Helium gas goes through a cycle $\mathrm{ABCDA}$ (consisting of two isochoric and two isobaric lines) as…
Helium gas goes through a cycle $\mathrm{ABCDA}$ (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)
$15.4 \%$
$9.1 \%$
$10.5 \%$
$12.5 \%$
Solution
Work done in complete cycle $=$ Area under $\mathrm{P-V}$ graph
$=\mathrm{P}_0 \mathrm{V}_0$
from $A$ to $B$, heat given to the gas
$=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}=\mathrm{n} \frac{3}{2} \mathrm{R} \Delta \mathrm{T}=\frac{3}{2} \mathrm{~V}_0 \Delta \mathrm{P}=\frac{3}{2} \mathrm{P}_0 \mathrm{V}_0$
from $B$ to $C$, heat given to the system
$=n C_{\mathrm{p}} \Delta \mathrm{T}=\mathrm{n}\left(\frac{5}{2} \mathrm{R}\right) \Delta \mathrm{T}$
$=\frac{5}{2}\left(2 \mathrm{P}_0\right) \Delta \mathrm{V}=5 \mathrm{P}_0 \mathrm{V}_0$
from $C$ to $D$ and $D$ to $A$, heat is rejected.
$\text { efficiency, } \eta=\frac{\text { work done by gas }}{\text { heat given to the gas }} \times 100$
$\eta=\frac{P_0 V_0}{\frac{3}{2} P_0 V_0+5 P_0 V_0}=15.4 \%$
