Question

Calculate the maximum work done in (in $kJ$) expanding $16g$ of oxygen gas at $300K$ and occupying a volume of $5d{m}^3$ isothermally until the volume becomes $25d{m}^3$.

• A. 7 kJ
• B. -5 kJ
• C. -2 kJ
• D. 9 kJ

Answer 1

The correct option is C -2 kJ

Maximum work is obtained when the process is reversible.
So, for a reversible isothermal process, the work done is given as,
$w=-2.303{\text{nRT log}}_{10}\left(\frac{V_2}{V_1}\right)$

$V_1$ = initial volume = $5d{m}^3$
$V_2$= final volume = $25d{m}^3$
moles of gas =$\frac{16}{32}=0.5$ mol

Substituting the values in the formula, we get,
$w=-2.303\times 0.5\times 8.314\times 300\text{log}\frac{25}{5}=-2007.5J$ or approximately $-2kJ$