Question

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

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

Answer 1

The correct option is C 26.6 R

Given,

$V=K{T}^{2/3}$

$dV=K\frac{2}{3}{T}^{-1/3}dT$

Work done, $dW=PdV$. . . . . . .(1)

from ideal gas equation,

$P=\frac{nRT}{V}$

$dW=\frac{nRT}{V}.K\frac{2}{3}{T}^{-1/3}dT$

$W=\frac{2nR}{3}\int dT$

$W=\frac{2R}{3}(T_2-T_1)$ (n=1 mole)

$W=\frac{2R}{3}\times 40=26.6R$

The correct option is C.