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 ${60}^{o}C$, then the work done by the gas is:

• A. $120R$
• B. $60R$
• C. $160R$
• D. $260kR$

Answer 1

The correct option is A $120R$

$V=k{T}^2$ derviative is $dV=2kTdT$

$W=pdV=\frac{RTdV}{V}$

substituing above value we have,

$W=\frac{RT2kT}{k{T}^2}dT=2R\times dT=120R$ ($dT={60}^o$)