frac{20}{5.2} \approx 3.846 \, atm \approx 3.86 \, atm$.

(A) The ideal gas equation in terms of density is given by $PV = nRT = \frac{m}{M}RT$,which implies $P = \frac{\rho RT}{M}$,where $\rho$ is the density.
Thus,$\frac{P}{\rho T} = \frac{R}{M} = \text{constant}$.
Therefore,$\frac{P_1}{\rho_1 T_1} = \frac{P_2}{\rho_2 T_2}$,which gives $\frac{\rho_1}{\rho_2} = \frac{P_1}{P_2} \times \frac{T_2}{T_1}$.
Given:
At the top: $P_{top} = 70 \, cm$ of $Hg$,$T_{top} = 7 + 273 = 280 \, K$.
At the bottom: $P_{bottom} = 76 \, cm$ of $Hg$,$T_{bottom} = 27 + 273 = 300 \, K$.
Substituting these values:
$\frac{\rho_{top}}{\rho_{bottom}} = \frac{70}{76} \times \frac{300}{280} = \frac{70}{76} \times \frac{30}{28} = \frac{70}{76} \times \frac{15}{14} = \frac{5 \times 15}{76} = \frac{75}{76}$.

(D) The number of moles of oxygen $(n_{O_2})$ is $n_{O_2} = \frac{8\,g}{32\,g/mol} = 0.25\,mol$.
The number of moles of nitrogen $(n_{N_2})$ is $n_{N_2} = \frac{7\,g}{28\,g/mol} = 0.25\,mol$.
According to Dalton's law of partial pressures,the total pressure $P_{total}$ is proportional to the total number of moles $n_{total} = n_{O_2} + n_{N_2} = 0.25 + 0.25 = 0.50\,mol$.
Given $P_{total} = 10\,atm$,the partial pressure of each gas is proportional to its mole fraction.
Since $n_{O_2} = n_{N_2}$,the partial pressure of nitrogen is $P_{N_2} = \frac{n_{N_2}}{n_{O_2} + n_{N_2}} \times P_{total} = \frac{0.25}{0.50} \times 10\,atm = 5\,atm$.
When oxygen is removed,only nitrogen remains in the vessel. Therefore,the pressure of the remaining gas is $5\,atm$.

(A) The number of moles $\mu$ is given by the mass of water divided by its molar mass.
$\mu = \frac{4.5 \,kg}{18 \times 10^{-3} \,kg/mol} = 250 \,mol$.
At standard temperature and pressure $(STP)$,$T = 273 \,K$ and $P = 10^5 \,Pa$.
Using the ideal gas equation $PV = \mu RT$:
$V = \frac{\mu RT}{P} = \frac{250 \times 8.314 \times 273}{10^5} \approx 5.67 \,m^3$.
Rounding to the nearest provided option,the correct value is $5.6 \,m^3$.

(D) Let the total volume of the cylinder be $V$. Let the volume occupied by the gas of mass $2m$ be $V_1$ and the volume occupied by the gas of mass $m$ be $V_2 = V - V_1$.
Since the piston is in equilibrium,the pressure $P$ on both sides is the same.
Since the piston is conducting,the temperature $T$ is the same for both gases.
Using the ideal gas equation $PV = nRT = \frac{m_{gas}}{M}RT$,where $M$ is the molar mass of the gas:
For the gas with mass $m$: $P(V - V_1) = \frac{m}{M}RT$ $(i)$
For the gas with mass $2m$: $P V_1 = \frac{2m}{M}RT$ $(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{P V_1}{P(V - V_1)} = \frac{\frac{2m}{M}RT}{\frac{m}{M}RT}$
$\frac{V_1}{V - V_1} = 2$
$V_1 = 2V - 2V_1$
$3V_1 = 2V$
$\frac{V_1}{V} = \frac{2}{3} \approx 0.67$
Thus,the larger mass occupies $0.67$ of the total volume.

(D) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$ ($m$ is the mass,$M$ is the molar mass).
Thus,$V = \left(\frac{m}{PM}\right) RT$.
The slope of the $V-T$ graph is $S = \frac{mR}{PM}$.
For the first case (line $D$): $S_1 = \frac{mR}{PM}$. From the graph,the slope of line $D$ is $2$.
For the second case (mass $2m$,pressure $P/2$): $S_2 = \frac{(2m)R}{(P/2)M} = 4 \left(\frac{mR}{PM}\right) = 4S_1$.
Since $S_1 = 2$,we have $S_2 = 4 \times 2 = 8$.
Looking at the graph,the line with a slope of $8$ is $A$.

(C) For an ideal gas,the equation of state is $PV = nRT$.
To compare the temperatures $T_1$ and $T_2$,we can draw a line of constant pressure $(P = \text{constant})$ on the $P-V$ graph.
At a constant pressure,the ideal gas equation gives $V \propto T$.
From the graph,for a fixed pressure,the volume corresponding to the curve $T_2$ (let it be $V_2$) is greater than the volume corresponding to the curve $T_1$ (let it be $V_1$),i.e.,$V_2 > V_1$.
Since $V \propto T$ at constant pressure,$V_2 > V_1$ implies $T_2 > T_1$ or $T_1 < T_2$.

(A) According to Charles's Law,for a fixed mass of a gas at constant pressure,the volume $V$ is directly proportional to the absolute temperature $T$ (in Kelvin),i.e.,$V = kT$.
However,when expressed in Celsius scale,$V = V_0(1 + \alpha t)$,where $t$ is the temperature in $^{\circ}C$.
This equation represents a straight line $y = mx + c$,where $y$ is volume $(V)$ and $x$ is temperature ($t$ in $^{\circ}C$).
In the given graph,the line intersects the horizontal axis at a negative value,which corresponds to $-273.15^{\circ}C$ (absolute zero).
Since volume cannot be negative,the vertical axis $(a)$ must represent Volume $(V)$ and the horizontal axis $(b)$ must represent temperature in $^{\circ}C$ because the temperature scale in $^{\circ}C$ can have negative values.
Therefore,$a =$ volume and $b = ^{\circ}C$ temperature.

(C) For an ideal gas at constant pressure,Charles's Law states that $V / T = \text{constant}.$
When the temperature changes from $T$ to $T + \Delta T,$ the volume changes from $V$ to $V + \Delta V.$
Thus,we have:
$\frac{V + \Delta V}{T + \Delta T} = \frac{V}{T}$
Cross-multiplying gives:
$T(V + \Delta V) = V(T + \Delta T)$
$VT + T \Delta V = VT + V \Delta T$
Subtracting $VT$ from both sides:
$T \Delta V = V \Delta T$
Rearranging to find the expression for $\delta$:
$\delta = \frac{\Delta V}{V \Delta T} = \frac{1}{T}$
Since $\delta = 1/T,$ the quantity $\delta$ is inversely proportional to the temperature $T.$ This relationship is represented by a rectangular hyperbola,which corresponds to the graph shown in option $C.$

(A) According to the ideal gas law,$PV = nRT$,which can be rearranged as $P = (nR/V)T$.
For a constant volume process,the pressure-temperature graph is a straight line passing through the origin with a slope equal to $nR/V$.
In the given graph,lines $1-2$ and $3-4$ represent processes at constant volumes. Since lines $1-2$ and $3-4$ are segments of lines passing through the origin,we have $V_1 = V_2$ and $V_3 = V_4$.
The slope of the line is $m = nR/V$. Since the slope is inversely proportional to volume $(m \propto 1/V)$,a smaller slope corresponds to a larger volume.
The slope of line $1-2$ is less than the slope of line $3-4$. Therefore,the volume $V_2$ must be greater than the volume $V_3$ $(V_2 > V_3)$.
Thus,the correct relation is $V_1 = V_2, V_3 = V_4$ and $V_2 > V_3$.

(A) The ideal gas equation is given by $PV = nRT$.
Since $n = \frac{m}{M}$ and density $\rho = \frac{m}{V}$,we can write $V = \frac{m}{\rho}$.
Substituting this into the ideal gas equation: $P \left( \frac{m}{\rho} \right) = \left( \frac{m}{M} \right) RT$.
Simplifying,we get $P = \left( \frac{RT}{M} \right) \rho$.
This equation represents a straight line passing through the origin,$P = m' \rho$,where the slope $m' = \frac{RT}{M}$.
Since the slope is directly proportional to temperature $T$ (for a given gas,$R$ and $M$ are constant),the line with the greater slope corresponds to the higher temperature.
In the given graph,the slope of the line for $T_1$ is greater than the slope of the line for $T_2$.
Therefore,$T_1 > T_2$.

(A) From the ideal gas equation,we have $PV = nRT$,which can be rewritten as $P = (nR/V)T$.
Comparing this with the equation of a straight line passing through the origin,$y = mx$,where $y = P$ and $x = T$,the slope $m = nR/V$.
In the given $P-T$ diagram,the line representing the process from state $1$ to state $2$ is a straight line that does not pass through the origin $(0,0)$.
Let the equation of the line be $P = mT + c$,where $c > 0$ (since the intercept on the $P$-axis is positive).
Substituting $P = nRT/V$,we get $nRT/V = mT + c$,or $V = nR / (m + c/T)$.
As the temperature $T$ increases from state $1$ to state $2$,the term $c/T$ decreases.
Consequently,the denominator $(m + c/T)$ decreases,which means the volume $V$ must increase.

(C) For an ideal gas,the equation of state is $PV = nRT$. Thus,the temperature $T$ is proportional to the product $PV$.
In the given $P-V$ diagram,the process is a straight line from state $1$ to state $2$.
Let the equation of the line be $P = -mV + c$,where $m$ is the slope and $c$ is the intercept.
The product $PV = (-mV + c)V = -mV^2 + cV$.
To find the variation of $T$ with $V$,we look at the function $f(V) = -mV^2 + cV$.
This is a downward-opening parabola. The value of $PV$ (and thus $T$) increases initially as $V$ increases,reaches a maximum at $V = c/(2m)$,and then decreases as $V$ increases further.
Therefore,the gas is heated in the beginning and cooled towards the end.

(A) From the ideal gas equation,$PV = \mu RT$,we have $V = (\frac{\mu R}{P})T$.
The slope of the $V-T$ graph is given by $m = \tan \theta = \frac{V}{T} = \frac{\mu R}{P}$.
Since the slope is inversely proportional to pressure $(m \propto \frac{1}{P})$,a smaller slope corresponds to a higher pressure.
From the graph,it is clear that $\theta_1 < \theta_2$,which implies $\tan \theta_1 < \tan \theta_2$.
Therefore,$(\frac{V}{T})_1 < (\frac{V}{T})_2$.
Since $(\frac{V}{T}) \propto \frac{1}{P}$,we have $(\frac{1}{P})_1 < (\frac{1}{P})_2$,which leads to $P_1 > P_2$.

(A) From the given $P-T$ graph,the slope of the line represents $\frac{T}{P}$.
Since the angle $\theta_2 > \theta_1$,we have $\tan \theta_2 > \tan \theta_1$.
This implies $\left( \frac{T}{P} \right)_2 > \left( \frac{T}{P} \right)_1$.
From the ideal gas equation $PV = \mu RT$,we can write $\frac{T}{P} = \frac{V}{\mu R}$.
Since $\mu$ and $R$ are constants,$\frac{T}{P} \propto V$.
Therefore,$\left( \frac{T}{P} \right)_2 > \left( \frac{T}{P} \right)_1$ implies $V_2 > V_1$.

(A) From the ideal gas equation,$PV = \mu RT = \frac{m}{M}RT$,where $m$ is the mass and $M$ is the molar mass.
Rearranging the equation,we get $\frac{PV}{T} = \frac{mR}{M}$.
Since the masses $m$ are equal and $R$ is a constant,we have $\frac{PV}{T} \propto \frac{1}{M}$.
From the graph,the slope of the lines is $\frac{PV}{T}$. The slope is greatest for line $C$ and smallest for line $A$,so $(\frac{PV}{T})_C > (\frac{PV}{T})_B > (\frac{PV}{T})_A$.
This implies $M_C < M_B < M_A$.
The molar masses are $M_{H_2} = 2 \text{ g/mol}$,$M_{He} = 4 \text{ g/mol}$,and $M_{O_2} = 32 \text{ g/mol}$.
Therefore,$M_{H_2} < M_{He} < M_{O_2}$.
Comparing the two relations,we get $C$ corresponds to $H_2$,$B$ to $He$,and $A$ to $O_2$.

(B) For an ideal gas,the equation of state is $PV = nRT$,where $n = \frac{m}{M}$ ($m$ is mass,$M$ is molar mass).
Thus,$PV = \frac{m}{M} RT$,which implies $P = \left( \frac{mR}{MV} \right) T$.
For the initial state,the slope of the line $A$ is $S_1 = \frac{mR}{MV}$.
In the new state,the mass $m' = 2m$ and the volume $V' = \frac{V}{2}$.
The new slope $S_2$ is given by $S_2 = \frac{m'R}{MV'} = \frac{(2m)R}{M(V/2)} = 4 \left( \frac{mR}{MV} \right) = 4S_1$.
Since the slope $S_2$ is greater than $S_1$,the new line must be steeper than line $A$. Looking at the graph,line $B$ has a greater slope than line $A$. Therefore,the correct line is $B$.

(C) The ideal gas equation is given by $pV = nRT = \frac{m}{M}RT$,where $m$ is the mass of the gas and $M$ is the molar mass.
For a fixed temperature $T$,we have $pV = \text{constant} \times m$,which implies $m = \frac{pVM}{RT}$.
For a constant pressure $p$,we can see from the graph that for the same pressure,the volume $V_2$ corresponding to mass $m_2$ is greater than the volume $V_1$ corresponding to mass $m_1$ (i.e.,$V_2 > V_1$).
Since $m = \frac{pVM}{RT}$,for a constant $p, T,$ and $M$,we have $m \propto V$.
Since $V_2 > V_1$,it follows that $m_2 > m_1$ or $m_1 < m_2$.

(A) From the ideal gas equation, $PV = nRT$, where $n = \frac{m}{M}$ ($M$ is the molar mass).
So, $PV = \frac{m}{M} RT$.
Rearranging for pressure $P$, we get $P = \left( \frac{mR}{MV} \right) T$.
Comparing this with the equation of a straight line $P = (\text{slope}) \cdot T$, the slope is given by $S = \frac{mR}{MV}$.
Since $R$, $M$, and $V$ are constants for both cases, the slope $S$ is directly proportional to the mass $m$ $(S \propto m)$.
For curve $A$, the mass is $m$, so $S_A \propto m$.
For curve $B$, the mass is $3m$, so $S_B \propto 3m$.
The ratio of the slopes of curves $B$ to $A$ is $\frac{S_B}{S_A} = \frac{3m}{m} = \frac{3}{1}$.
Thus, the ratio is $3:1$.

(C) According to Boyle's Law,at a constant temperature,the pressure $P$ of a given mass of gas is inversely proportional to its volume $V$.
Mathematically,$PV = k$ (where $k$ is a constant).
This can be rewritten as $P = k \cdot (1/V)$.
Comparing this to the equation of a straight line $y = mx + c$,where $y = P$,$x = 1/V$,$m = k$,and $c = 0$,we see that the graph of $P$ versus $1/V$ is a straight line passing through the origin.

(B) For an ideal gas, the equation of state is given by $PV = nRT$.
If the temperature $T$ is kept constant, then for a fixed amount of gas (fixed $n$), the product $nRT$ is a constant.
Therefore, $PV = \text{constant}$.
This implies that the value of $PV$ does not change with a change in volume $V$. The graph of $PV$ versus $V$ will be a horizontal straight line parallel to the $V$-axis. Thus, the correct graph is represented by option $B$.

(A) According to Charles's Law,for a gas at constant pressure,the volume $V_T$ at temperature $T$ $(^{\circ}C)$ is given by the relation:
$V_T = V_0 \left( 1 + \frac{T}{273} \right)$
$V_T = V_0 + \left( \frac{V_0}{273} \right) T$
Comparing this with the equation of a straight line $y = mx + c$,where $y = V_T$,$x = T$,$m = \frac{V_0}{273}$ (slope),and $c = V_0$ (y-intercept).
Since $V_0 > 0$,the graph is a straight line with a positive y-intercept and a positive slope. This corresponds to the graph shown in option $A$.

(B) According to Boyle's Law,for a gas at constant temperature,$P_1 V_1 = P_2 V_2$.
Here,the initial pressure $P_1 = 0.8 \, Pa$ and initial volume $V_1 = 5 \, L$.
The gas expands into an additional evacuated vessel of volume $3 \, L$,so the final volume $V_2 = V_1 + V_{evacuated} = 5 \, L + 3 \, L = 8 \, L$.
Substituting the values into the equation: $0.8 \times 5 = P_2 \times 8$.
$4 = P_2 \times 8$.
$P_2 = 4 / 8 = 0.5 \, Pa$.