Question

A gas $(\gamma=1.5)$ undergoes a cycle of adiabatic, isobaric and isochoric processes in an order. If the volume of the gas is doubled in the adiabatic process then the efficiency of the cycle is approximately,

• A. $18 \%$
• B. $46.4 \%$
• C. $38.5 \%$
• D. $9.25 \%$

Answer 1

Answer: (A)

According to the question, we drawn the following situation,
In isochoric process, $\frac{p_{A}}{T_{A}}=\frac{p_{B}}{T_{B}}$

$\left(\frac{p}{2 \sqrt{2}}, V, \frac{T}{2 \sqrt{2}}\right)$
As work done in adiabatic process,
$W =\frac{n R\left(T_{1}-T_{2}\right)}{\gamma-1}=\frac{R\left(T-\frac{T}{\sqrt{2}}\right)}{(1.5-1)} \,\,\,($ Taking, $ n=1)$
$V_{2} =2 V_{1} $
$\Rightarrow T_{2} =\frac{T_{1}}{\sqrt{2}} \text { or } T_{2}=\frac{T}{\sqrt{2}} $
$W_{1} =R T 2\left(1-\frac{1}{\sqrt{2}}\right)=0.58 \,\,\,(\because r=i $ Given $)$
Work done in isobaric process,
$W_{2}=p \Delta V=n R \Delta T$
$\Rightarrow W=n R\left(\frac{T}{2 \sqrt{2}}-\frac{T}{\sqrt{2}}\right)=-R T\left(\frac{1}{\sqrt{2}}-\frac{1}{2 \sqrt{2}}\right)=-0.35$
Now, heat given to system
$Q_{1}=n C_{p} \Delta T=\frac{7}{2} R\left(\frac{1}{\sqrt{2}}-\frac{1}{2 \sqrt{2}}\right)=-1.23\, J$
Negative, $Q_{1}$ shows that heat is given to the system.
Efficiency of the cycle,
$\eta =\frac{\text { net work done }}{\text { heat input }}=\frac{W_{1}+W_{2}}{Q_{1}}$
$=\frac{0.58-0.35}{1.23} $
$\Rightarrow \eta =\frac{0.23}{1.23} \times 100 \approx 18.6 \% \simeq 18 \%$
Hence, the efficiency of cycle is $18 \%$.