Question

Helium gas goes through a cycle $\text{ABCDA}$ (consisting of two isochoric and two isobaric lines) as shown in figure. Efficiency of this cycle is nearly:
[Assume the gas to be an ideal gas]

• A. $15.4\%$
• B. $9.1\%$
• C. $10.5\%$
• D. $12.5\%$

Answer 1

The correct option is A $15.4\%$

Work done during a complete cycle = Area under $P-V$ graph $W=2P_0{V}_0-P_0{V}_0=P_0{V}_0$
From $\text{A}$ to $\text{B}$, heat given to the gas
$Q_1=n{C}_V\Delta T=n\frac{3}{2}R\Delta T$
But from ideal gas equation, $\Delta \left(PV\right)=nR\Delta T$
$\Rightarrow Q_1=\frac{3}{2}(2P_0{V}_0-P_0{V}_0)=\frac{3}{2}{P}_0{V}_0$
From $\text{B}$ to $\text{C}$, heat given to the gas
$Q_2=n{C}_P\Delta T=n\left(\frac{5}{2}R\right)\Delta T=\frac{5}{2}\left(2P_0\right)\Delta V=5P_0{V}_0$
From $\text{C}$ to $\text{D}$, heat rejected by the gas
$Q_3=n{C}_V\Delta T=n\left(\frac{3}{2}R\right)\Delta T=\frac{3}{2}\left(P_0\right)\Delta V=-\frac{3}{2}{P}_0{V}_0$
and from $\text{D}$ to $\text{A}$, heat rejected by the gas is
$Q_4=n{C}_P\Delta T=n\left(\frac{5}{2}R\right)\Delta T=\frac{5}{2}\left(2P_0\right)\Delta V=-5P_0{V}_0$
Efficiency, $\eta =\frac{\text{Work done by gas}}{\text{Heat given to the gas}}\times 100$
$\eta =\frac{P_0{V}_0}{\frac{3}{2}{P}_0{V}_0+5P_0{V}_0}=15.4\%$
Thus, option (a) is the correct answer.