Question

A gas expands with temperature according to the relation $ V =KT^{2/3} $ . Work done when the temperature changes by $ 60\, K $ is

• A. $ 10 R $
• B. $ 30 R $
• C. $ 40 R $
• D. $ 20 R $

Answer 1

Answer: (C)

Here,
$ dW = PdV = \frac{RT}{V} dV \,\, (\because PV = RT)\quad ...(i) $
Given $ V = KT^{2/3} $
$ \therefore dV = K \frac{2}{3} T^{-1/3} dT $
$ \therefore \frac{dV}{V} = \frac{ K \frac{2}{3}T^{-1/3}dT}{KT^{2/3}} $
$ = \frac{2}{3}\frac{dT}{T} \quad ...(ii) $
From $ (i), W = \int\limits_{T_1}^{T_{2}} RT \frac{dV}{V} $
$ = \int\limits_{T_1}^{T_{2}} RT \frac{2}{3} \frac{dT}{T} \quad $ (Using $ (ii)) $
$ \therefore W = \frac{2}{3}R (T_2 -T_1) $
$ \frac{2}{3}R\times 60 = 40\,R $