Question

In a thermodynamic process on an ideal monoatomic 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 monoatomic ideal gas $X=\frac{3}{2}R\text{ln}\left(\frac{T}{T_A}\right)+R\text{ln}\left(\frac{V}{V_A}\right)$. Here, $R$ is gas constant, $T_A$ 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 qunatitites.
$\begin{array}{cc}\text{List - I}& \text{List - II}\\ \text{(i) Work done by the system in process}1\to 2\to 3& \left(P\right)\frac{1}{3}R{T}_0\text{ln}2\\ \text{(ii)Change in internal energy in process}1\to 2\to 3& \left(Q\right)\frac{1}{3}R{T}_0\\ \text{(iii)Heat absorbed by the system in process}1\to 2\to 3& \left(R\right)R{T}_0\\ \text{(iv) Heat absorbed by the system in process}1\to 2& \left(S\right)\frac{4}{3}R{T}_0\\ & \left(T\right)\Delta Q=R{T}_0+\frac{R{T}_0}{3}\text{ln}2\end{array}$
If the process carried out on one mole of monoatomic 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. $i\to P,ii\to T,iii-Q,iv\to T$
• B. $i\to S,ii\to T,iii-Q,iv\to U$
• C. $i\to P,ii\to R,iii-T,iv\to S$
• D. $i\to P,ii\to R,iii-T,iv\to P$

Answer 1

The correct option is D $i\to P,ii\to R,iii-T,iv\to P$

Process $1\to 2)$ (Isothermal)
Process $2\to 3$ (Isochoric)
W.D. $(1\to 2\to 3)$
$=\text{nRT ln}\left(\frac{V_f}{V_i}\right)+W_{2\to 3}$
$=\frac{R{T}_0}{3}\text{ln}2+0$
$\Delta U(1\to 2\to 3)$
$\Delta U=\frac{f}{2}nR(T_f-T_i)$
$=\frac{3}{2}R(T_0-\frac{T_0}{3})$
$\Delta U=R{T}_0$
For $1\to 2\to 3$
$\Delta Q=\Delta U+W$
$\Delta Q=R{T}_0+\frac{R{T}_0}{3}\text{ln}2$
For $1\to 2$
$\Delta Q=\Delta U_{1\to 2}+W_{1\to 2}$
$=\frac{f}{2}nR(T_f-T_i)+\text{nRT ln}\left(\frac{V_f}{V_i}\right)$
$=0+\frac{R{T}_0}{3}\text{ln}2$