A gas expands with temperature according to the relation, \(V=k T^{2 / 3}\), where \(k\) is a constant. Work…
A gas expands with temperature according to the relation, \(V=k T^{2 / 3}\), where \(k\) is a constant.
Work done when the temperature changes by \(60 \mathrm{~K}\) is (\(R=\) universal gas constant.)
\(10 \mathrm{R}\)
\(20 \mathrm{R}\)
\(50 \mathrm{R}\)
\(40 \mathrm{R}\)
Solution
Given, \(V=k T^{2 / 3}\)
From definition of work done,
\(\begin{aligned}
d W & =P d V=\frac{R T}{V} d V \\
& =\frac{R T}{k T^{2 / 3}} d V \quad \ldots (i)
\end{aligned}\)
Now, \(V=k T^{2 / 3}\)
Taking derivative on the both sides, we get
\(d V=K \frac{2}{3} T^{-1 / 3} d T \quad \ldots (ii)\)
Substituting the value of Eqs. (ii) in Eqs. (i) and taking integration on the both sides, we get
\(\begin{aligned}
W & =\frac{2}{3} R \int_{T_1}^{T_2} d T=\frac{2}{3} R\left(T_2-T_1\right) \\
& =\frac{2}{3} R(60-0)=\frac{2}{3} \times R \times 60=40 R
\end{aligned}\)