Question

When the gas expands with temperature using the relation $V=K{T}^{2/3}$ for the temperature change of $40K,$ the work done is

• A. $20.1R$
• B. $30.2R$
• C. $26.6R$
• D. $35.6R$

Answer 1

Answer: (C)

$W=\int PdV=\int \frac{\mu RT}{V}dV$
$=\int \frac{\mu RT}{K{T}^{\frac{2}{3}}}dV$ ...(i)
$\because V=K{T}^{\frac{2}{3}}$
$dV=K\frac{2}{3}{T}^{-1/3}dT$ ...(ii)
From Eqs. (i) and (ii)
$W=\int \frac{\mu RT}{K{T}^{\frac{2}{3}}}\times K\frac{2}{3}{T}^{-1/3}dT$
$=\frac{2}{3}R\int T^{1-\frac{1}{3}-\frac{2}{3}}dT$
$=\frac{2}{3}R{\int }_0^{40}{T}^{0}dT=\frac{2}{3}R(T{)}_0^{40}$
$=\frac{2}{3}R(40-0)=\frac{80R}{3}=26.6R$