Question

A monoatomic ideal gas goes through a cyclic process as shown in the figure. The efficiency of this process is

• A. $78\%$
• B. $42\%$
• C. $62\%$
• D. $19.04\%$

Answer 1

Answer: (D)

We know that,
$W=P\Delta V$
Now, $W_{AB}=0$
$W_{BC}=3P(2V-V)=3PV$
$W_{CD}=0$
$W_{DA}=-PV$
Total work done,
$W=W_{AB}+W_{BC}+W_{CD}+W_{DA}$
$=3PV-PV=2PV$
Heat given to the system from $A$ to $B$
$=n{C}_V\Delta T$
$=n\frac{3}{2}R\Delta T=\frac{3}{2}\times V\times \Delta P$
$=\frac{3}{2}\times V\times (3p-p)$
$=\frac{3}{2}\times V\times 2P$
$=3PV$
Similarly,
Heat given to the system from $B$ to $\bar{C}$
$=n{C}_P\Delta T$
$=n\left(\frac{5}{2}R\right)\wedge T$
$=\frac{5}{2}(3P)(\Delta V)$
$=\frac{5}{2}\times (3P)\cdot (2V-V)$
$=\frac{5}{2}\times 3P\times V$
$=\frac{15}{2}PV$
The heat is released from $C$ to $D$ and $D$ to $A$.
Efficiency $(\eta )=\frac{\text{ Work done by gas }}{\text{ Heat given to gas }}\times 100$
$=\frac{2PV}{3PV+\frac{15}{2}PV}\times 100$
$=\frac{2PV}{PV\left(3+\frac{15}{2}\right)}\times 100$
$=\frac{2}{\left(\frac{6+15}{2}\right)}\times 100$
$=\frac{2\times 2}{21}\times 100$
$=\frac{4}{21}\times 100=19.04\%$