Question

Calculate the maximum work done in expanding $16g$ of oxygen at $30K$ and occupying a volume of $5{dm}^3$ isothermally until the volume becomes $25{dm}^3$.

Answer 1

$n=\frac{16g}{32g}=0.5$ moles

$T=30K$

$V_1=5{dm}^3$

$V_2=25{dm}^3$

The work done will be maximum in case of isothermal reversible process.

$\therefore$ work done $=-nRTln\left(\frac{V_2}{V_1}\right)$

$=-0.5\times 8.314\times ln\left(\frac{25}{5}\right)$

$\therefore$ Max. work done $=-6.69J$