Question

$5$ mol of an ideal gas at $293K$ is expanded isothermally from an initial pressure of $0.4kPa$ to final pressure of $0.1kPa$ against a constant external pressure of $0.1kPa$.
Calculate $q,W,\Delta U$ and $\Delta H$.

Answer 1

Solution:-

As the gas is expanded isothermally, thus work done is given by-

$W=-2.303nRT\log \frac{P_i}{P_f}$

Given that:-

$n=5\text{moles}$

$R=8.314J/mol-K$

$T=293K$

$P_i=0.4kPa$

$P_f=0.1kPa$

$\therefore W=-2.303\times 5\times 8.314\times 293\times \log \frac{0.4}{0.1}$

$\Rightarrow W=-16.88kJ$

Hence the work done will be $-16.88kJ$.

As the process is isothermal, thus-

$\Delta U=\Delta H=0$

$\therefore q=-W=-(-16.88)=16.88kJ$