Question

An ideal monoatomic gas undergoes the cyclic process $\text{ABCDA}$ as shown in the figure. The efficiency of the cycle is

• A. $28.6\%$
• B. $19.04\%$
• C. $46.8\%$
• D. $62.3\%$

Answer 1

The correct option is B $19.04\%$

Total work done in the cyclic process can be calculated from the area under $P-V$ graph.

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

$\Rightarrow W_{Total}=2P_0{V}_0$

Now, from first law of thermodynamics, heat given to a gas is equal to the sum of the change in its internal energy and work done by it.

So, heat transferred during the $\text{process AB}:$
$Q_{AB}=W_{AB}+U_{AB}$
and $U_{AB}=n\times \frac{3}{2}R(T_B-T_A)=\frac{3}{2}(P_B{V}_B-P_A{V}_A)$ [for monoatomic gas]
$\Rightarrow Q_{AB}=0+\frac{3}{2}(3P_0{V}_0-P_0{V}_0)=3P_0{V}_0$

Heat transferred during the $\text{process BC}:$
$Q_{BC}=W_{BC}+U_{BC}$
$Q_{BC}=3P_0{V}_0+\frac{3}{2}(3P_0\times 2V_0-3P_0{V}_0)=7.5P_0{V}_0$

Heat transferred during the $\text{process CD}:$
$Q_{CD}=W_{CD}+U_{CD}$
$Q_{CD}=0+\frac{3}{2}(P_0\times 2V_0-3P_0\times 2V_0)=-6P_0{V}_0$

Heat transferred during the $\text{process DA}:$
$Q_{DA}=W_{DA}+U_{DA}$
$Q_{DA}=-P_0{V}_0+\frac{3}{2}(P_0\times V_0-P_0\times 2V_0)=-2.5P_0{V}_0$

So, net heat added during the cycle is
$Q_{net}=Q_{AB}+Q_{BC}=3P_0{V}_0+7.5P_0{V}_0$

& total work done in cycle $W_{Total}=W_{BC}+W_{DA}=2P_0{V}_0$

Therefore, efficiency of the cycle,
$\eta =\frac{W_{Total}}{Q_{net}}\times 100=\frac{2P_0{V}_0}{10.5P_0{V}_0}\times 100$

$\therefore \eta =19.04\%$

Hence, option (b) is the correct answer.