Question

Helium gas goes through a cycle ABCDA (consisting of two isochoric and two isobaric lines) as shown in figure. Efficiency of this cycle is nearly (Assume the gas to be close to ideal gas):-

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

Answer 1

The correct option is B $15.4\%$

Work done = $-P_{ext}dV$
For the process $A\to B$ and $C\to D$, there is no change in volume i.e. isochoric process. Hence the work done in these process in zero.
Work done in process $B\to C$ = $-2P_o(2V_o-V_o)$
$=-2P_o{V}_o$
Work done in process $D\to A$ = $-P_o(V_o-2V_o)$
$=P_o{V}_o$
Total work done $\left(w\right)$= $w_{A\to B}+w_{B\to C}+w_{C\to D}+w_{D\to A}$
$=0+-2P_o{V}_o+0+P_o{V}_o$
$\Rightarrow w=-P_o{V}_o$
Total heat taken $\left(Q_h\right)$ = $(n{C}_v\Delta T{)}_{A\to B}+(n{C}_p\Delta T{)}_{B\to C}$
Since $He$ is monoatomic gas hence $C_v=\frac{3R}{2}$ and $C_p=\frac{5R}{2}$
$Q_h={\left(n\frac{3R}{2}\Delta T\right)}_{A\to B}+{\left(n\frac{5R}{2}\Delta T\right)}_{B\to C}$
$Q_h={\left(\frac{3}{2}V\Delta P\right)}_{A\to B}+{\left(\frac{5}{2}P\Delta V\right)}_{B\to C}$
$Q_h=\frac{13}{2}{P}_o{V}_o$
Effeciency$\left(\eta \right)$ = $\frac{-w}{Q_h}\times 100$
= $\frac{P_o{V}_o}{\frac{13}{2}{P}_o{V}_o}\times 100$
$\Rightarrow \eta =15.38\%$