Question

One mole of a gas expands with temperature $T$ such that its volume, $V=k{T}^2$, where $k$ is a constant. If the temperature of the gas changes by ${60}^{\circ }C$ then the work done by the gas is

• A. $120R$
• B. $Rln60$
• C. $kRln60$
• D. $60kR$

Answer 1

The correct option is A $120R$

Given
$V=k{T}^2\therefore dV=2kT.dT$

We know that
$P=\frac{RT}{V}=\frac{RT}{k{T}^2}=\frac{R}{kT}$

Work done is $\int PdV=\int 2RdT$

$W=2R\int dT=2R\left(60\right)=120R$