An equimolar mixture of a mono atomic and diatomic gas is subjected to continuous expansion such that $\mathrm{dQ}=\frac{1}{3}\mathrm{dU}+\frac{1}{2}\mathrm{dW}$. The equation of the process in terms of the
variable P & V will be

$\mathrm{dQ}=\frac{1}{3}\mathrm{du}+\frac{1}{2}\mathrm{dw} ..........\left(1\right)$
From 1st law of thermodynamics, dQ = du + dw.   ……….$\left(2\right)$

From $\left(1\right) and \left(2\right)$,  4 dU = 3 dW  ……….$\left(3\right)$

Now U = $$\begin{gathered} \frac{3}{2}.1.RT +\frac{5}{2}.1.RT =4RT =4\left(PV\right) \\ \Rightarrow dU=4P dV+4V dP ........\left(4\right) \\ From \left(3\right) and \left(4\right) ,8V dP +11P dV =0 \\ Integrating , \ln \left(P^8 V^{11}\right)=\ln c \Rightarrow P{V}^{\frac{11}{8}}=cons\tan t . \end{gathered}$$