Question

A diatomic ideal gas goes through a cycle during which the absolute temperature varies four fold. Find the efficiency of the cycle.

• A. $AB$ and $CD$
• B. $AB$ and $BC$
• C. $BC$ and $CD$
• D. $CD$ and $DA$

Answer 1

The correct option is D $CD$ and $DA$

Using ideal gas equation, $PV=nRT$.
or $T\propto PV$
Since $PV$ is maximum at B and minimum at A,
Within the cycle, $T_{max}=T_B$ & $T_{min}=T_A$
Given that, $\frac{T_{max}}{T_{min}}=4$
$\therefore T_B=4T_A$


For $A\to B$: $P\propto V$
or $P{V}^{-1}=$ constant
Therefore, $A\to B$ is a polytropic process of the form $P{V}^x=k$ where $x=-1$
$C=C_V+\frac{R}{1-x}=\frac{R}{\gamma -1}+\frac{R}{2}$
$C=\frac{(\gamma +1)}{(\gamma -1)}\frac{R}{2}$
By using, $Q_{AB}=nC\Delta T$
$Q_{AB}=\frac{n(\gamma +1)}{(\gamma -1)}\frac{R}{2}[4T_0-T_0]=\frac{3}{2}nR{T}_0\left(\frac{\gamma +1}{\gamma -1}\right)$
[assuming $T_A=T_0$]
$Q_{BC}=\Delta U_{BC}=n{C}_V\Delta T=\frac{nR}{\gamma -1}[2T_0-4T_0]=-\frac{2nR{T}_0}{\gamma -1}$
$Q_{CA}=n{C}_P\Delta T=\frac{\gamma nR}{\gamma -1}[T_0-2T_0]=-\gamma nR\frac{T_0}{\gamma -1}$

Efficiency of a cycle is given by
$\eta =\frac{W}{Q_{input}}=\frac{Q_{AB}+Q_{BC}+Q_{CA}}{Q_{AB}}$
$\eta =1+\frac{Q_{BC}+Q_{CA}}{Q_{AB}}=1+\frac{\left(\frac{nR{T}_0}{\gamma -1}\right)\times (-2-\gamma )}{\left(\frac{nR{T}_0}{r-1}\right)\frac{3(\gamma +1)}{2}}=1+\frac{2\times (-2-\gamma )}{3(\gamma +1)}=1-\frac{2(2+\gamma )}{3\gamma +3}=\frac{\gamma -1}{3\gamma +3}$
Given, gas is ideal and diatomic in nature $\therefore \gamma =\frac{7}{5}$
Thus we get, $\eta =0.0556$ or $5.56\%$
Hence, option (b) is the correct.