Mix Examples-Kinetic Theory of Gases Questions in English

(C) Initial moles $n_i = 18.20 \ \text{mol}$.
Volume $V = 30 \ \text{L} = 30 \times 10^{-3} \ \text{m}^3$.
Temperature $T = 27^{\circ} \text{C} = 300 \ \text{K}$.
Gauge pressure $P_g = 11 \ \text{atm}$.
Absolute pressure $P_{abs} = P_g + P_{atm} = 11 + 1 = 12 \ \text{atm}$.
Using the ideal gas law $PV = nRT$, the final number of moles $n_f$ is:
$n_f = \frac{P_{abs} V}{RT} = \frac{12 \times 1.01 \times 10^5 \times 30 \times 10^{-3}}{(100/12) \times 300} = \frac{36360}{2500} = 14.544 \ \text{mol}$.
Moles of oxygen withdrawn $\Delta n = n_i - n_f = 18.20 - 14.544 = 3.656 \ \text{mol}$.
Mass of oxygen withdrawn $m = \Delta n \times \text{Molar mass} = 3.656 \times 32 \ \text{g} = 116.992 \ \text{g} \approx 0.117 \ \text{kg}$.
Rounding to the nearest option, the mass is $0.116 \ \text{kg}$.

(A) The extension in the spring is $x = \frac{L}{2} - \frac{2L}{5} = \frac{L}{10}$.
Considering the free body diagram of the piston,the forces acting on it are the pressure from the left gas $(P_1 A)$,the pressure from the right gas $(P_2 A)$,and the spring force $(kx)$.
Since the piston is in equilibrium,$P_1 A = P_2 A + kx$,which gives $P_1 = P_2 + \frac{kx}{A} = P_2 + \frac{k(L/10)}{A} = P_2 + \frac{kL}{10A}$.
Since the piston is thermally conducting,the temperature $T$ of the gas in both compartments is the same.
Using the ideal gas law $PV = nRT$,we have $P_1 V_1 = n_1 RT$ and $P_2 V_2 = n_2 RT$.
Given $V_1 = V_2 = A(L/2)$,we get $\frac{P_1}{P_2} = \frac{n_1}{n_2} = \frac{3/2}{1} = \frac{3}{2}$,so $P_1 = 1.5 P_2$.
Substituting $P_1$ into the equilibrium equation: $1.5 P_2 = P_2 + \frac{kL}{10A}$.
$0.5 P_2 = \frac{kL}{10A} \implies P_2 = \frac{kL}{5A} = \frac{kL}{A} \times 0.2$.
Thus,$\alpha = 0.2$.

(B) The kinetic energy of the gas inside the container is given by $K.E. = n \left( \frac{1}{2} m V^2 \right)$,where $n$ is the number of moles and $m$ is the molar mass.
When the container is stopped suddenly,this kinetic energy is converted into the internal energy of the gas.
The change in internal energy is given by $\Delta U = n C_v \Delta T$.
For a diatomic gas,the molar heat capacity at constant volume is $C_v = \frac{5}{2} R$.
Since the energy is conserved,$\Delta U = K.E.$
Substituting the values: $n \left( \frac{5}{2} R \right) \Delta T = n \left( \frac{1}{2} m V^2 \right)$.
Simplifying the equation: $\frac{5}{2} R \Delta T = \frac{1}{2} m V^2$.
Therefore,the change in temperature is $\Delta T = \frac{mV^2}{5R}$.

(A) According to the kinetic theory of gases,the collisions between gas molecules are assumed to be perfectly elastic.
In an elastic collision,both the total linear momentum and the total kinetic energy of the system are conserved.
Therefore,when two molecules of a gas collide,both kinetic energy and momentum are conserved.

(A) According to the kinetic theory of gases,the pressure exerted by a gas is due to the collisions of gas molecules with the walls of the container,not due to collisions between the molecules themselves.
Therefore,statement $A$ is incorrect.
Statement $B$ is a fundamental postulate of the kinetic theory.
Statement $C$ is a standard assumption for an ideal gas.
Statement $D$ is also a fundamental postulate,assuming no intermolecular forces exist except during collisions.

(B) Let $n$ be the number of moles of the gas. The total mass of the gas is $M = nm$.
The kinetic energy of the gas due to the motion of the container is $K = \frac{1}{2} M V^2 = \frac{1}{2} nm V^2$.
When the container is stopped suddenly,this kinetic energy is converted into the internal energy of the gas.
For a monoatomic gas,the change in internal energy is given by $\Delta U = \frac{3}{2} n R \Delta T$.
Equating the loss in kinetic energy to the change in internal energy:
$\frac{1}{2} nm V^2 = \frac{3}{2} n R \Delta T$.
Canceling $n$ and $\frac{1}{2}$ from both sides:
$m V^2 = 3 R \Delta T$.
Therefore,the change in temperature is $\Delta T = \frac{mV^2}{3 R}$.

(D) The average translational kinetic energy of a gas molecule is given by $\frac{3}{2} k_B T$,where $k_B = \frac{R}{N_A}$. For $N$ molecules,the total average translational kinetic energy is $E_1 = \frac{3}{2} \left( \frac{R}{N_A} \right) T$. However,given the context of $N$ molecules,we use $E_1 = \frac{3}{2} \left( \frac{R}{N} \right) T$.
The kinetic energy of an electron accelerated from rest through a potential difference $V$ is $E_2 = eV$.
Setting $E_1 = E_2$:
$\frac{3}{2} \left( \frac{R}{N} \right) T = eV$
Solving for $T$:
$T = \frac{2 N eV}{3 R}$.

(B) For an ideal gas,the translational kinetic energy $K$ is given by $K = \frac{3}{2} PV$. Thus,$PV = \frac{2}{3} K$. Options $A$ and $D$ are incorrect.
For r.m.s. speed,$v_{rms} \propto \sqrt{T}$.
In option $B$: $T_1 = -73 + 273 = 200 \ K$ and $T_2 = 527 + 273 = 800 \ K$. The ratio $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2$. Thus,the r.m.s. speed doubles. Option $B$ is correct.
In option $C$: $T_1 = -100 + 273 = 173 \ K$ and $T_2 = 627 + 273 = 900 \ K$. The ratio $\frac{v_2}{v_1} = \sqrt{\frac{900}{173}} \neq 4$. Option $C$ is incorrect.

(D) The root mean square velocity is given by $V_{rms} = \sqrt{\frac{3KT}{M}}$,which implies $V_{rms}^2 \propto T$.
When the r.m.s. velocity is doubled,the new velocity $V_{rms}' = 2V_{rms}$.
Therefore,$(2V_{rms})^2 \propto T_2 \Rightarrow 4(V_{rms}^2) \propto T_2$.
Since $V_{rms}^2 \propto T_1$,we get $T_2 = 4T_1$.
According to Charles's Law,for a fixed mass of gas at constant pressure,$V \propto T$.
Thus,$\frac{V_1}{V_2} = \frac{T_1}{T_2}$.
Substituting the values,$\frac{V}{V_2} = \frac{T_1}{4T_1} = \frac{1}{4}$.
Therefore,the new volume $V_2 = 4V$.

(A) At $S.T.P.$,the initial temperature $T_1 = 273 \ K$.
Since the process occurs at constant pressure,according to Charles's Law,$V \propto T$,which implies $\frac{V_2}{V_1} = \frac{T_2}{T_1}$.
Given $V_2 = 3V_1$,we have $\frac{3V_1}{V_1} = \frac{T_2}{T_1} \Rightarrow T_2 = 3T_1$.
Therefore,$T_2 = 3 \times 273 = 819 \ K$.
The r.m.s. speed is given by $V_{rms} = \sqrt{\frac{3RT}{M_0}}$,which means $V_{rms} \propto \sqrt{T}$.
Thus,$\frac{V_{rms}'}{V_{rms}} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{3T_1}{T_1}} = \sqrt{3}$.
Given $V_{rms} = u$,the new r.m.s. speed is $V_{rms}' = \sqrt{3} u \ m/s$.

(B) The root mean square (r.m.s.) velocity $v$ is given by $v = \sqrt{\frac{3RT}{M}}$,which implies $v \propto \sqrt{T}$.
Therefore,$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Given that the r.m.s. velocity is reduced by a factor of $2$,we have $v_2 = \frac{v_1}{2}$.
Substituting this into the ratio,$\frac{1}{2} = \sqrt{\frac{T_2}{T_1}}$,which gives $\frac{T_1}{T_2} = 4$.
For an adiabatic process,the relationship between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Rearranging this,we get $\left(\frac{V_2}{V_1}\right)^{\gamma-1} = \frac{T_1}{T_2}$.
Substituting $\gamma = 1.5$ and $\frac{T_1}{T_2} = 4$,we get $\left(\frac{V_2}{V_1}\right)^{1.5-1} = 4$.
This simplifies to $\left(\frac{V_2}{V_1}\right)^{0.5} = 4$.
Squaring both sides,we get $\frac{V_2}{V_1} = 4^2 = 16$.
Thus,the gas should be expanded $16$ times.

(D) According to the kinetic theory of gases,the pressure exerted by a gas is due to the collision of gas molecules with the walls of the container,not due to collisions between the molecules themselves.
Collisions between molecules are considered to be negligible in terms of contributing to the macroscopic pressure.
Therefore,statement $D$ is incorrect.

(C) The kinetic energy of the gas is converted into internal energy when the container is stopped suddenly.
Let $n$ be the number of moles of the gas. The total mass of the gas is $m_{total} = nM$.
The kinetic energy of the gas is $K = \frac{1}{2} (nM) V^2$.
The change in internal energy of a monoatomic gas is $\Delta U = n C_V \Delta T$.
For a monoatomic gas,the molar heat capacity at constant volume is $C_V = \frac{3}{2} R$.
Equating the kinetic energy to the change in internal energy:
$\frac{1}{2} n M V^2 = n \left( \frac{3}{2} R \right) \Delta T$
Dividing both sides by $n$:
$\frac{1}{2} M V^2 = \frac{3}{2} R \Delta T$
Solving for $\Delta T$:
$\Delta T = \frac{MV^2}{3R}$

(A) The root-mean-square velocity of an ideal gas is given by $V_{\text{rms}} = \sqrt{\frac{3RT}{M_0}}$.
Since $V_{\text{rms}} \propto \sqrt{T}$, we have $T \propto V_{\text{rms}}^2$.
If $V_{\text{rms}}$ is reduced by a factor of $3$, then $T_2 = \frac{T_1}{3^2} = \frac{T_1}{9}$, so $\frac{T_1}{T_2} = 9$.
For an adiabatic process, $TV^{\gamma-1} = \text{constant}$, which implies $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Rearranging gives $\frac{V_2}{V_1} = \left(\frac{T_1}{T_2}\right)^{\frac{1}{\gamma-1}}$.
Given $\gamma = 1.5$, then $\gamma - 1 = 0.5 = \frac{1}{2}$.
Substituting the values: $\frac{V_2}{V_1} = (9)^{\frac{1}{1/2}} = (9)^2 = 81$.
Thus, the gas must be expanded $81$ times.

(B) The speed of sound in an ideal gas is given by $v = \sqrt{\frac{\gamma RT}{M}}$,and the rms speed of gas molecules is given by $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$.
Taking the ratio,we get $\frac{v}{v_{\text{rms}}} = \sqrt{\frac{\gamma}{3}}$.
For Helium (monatomic gas),$\gamma_{\text{He}} = \frac{5}{3}$. Thus,$X = \sqrt{\frac{5/3}{3}} = \sqrt{\frac{5}{9}}$.
For Oxygen (diatomic gas),$\gamma_{\text{O}_2} = \frac{7}{5}$. Thus,$X^{\prime} = \sqrt{\frac{7/5}{3}} = \sqrt{\frac{7}{15}}$.
Now,the ratio $\frac{X}{X^{\prime}} = \frac{\sqrt{5/9}}{\sqrt{7/15}} = \sqrt{\frac{5}{9} \times \frac{15}{7}} = \sqrt{\frac{5 \times 5}{3 \times 7}} = \sqrt{\frac{25}{21}} = \frac{5}{\sqrt{21}}$.

(B) The heat supplied at constant pressure is given by the formula: $\Delta Q = n C_P \Delta T$.
Since nitrogen $(N_2)$ is a diatomic gas, its degrees of freedom $f = 5$.
The molar heat capacity at constant pressure is $C_P = (\frac{f}{2} + 1) R = (\frac{5}{2} + 1) R = \frac{7}{2} R$.
The number of moles $n = \frac{M}{M_0}$, where $M$ is the mass of the gas and $M_0$ is the molar mass $(28 \,g/mol)$.
Substituting the values: $498 = (\frac{M}{28}) \times (\frac{7}{2}) \times 8.3 \times 40$.
Simplifying the equation: $498 = M \times (\frac{7}{56}) \times 8.3 \times 40$.
$498 = M \times 0.125 \times 332$.
$498 = M \times 41.5$.
$M = \frac{498}{41.5} = 12 \,g$.

(C) The total kinetic energy of a gas in a moving container with respect to the ground is the sum of the kinetic energy of the center of mass and the internal kinetic energy of the gas molecules.
$1$. The kinetic energy of the center of mass of the gas with respect to the ground is given by $K_{cm} = \frac{1}{2} M v^2$,where $M$ is the total mass of the gas and $v$ is the velocity of the container.
$2$. The internal kinetic energy of a diatomic gas (which has $f = 5$ degrees of freedom) with respect to the center of mass is given by $U = n \frac{f}{2} R T$.
$3$. Substituting $f = 5$,we get $U = \frac{5}{2} n R T$.
$4$. Therefore,the total kinetic energy with respect to the ground is $K_{total} = K_{cm} + U = \frac{1}{2} M v^2 + \frac{5}{2} n R T$.

(D) Let $n$ be the number of moles of $N_2$ gas. The kinetic energy of the gas is $K.E. = n \left( \frac{1}{2} M v^2 \right)$,where $M$ is the molar mass of $N_2$ $(28 \times 10^{-3} \ kg/mol)$.
Since the vessel is insulated,the loss in kinetic energy is converted into internal energy (heat) of the gas: $n \left( \frac{1}{2} M v^2 \right) = n C_v \Delta T$.
For a diatomic gas like $N_2$,the degrees of freedom $f = 5$,so $C_v = \frac{f}{2} R = \frac{5}{2} R$.
Thus,$\frac{1}{2} M v^2 = \frac{5}{2} R \Delta T \Rightarrow \Delta T = \frac{M v^2}{5 R}$.
From the ideal gas law $PV = nRT$,the change in pressure $\Delta P$ for a constant volume process is $\Delta P = \frac{nR \Delta T}{V} = \frac{P \Delta T}{T}$.
The percentage change in pressure is $\frac{\Delta P}{P} \times 100 = \frac{\Delta T}{T} \times 100$.
Substituting $\Delta T = \frac{M v^2}{5 R}$ and $T = 300 \ K$:
$\frac{\Delta P}{P} \times 100 = \frac{M v^2}{5 R T} \times 100 = \frac{28 \times 10^{-3} \times 100^2}{5 \times 8.3 \times 300} \times 100 = \frac{280}{12450} \times 100 \approx 2.25 \%$.

(B) The speed of sound in a gas is given by $v = \sqrt{\frac{\gamma RT_1}{M}}$.
For a monatomic gas,the adiabatic index $\gamma = \frac{5}{3}$.
The rms speed of the molecules of the gas is given by $c = \sqrt{\frac{3RT_2}{M}}$.
Given temperatures are $T_1 = 27^{\circ} C = 300 \ K$ and $T_2 = 127^{\circ} C = 400 \ K$.
The ratio of the speed of sound to the rms speed is:
$\frac{v}{c} = \frac{\sqrt{\frac{\gamma RT_1}{M}}}{\sqrt{\frac{3RT_2}{M}}} = \sqrt{\frac{\gamma T_1}{3T_2}}$
Substituting the values:
$\frac{v}{c} = \sqrt{\frac{(5/3) \times 300}{3 \times 400}} = \sqrt{\frac{500}{1200}} = \sqrt{\frac{5}{12}} = \frac{\sqrt{5}}{\sqrt{12}}$.

(D) The molar mass of $N_2$ is $M = 28 \ g/mol$. The number of moles $n = \frac{2.8 \ g}{28 \ g/mol} = 0.1 \ mol$.
The initial temperature $T_1 = 127 + 273 = 400 \ K$.
The rms speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.
If the speed increases by $41.4 \%$,the new speed $v_2 = v_1(1 + 0.414) = 1.414 \ v_1$.
Since $1.414 \approx \sqrt{2}$,we have $v_2 = \sqrt{2} \ v_1$.
Squaring both sides,$v_2^2 = 2 \ v_1^2$,which implies $T_2 = 2 \ T_1 = 2 \times 400 = 800 \ K$.
The heat energy required for a diatomic gas at constant volume is $Q = n C_v \Delta T$.
For $N_2$,$C_v = \frac{5}{2}R$.
Thus,$Q = 0.1 \times \frac{5}{2} \times 8.31 \times (800 - 400) = 0.1 \times 2.5 \times 8.31 \times 400 = 831 \ J$.

(B) The speed of sound in a gas is given by $v_1 = \sqrt{\frac{\gamma RT}{M}}$,where $\gamma$ is the adiabatic index,$R$ is the gas constant,$T$ is the temperature in Kelvin,and $M$ is the molar mass.
For a diatomic gas,$\gamma = 1.4 = \frac{7}{5}$.
At $273^{\circ}C$,$T_1 = 273 + 273 = 546 \ K$.
So,$v_1 = \sqrt{\frac{7RT_1}{5M}} = \sqrt{\frac{7R(546)}{5M}}$.
The r.m.s. speed of gas molecules is given by $v_2 = \sqrt{\frac{3RT_2}{M}}$.
At $273 \ K$,$T_2 = 273 \ K$.
So,$v_2 = \sqrt{\frac{3R(273)}{M}}$.
Now,the ratio $\frac{v_1}{v_2} = \sqrt{\frac{7R(546)}{5M} \cdot \frac{M}{3R(273)}} = \sqrt{\frac{7 \cdot 546}{5 \cdot 3 \cdot 273}}$.
Since $546 = 2 \cdot 273$,we have $\frac{v_1}{v_2} = \sqrt{\frac{7 \cdot 2}{5 \cdot 3}} = \sqrt{\frac{14}{15}}$.

(B) For a monatomic gas,the degrees of freedom $f_1 = 3$. Given $n_1 = 2$ moles and $T_1 = 27^{\circ} C = 300 \ K$. The internal energy $U$ is given by $U = \frac{n_1 f_1 R T_1}{2} = \frac{2 \times 3 \times R \times 300}{2} = 900 R$.
For a diatomic gas,the degrees of freedom $f_2 = 5$. Given $n_2 = 3$ moles and $T_2 = 127^{\circ} C = 400 \ K$. The internal energy $U'$ is given by $U' = \frac{n_2 f_2 R T_2}{2} = \frac{3 \times 5 \times R \times 400}{2} = 3000 R$.
Now,calculating the ratio: $\frac{U'}{U} = \frac{3000 R}{900 R} = \frac{30}{9} = \frac{10}{3}$.
Therefore,$U' = \frac{10 U}{3}$.

(A) Initial state: $4$ moles of diatomic gas at temperature $T$. Internal energy $U_i = 4 \times \frac{5}{2} RT = 10 RT$.
Final state: After dissociation,$2$ moles of diatomic gas remain,and $2$ moles of diatomic gas dissociate into $4$ moles of monatomic gas (since $1$ mole of $X_2$ gives $2$ moles of $X$).
Final internal energy $U_f = (2 \times \frac{5}{2} RT) + (4 \times \frac{3}{2} RT) = 5 RT + 6 RT = 11 RT$.
Since the temperature remains constant,the heat supplied $Q$ is equal to the change in internal energy $\Delta U$.
$Q = U_f - U_i = 11 RT - 10 RT = RT$.

(D) The heat energy required at constant volume is given by the formula $Q = n C_v \Delta T$.
For a monatomic gas,the molar heat capacity at constant volume is $C_v = \frac{3}{2} R$.
Given $n_1 = 2 \text{ moles}$,$\Delta T_1 = 40^{\circ} C - 30^{\circ} C = 10 \text{ K}$.
So,$Q = 2 \times \frac{3}{2} R \times 10 = 30 R$.
For a diatomic gas,the molar heat capacity at constant volume is $C_v = \frac{5}{2} R$.
Given $n_2 = 4 \text{ moles}$,$\Delta T_2 = 33^{\circ} C - 28^{\circ} C = 5 \text{ K}$.
Let the required heat be $Q'$.
$Q' = n_2 C_v \Delta T_2 = 4 \times \frac{5}{2} R \times 5 = 50 R$.
Now,taking the ratio: $\frac{Q'}{Q} = \frac{50 R}{30 R} = \frac{5}{3}$.
Therefore,$Q' = \frac{5}{3} Q$.

(C) Statement $(I)$ is false because gas thermometers are more sensitive than liquid thermometers due to the higher coefficient of thermal expansion of gases.
Statement $(II)$ is true. Boltzmann's constant $k_B$ is defined as the ratio of the universal gas constant $R$ to Avogadro's number $N_A$,i.e.,$k_B = \frac{R}{N_A}$.
Statement $(III)$ is true. From the ideal gas equation $PV = nRT$,we have $PV = \frac{m}{M}RT$,where $m$ is mass and $M$ is molar mass. Since density $\rho = \frac{m}{V}$,we get $P = \frac{\rho RT}{M}$. At constant pressure $P$,$\rho \propto \frac{1}{T}$.
Therefore,statements $(II)$ and $(III)$ are true,but statement $(I)$ is false.

(A) From the ideal gas equation,$pV = nRT = (m/M)RT$,where $m$ is the mass of the gas and $M$ is the molar mass. Since $\rho = m/V$,we have $p = (\rho/M)RT$,or $\rho = pM / (RT)$.
Initially,the mass of the gas in the vessel is $m_1 = \rho V$.
After some gas is released,the pressure becomes $p' = p - \Delta p$. Since the volume $V$ and temperature $T$ remain constant,the new density $\rho'$ is given by $\rho' = p'M / (RT) = (p - \Delta p)M / (RT)$.
The new mass of the gas in the vessel is $m_2 = \rho' V = \frac{(p - \Delta p)M}{RT} V = \frac{(p - \Delta p)}{p} \rho V$.
The mass of the released gas $\Delta m$ is $m_1 - m_2 = \rho V - \frac{(p - \Delta p)}{p} \rho V$.
$\Delta m = \rho V \left(1 - \frac{p - \Delta p}{p}\right) = \rho V \left(\frac{p - p + \Delta p}{p}\right) = \frac{\rho V \Delta p}{p}$.

(C) The rms speed of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $v_{rms} \propto \sqrt{T}$,if the speed increases from $v$ to $\sqrt{3}v$,the temperature increases from $T_1$ to $T_2$ such that $\frac{v_{rms2}}{v_{rms1}} = \sqrt{\frac{T_2}{T_1}} = \sqrt{3}$.
Thus,$\frac{T_2}{T_1} = 3$,which implies $T_2 = 3T_1$.
The heat required for a diatomic gas at constant volume is given by $Q = n C_v \Delta T$.
For a diatomic gas,the degrees of freedom $f = 5$,so $C_v = \frac{5}{2}R$.
Given $n = 4 \ mol$,$Q = 83.1 \ kJ = 83100 \ J$,and $R = 8.31 \ J \ mol^{-1} \ K^{-1}$.
Substituting the values: $83100 = 4 \times \frac{5}{2} \times 8.31 \times (3T_1 - T_1)$.
$83100 = 10 \times 8.31 \times 2T_1$.
$83100 = 166.2 \times T_1$.
$T_1 = \frac{83100}{166.2} = 500 \ K$.
Converting to Celsius: $T_1(^{\circ}C) = 500 - 273 = 227^{\circ}C$.

(B) The kinetic energy of the gas molecules due to the bulk motion of the vessel is converted into internal energy of the gas when the vessel stops suddenly.
Let $m$ be the molar mass of the gas and $v$ be the velocity of the vessel.
The kinetic energy per mole of the gas is $K.E. = \frac{1}{2} M v^2$, where $M$ is the molar mass in $kg/mol$.
$M = 83 \, g/mol = 0.083 \, kg/mol$.
$K.E. = \frac{1}{2} \times 0.083 \times (30)^2 = \frac{1}{2} \times 0.083 \times 900 = 0.083 \times 450 = 37.35 \, J/mol$.
For a monatomic gas, the change in internal energy is $\Delta U = n C_v \Delta T$.
Since $n=1$ mole and $C_v = \frac{3}{2} R$ for a monatomic gas:
$\Delta U = 1 \times \frac{3}{2} \times 8.3 \times \Delta T = 12.45 \Delta T$.
Equating the kinetic energy to the change in internal energy:
$37.35 = 12.45 \Delta T$.
$\Delta T = \frac{37.35}{12.45} = 3 \, K$.

(A) Given: Initial temperature $T_1 = 27^{\circ} C = 300 \ K$,Initial pressure $P_1 = 600 \ kPa$.
After releasing one-fourth of the gas,the remaining amount of gas is $n_2 = \frac{3}{4} n_1$.
The volume $V$ of the tank remains constant,so $V_1 = V_2 = V$.
The final temperature is $T_2 = 327^{\circ} C = 600 \ K$.
Using the ideal gas law $PV = nRT$,we have $\frac{P_1 V}{n_1 T_1} = \frac{P_2 V}{n_2 T_2}$.
Rearranging for $P_2$: $P_2 = P_1 \times \left( \frac{n_2}{n_1} \right) \times \left( \frac{T_2}{T_1} \right)$.
Substituting the values: $P_2 = 600 \times \left( \frac{3}{4} \right) \times \left( \frac{600}{300} \right)$.
$P_2 = 600 \times 0.75 \times 2 = 900 \ kPa$.

(C) According to the kinetic theory of gases,the average kinetic energy of a gas molecule is given by $K.E. = \frac{3}{2} k_B T$. This shows that the temperature $T$ is a direct measure of the average kinetic energy of the molecules. Thus,statement $(a)$ is true.
Temperature is a state function that depends only on the average kinetic energy of the particles and is independent of the nature of the gas. Thus,statement $(b)$ is false.
The average speed of gas molecules is given by $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$,where $M$ is the molar mass. Since $v_{avg} \propto \frac{1}{\sqrt{M}}$,heavier molecules (larger $M$) have a lower average speed. Thus,statement $(c)$ is true and statement $(d)$ is false.
Therefore,statements $(a)$ and $(c)$ are true.

(D) For an ideal gas,the equation of state is $PV = nRT$.
At constant pressure $P$,differentiating with respect to temperature $T$,we get $P \frac{dV}{dT} = nR$,which implies $\frac{dV}{dT} = \frac{nR}{P}$.
The coefficient of volume expansion $\gamma$ is defined as $\gamma = \frac{1}{V} \frac{dV}{dT}$.
Substituting $\frac{dV}{dT} = \frac{nR}{P}$,we get $\gamma = \frac{1}{V} \left( \frac{nR}{P} \right) = \frac{nR}{PV}$.
Since $PV = nRT$,we have $\gamma = \frac{nR}{nRT} = \frac{1}{T}$.
Given temperature $T = 27^{\circ} C = 27 + 273 = 300 \ K$.
Therefore,$\gamma = \frac{1}{300} \ K^{-1} \approx 0.00333 \ K^{-1} = 33 \times 10^{-4} \ K^{-1}$.

(D) For an ideal gas,the volume expansion coefficient $(\gamma_V)$ and the pressure coefficient $(\gamma_P)$ are defined as follows:
$\gamma_V = \frac{1}{V} (\frac{\partial V}{\partial T})_P$
$\gamma_P = \frac{1}{P} (\frac{\partial P}{\partial T})_V$
For an ideal gas,the equation of state is $PV = nRT$.
At constant pressure,$V = (\frac{nR}{P})T$,so $(\frac{\partial V}{\partial T})_P = \frac{nR}{P}$. Thus,$\gamma_V = \frac{1}{V} \cdot \frac{nR}{P} = \frac{1}{T}$.
At constant volume,$P = (\frac{nR}{V})T$,so $(\frac{\partial P}{\partial T})_V = \frac{nR}{V}$. Thus,$\gamma_P = \frac{1}{P} \cdot \frac{nR}{V} = \frac{1}{T}$.
Since $\gamma_V = \gamma_P = \frac{1}{T}$,their difference is $\gamma_V - \gamma_P = 0$.

(D) The heat supplied at constant pressure is given by $Q = n C_p \Delta T$. Since $Q$ and $n$ are equal,$C_p \Delta T$ is constant for both gases.
For a monatomic gas,$C_{p,1} = \frac{5}{2}R$ and the change in internal energy is $\Delta U_1 = n C_{v,1} \Delta T_1 = n (\frac{3}{2}R) \Delta T_1$.
For a diatomic gas,$C_{p,2} = \frac{7}{2}R$ and the change in internal energy is $\Delta U_2 = n C_{v,2} \Delta T_2 = n (\frac{5}{2}R) \Delta T_2$.
Since $Q = n C_{p,1} \Delta T_1 = n C_{p,2} \Delta T_2$,we have $\frac{5}{2}R \Delta T_1 = \frac{7}{2}R \Delta T_2$,which implies $\Delta T_1 = \frac{7}{5} \Delta T_2$.
The ratio of internal energy changes is $\frac{\Delta U_1}{\Delta U_2} = \frac{n (\frac{3}{2}R) \Delta T_1}{n (\frac{5}{2}R) \Delta T_2} = \frac{3}{5} \times \frac{\Delta T_1}{\Delta T_2} = \frac{3}{5} \times \frac{7}{5} = \frac{21}{25}$.

(C) For $1 \text{ mole}$ of an ideal gas,$pV = RT \Rightarrow p = \frac{RT}{V}$.
Given the process equation: $p = p_0 \left(1 - \alpha \frac{V^3}{V_0^3}\right)$.
Equating the two expressions for $p$: $\frac{RT}{V} = p_0 - \frac{\alpha p_0 V^3}{V_0^3} \Rightarrow T = \frac{p_0 V}{R} - \frac{\alpha p_0 V^4}{R V_0^3}$.
For maximum temperature,$\frac{dT}{dV} = 0$: $\frac{dT}{dV} = \frac{p_0}{R} - \frac{4 \alpha p_0 V^3}{R V_0^3} = 0$.
This gives $V^3 = \frac{V_0^3}{4 \alpha}$.
Substituting $V^3$ back into the expression for $T$: $T_{\max} = \frac{p_0}{R} \left( \frac{V_0}{(4 \alpha)^{1/3}} \right) - \frac{\alpha p_0}{R V_0^3} \left( \frac{V_0^3}{4 \alpha} \right) \left( \frac{V_0}{(4 \alpha)^{1/3}} \right) = \frac{p_0 V_0}{R (4 \alpha)^{1/3}} \left( 1 - \frac{1}{4} \right) = \frac{3}{4} \frac{p_0 V_0}{R (4 \alpha)^{1/3}}$.
Given $T_{\max} = \frac{3}{4} \frac{p_0 V_0}{R}$,we have $\frac{3}{4} \frac{p_0 V_0}{R} = \frac{3}{4} \frac{p_0 V_0}{R (4 \alpha)^{1/3}}$.
Thus,$(4 \alpha)^{1/3} = 1 \Rightarrow 4 \alpha = 1 \Rightarrow \alpha = \frac{1}{4}$.

(C) Initial state: The tube length is $100 \ cm$, with a $10 \ cm$ mercury column in the middle. The length of air on each side is $(100 - 10) / 2 = 45 \ cm$. Initial pressure $P_0 = 76 \ cm$ of $Hg$, initial temperature $T_0 = 31 + 273 = 304 \ K$.
Applying the ideal gas law $\frac{PV}{T} = \text{constant}$, for the initial state: $\frac{P_0 V_0}{T_0} = \frac{76 \times 45}{304}$.
Final state: Let the new length of the air column at $0^{\circ} C$ $(273 \ K)$ be $l$, and the length of the air column at $273^{\circ} C$ $(546 \ K)$ be $(90 - l)$. Let the new pressure be $P'$.
Since the mercury column is at rest, the pressure on both sides must be equal: $P_2 = P_3 = P'$.
Using the gas law for both sides:
$\frac{P' l}{273} = \frac{P' (90 - l)}{546} = \frac{76 \times 45}{304}$.
From $\frac{P' l}{273} = \frac{P' (90 - l)}{546}$, we get $2l = 90 - l$, which implies $3l = 90$, so $l = 30 \ cm$.
Now, substitute $l = 30 \ cm$ into the equation: $\frac{P' \times 30}{273} = \frac{76 \times 45}{304}$.
$P' = \frac{76 \times 45 \times 273}{304 \times 30} = \frac{76 \times 45 \times 273}{9120} = 102.375 \approx 102.4 \ cm$ of $Hg$.

(D) From the ideal gas equation,we have $PV = nRT$.
Since the number of moles $n = \frac{m}{M}$,where $m$ is the mass and $M$ is the molar mass,we can write $PV = \frac{m}{M}RT$.
Dividing both sides by volume $V$,we get $P = \frac{m}{V} \cdot \frac{RT}{M} = \frac{\rho RT}{M}$,where $\rho = \frac{m}{V}$ is the density.
For two gases $A$ and $B$ at the same temperature $T$,the ratio of their partial pressures is given by:
$\frac{P_A}{P_B} = \frac{\rho_A R T / M_A}{\rho_B R T / M_B} = \frac{\rho_A}{\rho_B} \times \frac{M_B}{M_A}$.
Rearranging to find the ratio of densities:
$\frac{\rho_A}{\rho_B} = \frac{P_A}{P_B} \times \frac{M_A}{M_B}$.
Given $\frac{M_A}{M_B} = \frac{3}{4}$ and $\frac{P_A}{P_B} = \frac{2}{3}$.
Substituting these values:
$\frac{\rho_A}{\rho_B} = \frac{2}{3} \times \frac{3}{4} = \frac{6}{12} = 0.5$.

(B) In a uniform gravitational field,the center of mass $(COM)$ and the center of gravity $(COG)$ coincide.
The height of the center of mass $y_{\text{cen}}$ is given by the formula: $y_{\text{cen}} = \frac{\int_{0}^{\infty} y \, dm}{\int_{0}^{\infty} dm} = \frac{\int_{0}^{\infty} y \rho(y) \, dy}{\int_{0}^{\infty} \rho(y) \, dy}$.
According to the barometric formula,the density of the gas at height $y$ is given by: $\rho(y) = \rho_{0} e^{-Mgy / RT}$.
Substituting this into the integral:
$y_{\text{cen}} = \frac{\int_{0}^{\infty} y e^{-Mgy / RT} \, dy}{\int_{0}^{\infty} e^{-Mgy / RT} \, dy}$.
Using the standard integral $\int_{0}^{\infty} x e^{-ax} \, dx = \frac{1}{a^2}$ and $\int_{0}^{\infty} e^{-ax} \, dx = \frac{1}{a}$,where $a = \frac{Mg}{RT}$:
$y_{\text{cen}} = \frac{1/a^2}{1/a} = \frac{1}{a} = \frac{RT}{Mg}$.

(C) For a process,the molar heat capacity is given by $C = C_{V} + \frac{P}{n} \frac{dV}{dT}$.
For a monoatomic gas,$C_{V} = \frac{3}{2} R$.
The process is a straight line in the $V-T$ diagram passing through $(V_{0}, T_{0})$ and $(2V_{0}, 3T_{0})$.
The equation of the line is $V - V_{0} = \frac{2V_{0} - V_{0}}{3T_{0} - T_{0}} (T - T_{0}) = \frac{V_{0}}{2T_{0}} (T - T_{0})$.
Thus,$\frac{dV}{dT} = \frac{V_{0}}{2T_{0}}$.
Using the ideal gas law $PV = nRT$,we have $P = \frac{nRT}{V}$.
At the point $(V_{0}, T_{0})$,$P = \frac{nRT_{0}}{V_{0}}$.
Substituting these into the heat capacity formula:
$C = \frac{3}{2} R + \left( \frac{nRT_{0}}{V_{0}} \right) \frac{1}{n} \left( \frac{V_{0}}{2T_{0}} \right) = \frac{3}{2} R + \frac{R}{2} = 2R$.

(B) The total pressure inside the container is the sum of the partial pressure of dry air $(p)$ and the saturated vapour pressure of water $(\bar{p})$,so $P_{total} = p + \bar{p}$.
When the volume is compressed to half $(V' = V/2)$ at a constant temperature,the partial pressure of the dry air follows Boyle's Law $(P_1V_1 = P_2V_2)$.
Thus,the new pressure of dry air becomes $p' = p \times (V / (V/2)) = 2p$.
Since the temperature remains constant,the saturated vapour pressure of water $(\bar{p})$ remains unchanged because it depends only on temperature.
Therefore,the new total pressure is $P'_{total} = p' + \bar{p} = 2p + \bar{p}$.

(B) The collision frequency $(Z)$ is given by the formula: $Z = \sqrt{2} \pi d^2 N \bar{v}$,where $\bar{v} = \sqrt{\frac{8RT}{\pi M}}$.
Since temperature $(T)$ and number density $(N)$ are the same for both gases,we have $Z \propto d^2 \sqrt{\frac{1}{M}}$.
Given: $d_A = \frac{d_B}{2}$ and $M_A = 4M_B$.
Therefore,the ratio of collision frequencies is:
$\frac{Z_A}{Z_B} = \left( \frac{d_A}{d_B} \right)^2 \sqrt{\frac{M_B}{M_A}}$
$\frac{Z_A}{Z_B} = \left( \frac{1}{2} \right)^2 \sqrt{\frac{M_B}{4M_B}} = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$.
Given $Z_B = 32 \times 10^{18} /s$,we get:
$Z_A = \frac{32 \times 10^{18}}{8} = 4 \times 10^{18} /s$.

(A) Consider a small element of gas of length $dx$ at a distance $x$ from the axis of rotation $A$. Let $A$ be the cross-sectional area of the tube.
The net force on this element providing the centripetal acceleration is $(P+dP)A - PA = (dm) \omega^2 x$.
$AdP = (dm) \omega^2 x$.
Since $dm = \rho A dx$,we have $dP = \rho \omega^2 x dx$.
Using the ideal gas law $PM = \rho RT$,we have $\rho = \frac{PM}{RT}$.
Substituting $\rho$,we get $dP = \left(\frac{PM}{RT}\right) \omega^2 x dx$.
Rearranging the terms,$\frac{dP}{P} = \frac{M \omega^2}{RT} x dx$.
Integrating from $x=0$ to $x=l$ and $P=P_A$ to $P=P_B$:
$\int_{P_A}^{P_B} \frac{dP}{P} = \frac{M \omega^2}{RT} \int_0^l x dx$.
$\ln\left(\frac{P_B}{P_A}\right) = \frac{M \omega^2}{RT} \left[\frac{x^2}{2}\right]_0^l = \frac{M \omega^2 l^2}{2RT}$.
Therefore,$P_B = P_A \exp\left(\frac{M \omega^2 l^2}{2RT}\right)$.

(A) Statement $I$ is correct. The change in internal energy for an ideal gas is given by $\Delta U = nC_v \Delta T$. Since $C_v = \frac{R}{\gamma - 1}$,substituting this gives $\Delta U = \frac{nR}{\gamma - 1}(T_f - T_i)$.
Statement $II$ is also correct. The molar heat capacity at constant volume is $C_v = \frac{fR}{2}$ and at constant pressure is $C_p = C_v + R = (\frac{f}{2} + 1)R$. Therefore,$\gamma = \frac{C_p}{C_v} = \frac{(\frac{f}{2} + 1)R}{\frac{fR}{2}} = \frac{f+2}{f} = 1 + \frac{2}{f}$.
Since both statements are true and the definition of $\gamma$ in terms of $f$ is a fundamental property used to derive the expression in Statement $I$,$R$ is the correct explanation of $A$.

(C) The average kinetic energy of an ideal gas molecule is given by the formula $KE_{avg} = \frac{3}{2}kT$,where $k$ is the Boltzmann constant and $T$ is the absolute temperature.
Since $KE_{avg}$ depends only on the temperature $T$,if the average kinetic energy of $H_2$ and $O_2$ molecules is the same,their temperatures must be the same. Thus,Assertion $A$ is true.
The root mean square (r.m.s.) speed of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the temperature,and $M$ is the molar mass.
Since the molar mass $M$ of $H_2$ $(2 \ g/mol)$ and $O_2$ $(32 \ g/mol)$ are different,their $v_{rms}$ values will be different even at the same temperature. Thus,Reason $R$ is false.

(A) Let $V$ be the volume of each vessel.
Initially,the total number of moles $n = n_1 + n_2 = \frac{P_0 V}{RT_0} + \frac{P_0 V}{RT_0} = \frac{2P_0 V}{RT_0}$.
Given $P_0 = 90 \text{ kPa}$ and $T_0 = 400 \text{ K}$,so $n = \frac{2 \cdot 90 \cdot V}{R \cdot 400} = \frac{180V}{400R}$.
Finally,let the pressure in both vessels be $P'$ because they are connected.
The total number of moles $n'$ remains constant,so $n' = n$.
$n' = \frac{P' V}{RT_1} + \frac{P' V}{RT_2} = \frac{P' V}{R} (\frac{1}{400} + \frac{1}{500}) = \frac{P' V}{R} (\frac{5+4}{2000}) = \frac{9P' V}{2000R}$.
Equating $n = n'$:
$\frac{180V}{400R} = \frac{9P' V}{2000R}$.
$P' = \frac{180}{400} \cdot \frac{2000}{9} = \frac{180}{9} \cdot \frac{2000}{400} = 20 \cdot 5 = 100 \text{ kPa}$.

(C) We know the ideal gas equation is $PV = nRT$,which implies $P = \frac{nRT}{V}$.
Given the process equation is $PT^3 = C$,where $C$ is a constant.
Substituting the expression for $P$ into the process equation,we get $\left(\frac{nRT}{V}\right) T^3 = C$.
This simplifies to $\frac{T^4}{V} = \frac{C}{nR} = C'$,where $C'$ is another constant.
Therefore,$V = \frac{1}{C'} T^4$,which means $V \propto T^4$.
Taking the logarithmic derivative of $V = kT^4$,we get $\frac{dV}{V} = 4 \frac{dT}{T}$.
The coefficient of volume expansion $\beta$ is defined as $\beta = \frac{1}{V} \frac{dV}{dT}$.
Substituting the derivative,we get $\beta = \frac{1}{V} \left( \frac{4V}{T} \right) = \frac{4}{T}$.