Explanation

(c) From ideal gas equation $PV = \mu\ RT \Rightarrow P = \frac{\mu\ RT}{V}$
Given $PT^{2} = K \Rightarrow \frac{\mu\ RT}{V}.T^{2} = K \Rightarrow \mu\ RT^{3} = KV\ldots(i)$
Differentiating both sides, we get $3\mu\ RT^{2}dT = K\ dV\ldots(ii)$
Dividing equation (ii) by (i), we get $\frac{3}{T}dT = \frac{dV}{V}$
Coefficient of volume expansion $= \frac{dV}{V\ dT} = \frac{3}{T}$