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 charges by ${60}^{0}C$ then the work done by the gas is

• A. $120R$
• B. $60R$
• C. $40R$
• D. $80R$

Answer 1

Answer: (A)

$120R$

$\text{Solution}:$

$V=k{T}^2$

$\therefore 2kTdT=dV$

and $P=\frac{RT}{V}=\frac{RT}{k{T}^2}=\frac{R}{kT}$

Now, $dW=PdV=2RdT$,

Also,

$W={\int }_0^{60}2RdT=[2RT{]}_0^{60}=2R(60-0)=120R$

$\text{Hence A is the correct option}$