JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Heat & Thermodynamics

In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by $\mathrm{T}\Delta \mathrm{X}$, where T is temperature of the system and $\Delta \mathrm{X}$ is the infinitesimal change in a thermodynamic quantity X of the system. For a mole of monatomic ideal gas $\mathrm{X}=\frac{3}{2}\mathrm{R}\ln \left(\frac{\mathrm{T}}{{\mathrm{T}}_{\mathrm{A}}}\right)+\mathrm{R}\ln \left(\frac{\mathrm{V}}{{\mathrm{V}}_{\mathrm{A}}}\right)$. Here, R is gas constant, V is volume of gas, ${\mathrm{T}}_{\mathrm{A}}$ and ${\mathrm{V}}_{\mathrm{A}}$ 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
(I) Work done by the system in process $1\to 2\to 3$
(II) Change in internal energy in process $1\to 2\to 3$
(III) Heat absorbed by the system in process $1\to 2\to 3$
(IV) Heat absorbed by the system in process $1\to 2$

List-II
(P) $\frac{1}{3}{\mathrm{RT}}_0\ln 2$
(Q) $\frac{1}{3}{\mathrm{RT}}_0$
(R) ${\mathrm{RT}}_0$
(S) $\frac{4}{3}{\mathrm{RT}}_0$
(T) $\frac{1}{3}{\mathrm{RT}}_0(3+\ln 2)$
(U) $\frac{5}{6}R{T}_0$

If the process on one mole of monatomic ideal gas is an shown is as shown in the TV-diagram with ${\mathrm{P}}_0{\mathrm{V}}_0=\frac{1}{3}{\mathrm{RT}}_0$, the correct match is