Question

Some gas $(\frac{C_p}{C_v}$ = $\gamma$ = 1.25) follows the cycle ABCDA as shown in the figure. The ratio of the energy given out by the gas to its surroundings during the isochoric section of the cycle to expansion work done during the isobaric section of the cycle is:

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (B)

4

Let $n$ moles of gas follows the cycle $ABCDA$

$Q_{\text{isoch or ic}}=n{C}_v\Delta T=\frac{nR}{\gamma -1}(T_B-T_C)$

$W_{isobaric}=nR\Delta T=nR(T_A-T_D)=nR\left(T_{B-T_C}\right)$

$\therefore$ Required ratio $=\frac{Q_{\text{isoch or ic}}}{W_{isobaric}}=\frac{1}{\gamma -1}=4$