Question

An ideal monoatomic gas is carried around the cycle ABCDA as shown in the figure. The efficiency of the gas cycle is

• A. $\frac{2}{31}$
• B. $\frac{4}{31}$
• C. $\frac{2}{21}$
• D. $\frac{4}{21}$

Answer 1

The correct option is D

$\frac{4}{21}$

Net work done per cycle,

$W_{net}=(3P_0-P_0)\times (2V_0-V_0)$

$=2P_0{V}_0$

Process $A\to B$ :

$W_1=0;\Delta Q=\Delta U=n{C}_v.\Delta T$

$=\frac{3R}{2}[\frac{3P_0{V}_0}{R}-\frac{P_0{V}_0}{R}]=+3P_0{V}_0$

Process $B\to C$∶

$\Delta Q=n{C}_p\Delta T$

$=\frac{5R}{2}[\frac{6P_0{V}_0}{R}-\frac{3P_0{V}_0}{R}]$

$=+\frac{15}{2}{P}_0{V}_0$

In the processes $C\to D$ and $D\to A$, heat is lost (i.e, $\Delta Q$ is-ve)

Total heat absorbed per cycle is,

$Q_{\text{absorbed}}=3P_0{V}_0+\frac{15P_0{V}_0}{2}$

$=\frac{21P_0{V}_0}{2}$

$\therefore Efficiency=\frac{W_{net}\text{Per cycle}}{Q_{\text{absorbed}}}$

$=\frac{2P_0{V}_0}{\left(\frac{21P_0{V}_0}{2}\right)}=\frac{4}{21}$