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,
$W=\frac{nR\left(T_1-T_2\right)}{\gamma -1}=\frac{R\left(T-\frac{T}{\sqrt{2}}\right)}{(1.5-1)}($ Taking, $n=1)$
$V_2=2V_1$
$\Rightarrow T_2=\frac{T_1}{\sqrt{2}}\text{ or }{T}_2=\frac{T}{\sqrt{2}}$
$W_1=RT2\left(1-\frac{1}{\sqrt{2}}\right)=0.58(\because r=i$ Given $)$
Work done in isobaric process,
$W_2=p\Delta V=nR\Delta T$
$\Rightarrow W=nR\left(\frac{T}{2\sqrt{2}}-\frac{T}{\sqrt{2}}\right)=-RT\left(\frac{1}{\sqrt{2}}-\frac{1}{2\sqrt{2}}\right)=-0.35$
Now, heat given to system
$Q_1=n{C}_p\Delta T=\frac{7}{2}R\left(\frac{1}{\sqrt{2}}-\frac{1}{2\sqrt{2}}\right)=-1.23J$
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 $18\%$.