Question

Helium gas goes through a cycle $A B C D A$ (consisting of two isochoric and isobaric lines) as shown in figure. Find efficiency of this cycle. (Assume the gas to be close to an ideal gas)

Answer 1

If $T _0$ be the temperature at $A$, then temperature at $B$,

$\frac{P_0 V_0}{T_0}=\frac{2 P_0 \times V_0}{T_B} \text { or } T_B=2 T_0$
And temperature at $C$ becomes, $4 T_0$
Heat is extracted by the system in process $A \rightarrow B$ and $B \rightarrow$ $C$ so,
$Q =Q_v+Q_P $
$=n C v \Delta T+n C_P \Delta T$
$=n \times \frac{3 R}{2} \times\left(2 T_0-T_0\right)+ n \times \frac{5 R}{2}\left(4 T_0-2 T_0\right) $
$=\frac{13}{2} n R T_0$
Work done in the complete cycle
$W=P_0 V_0=n R T_0$
Now efficiency, $\eta =\frac{W}{Q} $
$=\frac{n R T_0}{\frac{13}{2} n R T_0} \times 100$
$=15.4 \%$