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:
$$ \begin{aligned} & d W=p d V=\frac{R T}{V} d V . . \text { (i) } \\ & \text { As, } V=k T^{2 / 3}, d V=k \frac{2}{3} T^{-1 / 3} d T \\ & \frac{d V}{V}=\frac{k \frac{2}{3} T^{-1 / 3} d T}{k T^{2 / 3}}=\frac{2}{3} \frac{d T}{T} \\ & \text { From Eq. (i), } W=\int_{T_1}^{T_2} R T \frac{d V}{V}=\int_{T_1}^{T_2} R T \frac{2}{3} \frac{d T}{T} \\ & W=\frac{2}{3} R\left(T_2-T_1\right) \\ & =\frac{2}{3} R \times 60=40 R \end{aligned} $$
BETA
