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 carried out on one mole of monatomic ideal gas is as shown in figure in the PV-diagram with $P_{0} V_{0} = \frac{1}{3}RT_{0},$ the correct match is.

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

Answer 1

Answer: (B)

Degree of freedom f = 3
Work done in any process = Area under P-V graph
$\Rightarrow $ Work done in $1 \to 2 \to 3 = P_{0} \,V_{0}$
$= \frac{RT_{0}}{3}\Rightarrow \left(Q\right)$
(II) Change in internal energy $1 \to 2 \to 3$
$\Delta U = nC_{v}\Delta T$
$= \frac{f}{2}nR\Delta T$
$= \frac{f}{2}\left(P_{f} V_{f} - P_{i} V_{i}\right)$
$= \frac{3}{2}\left(\frac{3P_{0}}{2}2V_{0}-P_{0}V_{0}\right)$
$= 3P_{0}V_{0}$
$\Delta U = RT_{0} \Rightarrow \left(R\right)$
(III) Heat absorbed in $1 \to 2 \to 3$
for any process, $I^{st}$ law of thermodynamics
$\Delta Q = \Delta W+ \omega$
$\Delta Q = RT_{0} +\frac{RT_{0}}{3}$
$\Delta Q = \frac{4RT_{0}}{3}\Rightarrow \left(S\right)$
(IV) Heat absorbed in process $1 \to 2$
$\Delta Q = \Delta U+ W$
$= \frac{f}{2}\left(P_{f} V_{f} - P_{i} V_{i}\right)+W$
$= \frac{3}{2}\left(P_{0} 2V_{0}-P_{0} V_{0}\right)+P_{0}V_{0}$
$= \frac{5}{2} P_{0} V_{0}$
$= \frac{5}{2}\left(\frac{RT_{0}}{3}\right)$
$\Delta Q = \frac{5RT_{0}}{6} \Rightarrow \left(U\right)$