Question

One mole gas expands with temp. as $V=K{T}^{2/3}$.What is the work done when temp.change by ${30}^{\circ }$c?

• A. 10R
• B. 20R
• C. 30R
• D. 40R

Answer 1

The correct option is B 20R

Given,

$dT={30}^0$

$V=K{T}^{2/3}$. . . . . . . .(1)

$n=1mole$

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

Work done, $W=\int PdV$

$W=\int \frac{nRT}{V}dV$ ($PV=nRT$)

Substitute equation (1), and equation (2) in the above equation, we get

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

$W=\frac{2R}{3}\times 30=20R$

The correct option is B.