# 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?

• Since the process is isothermal ($q=0$) and isobaric, then we can say that $\Delta H = 0$ since at constant pressure $\Delta H = q$. Commented Feb 11, 2015 at 2:52
• i also thought ike that but then i statred to consider that atually isothermal mean dT=0 and not Dq/ Commented Feb 11, 2015 at 5:19
• Ran, you are right that isothermal means that dT = 0. Heat transfer generally occurs in isothermal processes, so Q is not zero. Commented Feb 11, 2015 at 5:25
• The "constant pressure" in the question is probably meant to refer to the atmosphere, not necessarily the helium. If the helium were at constant temperature, constant number of moles, and constant pressure, how would it be expanding??? Commented Feb 11, 2015 at 5:53
• actually it mention constant external pressure of 2 atm and isothermal expand under 320k... Commented Feb 11, 2015 at 7:19

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 $$\Delta H = \Delta U + \Delta(PV) = \Delta U + nR\Delta T = 0$$.