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 $X=\frac{3}{2}\text{R ln}\left(\frac{T}{T_A}\right)+\text{R ln}\left(\frac{V}{V_A}\right)$. Here, $V$ 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.
$\begin{array}{cc}\text{List - I}& \text{List - II}\\ \text{(i) Work done by the system in process}1\to 2\to 3& \left(P\right)\frac{1}{3}R{T}_0\text{ln}2\\ \text{(ii)Change in internal energy in process}1\to 2\to 3& \left(Q\right)\frac{1}{3}R{T}_0\\ \text{(iii)Heat absorbed by the system in process}1\to 2\to 3& \left(R\right)R{T}_0\\ \text{(iv) Heat absorbed by the system in process}1\to 2& \left(S\right)\frac{4}{3}R{T}_0\\ & \left(T\right)\frac{1}{3}R{T}_0(3+\text{ln}2)\\ & \left(U\right)\frac{5}{6}R{T}_0\end{array}$

If the process carried out on one mole of monoatomic ideal gas is as shown in figure in the PV diagram with $P_0{V}_0=\frac{1}{3}R{T}_0$. The correct match is-

• A. $i\to Q,ii\to R,iii\to S,iv\to U$
• B. $i\to Q,ii\to S,iii-R,iv\to U$
• C. $i\to Q,ii\to R,iii\to P,iv\to U$
• D. $i\to S,ii\to R,iii\to Q,iv\to T$

Answer 1

The correct option is A $i\to Q,ii\to R,iii\to S,iv\to U$

Degree of freedom = 3
W.D. = Area under PV graph $=\frac{R{T}_0}{3}$
Change in internal Energy
$\Delta U=n{C}_v\Delta T$
$=\frac{f}{2}nR\Delta T=\frac{f}{2}(p_f{v}_f-P_i{V}_i)$
$=\frac{3}{2}(\frac{3P_0}{2}2V_0-P_0{V}_0)=3P_0{V}_0$
$\Delta U=R{T}_0$
Heat absorbed $(1\to 2\to 3)$
$\Delta Q=\Delta U+W$
$\Delta Q=R{T}_0+\frac{R{T}_0}{3}$
$\Delta Q=\frac{4T_{0}R}{3}$
Heat absorbed $(1\to 2)$
$\Delta Q=\Delta U+W$
$=\frac{f}{2}(P_f{V}_f-P_i{V}_i)+W$
$=\frac{3}{2}(P_{0}2V_0-P_0{V}_0)+P_0{V}_0$
$=\frac{5}{2}{P}_0{V}_0=\frac{5}{2}\left(\frac{R{T}_0}{3}\right)$