Question

The maximum work of expansion of $2$ moles of an ideal gas at $300K$, occupying a volume of $20d{m}^3$,isothermally and reversibly until the volume becomes $40d{m}^3$ is:
$(R=8.314J{K}^{-1}mo{l}^{-1})$

• A. $-34.57J$
• B. $-3.457J$
• C. $-37.92kJ$
• D. $-34.57kJ$

Answer 1

The correct option is D $-34.57kJ$

As we know that work done in an isothermally reversible process is given as-

$W=-2.303nRT\log \frac{V_f}{V_i}.....\left(1\right)$

Given that:-

$n=2\text{moles}$

$T=300K$

$V_f=40{dm}^3$

$V_i=20{dm}^3$

$R=8.314J/mol-K$

Substituting all these values in ${eq}^n\left(1\right)$, we have

$W=-2.303\times 2\times 8.314\times 300\times \log \frac{40}{20}$

$\Rightarrow W=-3.458kJ$

Hence the work done for the given process is $-3.458kJ$.