Question

In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by T$\Delta$X, where T is temperature of the system and $\Delta$X is the infinitesimal change in a thermodynamic quantity X of the system. For a mole of monatomic ideal gas $X = \frac{3}{2} R\, In \left(\frac{T}{T_{A}}\right)+R\,In \left(\frac{V}{V_{A}}\right)$. Here, R is gas constant, V is volume of gas, $T_{A}$ and $V_{A}$ are constants.
The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities.

List-I	List-II
(I) Work done by the system in process 1$\to$2$\to$3	(p) $\frac{1}{3}RT_{0} \,In\,2$
(II) Change in internal energy in process 1$\to$2$\to$3	(Q) $\frac{1}{3}RT_{0}$
(III) Heat absorbed by the system in process 1$\to$2$\to$3	(R) $RT_{0}$
(IV) Heat absorbed by the system in process 1$\to$2	(S) $\frac{4}{3}RT_{0}$
	(T) $\frac{1}{3}RT_{0} \left(3+In\,2\right)$
	(U) $\frac{5}{6}RT_{0} $

Question: If the process on one mole of monatomic ideal gas is as shown in the TV-diagram with $P_{0}V_{0} = \frac{1}{3}RT_{0},$ the correct match is.

• A. I $\to$ P, II $\to$ R, III $\to$ T, IV $\to$ S
• B. I $\to$ P, II $\to$ T, III $\to$ Q, IV $\to$ T
• C. I $\to$ P, II $\to$ R, III $\to$ T, IV $\to$ P
• D. I $\to$ S, II $\to$ T, III $\to$ Q, IV $\to$ U

Answer 1

Answer: (C)

Process $1 \to 2$ is isothermal (temperature constant)
Process $2 \to 3$ is isochoric (volume constant)
(I) Work done in $1 \to 2 \to 3$
$W = W_{1\to2} + W_{2 \to 3}$
$= nRT\,ln \left(\frac{V_{f}}{V_{i}}\right)+W_{2 \to 3}$
$= \frac{RT_{0}}{3}ln\left(\frac{2V_{0}}{V_{0}}\right)+0$
$W = \frac{RT_{0}}{3}ln 2\quad\Rightarrow \left(P\right)$
(II) $\Delta U\, in\,1 \to 2 \to 3$
$\Delta U = \frac{f}{2}nR\left(T_{f} - T_{i}\right)$
$= \frac{3}{2}R\left(T_{0}-\frac{T_{0}}{3}\right)$
$= \frac{3}{2}R\left(\frac{2T_{0}}{3}\right)$
$\Delta U = RT_{0}\quad\Rightarrow \left(R\right)$
(III) For any system, first law of thermodynamics
for $1 \to 2 \to 3$
$\Delta Q = \Delta U + W$
$\Delta Q = RT_{0} + \frac{RT_{0}}{3}ln2$
$\Delta Q = \frac{RT_{0}}{3}\left(3+ln2\right) \Rightarrow \left(T\right)$
For process $1 \Rightarrow 2$ (isothermal)
$\Delta Q = \Delta U + W$
$= \frac{f}{2}nR\left(T_{f} - T_{i}\right)+nRT ln\left(V_{f} /V_{i}\right)$
$= 0 +R\left(\frac{T_{0}}{3}\right)ln\left(\frac{2v_{0}}{v_{0}}\right)$
$\Delta Q = \frac{RT_{0}}{3}ln 2\,\Rightarrow \, \left(P\right)$