| Column $I$ | Column $II$ |
| $(p)$ isobaric | $(x)$ $\frac{{PV(1 - {2^{1 - \gamma }})}}{{\gamma - 1}}$ |
| $(q)$ isothermal | $(y)$ $PV$ |
| $(r)$ adiabatic | (z) $PV\,\iota n\,2$ |
The correct matching of column $I$ and column $II$ is given by
$P_{i}=P$
$(P) \rightarrow(y)$ isobaric process
$\mathrm{W}=\mathrm{P} \Delta \mathrm{V}=\mathrm{PV}$
$(q) \rightarrow(z)$ isothermal
$\mathrm{W}=\mathrm{nRT} \ln \frac{\mathrm{V}_{\mathrm{f}}}{\mathrm{V}_{\mathrm{i}}}=\mathrm{PV} \ln 2$
$(r) \rightarrow(x)$ Adiabatic
$P_{f}=\left(\frac{V_{i}}{V_{f}}\right)^{\gamma} P_{i}=2^{\gamma} P$
$W=\frac{P_{f} V_{f}-P_{i} V_{i}}{1-\gamma}=\frac{\left(2^{-\gamma} P\right)(2 V)-P V}{1-\gamma}$
$\Rightarrow w=\frac{P V\left(1-2^{1-\gamma}\right)}{\gamma-1}$

$(A)$ If $V_2=2 V_1$ and $T_2=3 T_1$, then the energy stored in the spring is $\frac{1}{4} P_1 V_1$
$(B)$ If $V_2=2 V_1$ and $T_2=3 T_1$, then the change in internal energy is $3 P_1 V_1$
$(C)$ If $V_2=3 V_1$ and $T_2=4 T_1$, then the work done by the gas is $\frac{7}{3} P_1 V_1$
$(D)$ If $V_2=3 V_1$ and $T_2=4 T_1$, then the heat supplied to the gas is $\frac{17}{6} P_1 V_1$
| Column $I$ | Column $II$ |
| $(A)$ Process $A \rightarrow B$ | $(p)$ Internal energy decreases. |
| $(B)$ Process $B \rightarrow C$ | $(q)$ Internal energy increases. |
| $(C)$ Process $C \rightarrow D$ | $(r)$ Heat is lost. |
| $(D)$ Process $D \rightarrow A$ | $(s)$ Heat is gained. |
| $(t)$ Work is done on the gas. |

