Question

In a thermodynamic process on an ideal monoatomic gas, the infinitesimal heat absorbed by the gas is given by TX, where T is temperature of the system and X is the infinitesimal change in a thermodynamic quantity X of the system. For a mole of monatomic ideal gas           A A 3 T V X R ln R ln 2 T V . Here, R is gas constant, V is volume of gas. TA and VA are constants. The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities

Answer 1

X is the entropy (up to an arbitrary additive constant).

Answer 2

For a reversible process the heat absorbed is given by

$$\delta Q_{\rm rev}=T\,dS.$$

In the problem, the heat is given by

$$\delta Q = T\,dX.$$

Thus, by comparing the two, we see that

$$dX = dS,$$

which means that $X$ and $S$ differ only by a constant (i.e. $X=S+$ constant).

For one mole of a monatomic ideal gas the change in entropy is

$$\Delta S = \frac{3}{2}R \ln\left(\frac{T}{T_A}\right)+R\ln\left(\frac{V}{V_A}\right) = R\ln\left[\left(\frac{T}{T_A}\right)^{\frac{3}{2}}\frac{V}{V_A}\right].$$

This exactly matches the given expression for $X$ (apart from the constant which is arbitrary).

Thus, $X$ is essentially the entropy of the system.