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?

  • $\begingroup$ Since the process is isothermal ($q=0$) and isobaric, then we can say that $\Delta H = 0$ since at constant pressure $\Delta H = q$. $\endgroup$ – Ben Norris Feb 11 '15 at 2:52
  • $\begingroup$ i also thought ike that but then i statred to consider that atually isothermal mean dT=0 and not Dq/ $\endgroup$ – Ran shachar Feb 11 '15 at 5:19
  • $\begingroup$ Ran, you are right that isothermal means that dT = 0. Heat transfer generally occurs in isothermal processes, so Q is not zero. $\endgroup$ – Curt F. Feb 11 '15 at 5:25
  • $\begingroup$ 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??? $\endgroup$ – Curt F. Feb 11 '15 at 5:53
  • $\begingroup$ actually it mention constant external pressure of 2 atm and isothermal expand under 320k... $\endgroup$ – Ran shachar Feb 11 '15 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 $$\Delta H = \Delta U + \Delta(PV) = \Delta U + nR\Delta T = 0$$.

  • $\begingroup$ thanks. if helium no consider ideal gas i assume cannot calculate dH. correct? $\endgroup$ – Ran shachar Feb 12 '15 at 1:38
  • $\begingroup$ Both ΔU and ΔH will be small but nonzero if the gas isn't ideal; you can compute them if you use the equation of state for the gas. $\endgroup$ – Fred Senese Feb 12 '15 at 6:21

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