Question

Helium gas goes through a cycle $ABCDA$ (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

Answer: 15.4

If $T_0$ be the temperature at $A$, then temperature at $B$,

$\frac{P_0{V}_0}{T_0}=\frac{2P_0\times V_0}{T_B}\text{ or }{T}_B=2T_0$
And temperature at $C$ becomes, $4T_0$
Heat is extracted by the system in process $A\to B$ and $B\to$ $C$ so,
$Q=Q_v+Q_P$
$=nCv\Delta T+n{C}_P\Delta T$
$=n\times \frac{3R}{2}\times \left(2T_0-T_0\right)+n\times \frac{5R}{2}\left(4T_0-2T_0\right)$
$=\frac{13}{2}nR{T}_0$
Work done in the complete cycle
$W=P_0{V}_0=nR{T}_0$
Now efficiency, $\eta =\frac{W}{Q}$
$=\frac{nR{T}_0}{\frac{13}{2}nR{T}_0}\times 100$
$=15.4\%$