1 mole of rigid diatomic gas performs a work of $$\frac{Q}{5}$$ when heat $$Q$$ is supplied to it. The molar heat capacity of the gas during this transformation is $$\frac{xR}{8}$$. The value of $$x$$ is
[$$R$$ universal gas constant]

We are given 1 mole of a rigid diatomic gas that performs work $$W = \frac{Q}{5}$$ when heat $$Q$$ is supplied to it. We need to find the molar heat capacity during this process.

For a rigid diatomic gas, there are no vibrational degrees of freedom, so $$C_v = \frac{5R}{2}$$.

By the first law of thermodynamics, $$Q = \Delta U + W$$. Substituting $$W = \frac{Q}{5}$$, we get $$Q = \Delta U + \frac{Q}{5}$$, which gives $$\Delta U = \frac{4Q}{5}$$.

For 1 mole of gas, $$\Delta U = n C_v \Delta T = \frac{5R}{2} \Delta T$$. So $$\frac{4Q}{5} = \frac{5R}{2} \Delta T$$.

The molar heat capacity of the process is defined as $$C = \frac{Q}{n \Delta T} = \frac{Q}{\Delta T}$$ (since $$n = 1$$). From the above equation, $$\Delta T = \frac{8Q}{25R}$$.

Therefore, $$C = \frac{Q}{\frac{8Q}{25R}} = \frac{25R}{8}$$.

Comparing with $$\frac{xR}{8}$$, we get $$x = 25$$.