Helium gas expand isobaric isothermal

If $4\ \mathrm{mol}$ helium gas expand against constant pressure in isothermal process, how can calculate $W$?


Is it $W=-nRT\ln\frac{V\_2}{V\_1}$ as defined by isothermal or is it $W=-p\left(V\_2-V\_1\right)$ as define from isobaric?


Also, is enthalpy equal to $0$ in such process?




If the helium is expanding isothermally against a constant external pressure $P\_{ext}$, the work is calculated as
$$w = -P\_{ext}(V\_2 - V\_1)$$
This process is irreversible. The first expression you gave for work is (sort of) the expression for work for a *reversible* isothermal ideal gas expansion.


**If helium is assumed to be an ideal gas**, then $\Delta U = 0$ because the process is isothermal. You have $q = -w$ (and not zero) for this process. $\Delta H$ will be zero too, since $$\Delta H = \Delta U + \Delta(PV) = \Delta U + nR\Delta T = 0$$.


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