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 $X=\frac{3}{2}RIn\left(\frac{T}{T_A}\right)+RIn\left(\frac{V}{V_A}\right)$. 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) $\frac{1}{3}R{T}_{0}In2$
(II) Change in internal energy in process 1$\to$2$\to$3	(Q) $\frac{1}{3}R{T}_0$
(III) Heat absorbed by the system in process 1$\to$2$\to$3	(R) $R{T}_0$
(IV) Heat absorbed by the system in process 1$\to$2	(S) $\frac{4}{3}R{T}_0$
	(T) $\frac{1}{3}R{T}_0\left(3+In2\right)$
	(U) $\frac{5}{6}R{T}_0$

Question: If the process on one mole of monatomic ideal gas is as shown in the TV-diagram with $P_0{V}_0=\frac{1}{3}R{T}_0,$ 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 $1\to 2\to 3$
$W=W_{1\to 2}+W_{2\to 3}$
$=nRTln\left(\frac{V_f}{V_i}\right)+W_{2\to 3}$
$=\frac{R{T}_0}{3}ln\left(\frac{2V_0}{V_0}\right)+0$
$W=\frac{R{T}_0}{3}ln2\Rightarrow \left(P\right)$
(II) $\Delta Uin1\to 2\to 3$
$\Delta U=\frac{f}{2}nR\left(T_f-T_i\right)$
$=\frac{3}{2}R\left(T_0-\frac{T_0}{3}\right)$
$=\frac{3}{2}R\left(\frac{2T_0}{3}\right)$
$\Delta U=R{T}_0\Rightarrow \left(R\right)$
(III) For any system, first law of thermodynamics
for $1\to 2\to 3$
$\Delta Q=\Delta U+W$
$\Delta Q=R{T}_0+\frac{R{T}_0}{3}ln2$
$\Delta Q=\frac{R{T}_0}{3}\left(3+ln2\right)\Rightarrow \left(T\right)$
For process $1\Rightarrow 2$ (isothermal)
$\Delta Q=\Delta U+W$
$=\frac{f}{2}nR\left(T_f-T_i\right)+nRTln\left(V_f/V_i\right)$
$=0+R\left(\frac{T_0}{3}\right)ln\left(\frac{2v_0}{v_0}\right)$
$\Delta Q=\frac{R{T}_0}{3}ln2\Rightarrow \left(P\right)$