
Therefore, temperature and hence, internal energy of the gas will decrease ($T \propto PV$) or $\Delta {U_{A \to B}} = $negative.
Further $\Delta {W_{A \to B}}$ is also negative as the volume of the gas is decreasing. Thus $\Delta {Q_{A \to B}}$ is negative.
In process $B$ to $C$ pressure of the gas is constant while volume is increasing.
Hence temperature should increase or $\Delta {U_{B \to C}}$= positive.
During $C$ to $A$ volume is constant while pressure is increasing.
Therefore, temperature and hence, internal energy of the gas should increase or $\Delta {U_{C \to A}}$= positive.
During process $CAB$ volume of the gas is decreasing. Hence, work done by the gas is negative.
| 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. |


($1$) The value of $\frac{T_R}{T_0}$ is
$(A)$ $\sqrt{2}$ $(B)$ $\sqrt{3}$ $(C)$ $2$ $(D)$ $3$
($2$) The value of $\frac{Q}{R T_0}$ is
$(A)$ $4(2 \sqrt{2}+1)$ $(B)$ $4(2 \sqrt{2}-1)$ $(C)$ $(5 \sqrt{2}+1)$ $(D)$ $(5 \sqrt{2}-1)$
Give the answer or qution ($1$) and ($2$)

