Thermodynamics Question 333
Question: A gas expands with temperature according to the relation $ V=k{{T}^{2/3}} $ . Calculate work done when the temperature changes by 60 K?
Options:
A) 10 R
B) 30 R
C) 40 R
D) 20 R
Show Answer
Answer:
Correct Answer: C
Solution:
[c]$ dW=pdV=\frac{RT}{V}dV $ .. (i)
As, $ V=k{{T}^{2/3}},dV=k\frac{2}{3}{{T}^{-2/3}}dT $
$ \frac{dV}{V}=\frac{k\frac{2}{3}{{T}^{-2/3}}dT}{k{{T}^{2/3}}}=\frac{2}{3}\frac{dT}{T} $
From Eq. (i), $ W=\int _{T _{1}}^{T _{2}}{RT}\frac{dV}{V}=\int _{T _{1}}^{T _{2}}{RT\frac{2}{3}\frac{dT}{T}} $
$ W=\frac{2}{3}R(T _{2}-T _{1}) $
$ =\frac{2}{3}R\times 60=40R $
