2 mole of ideal gas expands isothermally and reversibly from 1 L to 10 L at 300 K, Then $\mathrm{\Delta }H$ is:

n = 2 moles
T = 300 K

$\begin{array}{l}{V}_f=10\mathrm{L}\\ V_1=1\mathrm{L}\end{array}$

R = Gas constant $=8.314\mathrm{J}/\mathrm{K}-\mathrm{mol}$

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

$\begin{array}{l}\mathrm{W}=-2.303\times 2\times 8.314\times 300\times \log \frac{10}{1}\\ \Rightarrow \mathrm{W}=-11488.285\mathrm{J}=-11.4\mathrm{kJ}\end{array}$

Now as we know that,

$\mathrm{\Delta H}=\mathrm{\Delta U}+\mathrm{W}$

For an isothermal process,

$\begin{array}{l}\mathrm{\Delta U}=0\\ \therefore \mathrm{\Delta H}=\mathrm{W}=-11.4\mathrm{kJ}\end{array}$

Hence the enthalpy for the given process is $-11.47\mathrm{kJ}$.