In a certain thermodynamical process, the pressure of a gas depends on its volume as $$kV^3$$. The work done when the temperature changes from 100°C to 300°C will be $$xnR$$ where $$n$$ denotes number of moles of a gas, find $$x$$.

We are given that the pressure of a gas depends on its volume as $$P = kV^3$$. For an ideal gas, $$PV = nRT$$, so substituting gives $$kV^3 \cdot V = nRT$$, which means $$kV^4 = nRT$$.

To find the work done, we use $$W = \int P\,dV$$. Since $$P = kV^3$$, we have $$dP = 3kV^2\,dV$$, so $$V\,dP = 3kV^3\,dV = 3P\,dV$$. From the product rule, $$d(PV) = P\,dV + V\,dP = P\,dV + 3P\,dV = 4P\,dV$$. Since $$PV = nRT$$, we have $$d(PV) = nR\,dT$$. Therefore, $$4P\,dV = nR\,dT$$, giving $$P\,dV = \frac{nR}{4}\,dT$$.

Integrating from $$T_1 = 100°C = 373\,\text{K}$$ to $$T_2 = 300°C = 573\,\text{K}$$:

$$W = \int_{T_1}^{T_2} \frac{nR}{4}\,dT = \frac{nR}{4}(573 - 373) = \frac{nR}{4} \times 200 = 50\,nR$$

Comparing with $$W = xnR$$, we get $$x = 50$$.