JEE Main 2016 — Thermodynamics Question with Solution

Work done by an ideal daitomic gas is given by, $$\begin{gathered} W=∆Q-∆U \\ ∆Q=n{C}_p∆T where C_p for diatomic gas is \frac{5}{2}R \\ ∆U=n{C}_v∆T where C_v for diatomic gas is \frac{3}{2}R \\ hence W=n\frac{5}{2}R∆T-n\frac{3}{2}R∆T=nR∆T\left(\frac{5}{2}-\frac{3}{2}\right)=nR∆T \end{gathered}$$

And heat supplies is given by, $(∆Q{)}_p=n{C}_p∆T=n\frac{5}{2}R∆T$

Hence the ratio is given by$\frac{∆\mathrm{W}}{{\left(∆\mathrm{Q}\right)}_{\mathrm{P}}}=\frac{nR∆T}{n{\displaystyle \frac{5}{2}}R∆T}$

                                           $\frac{∆\mathrm{W}}{{\left(∆\mathrm{Q}\right)}_{\mathrm{P}}}=\frac{2}{5}$