Helium gas expand isobaric isothermal
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 $$\Delta H = \Delta U + \Delta(PV) = \Delta U + nR\Delta T = 0$$.