An ideal gas undergoes isothermal expansion at constant pressure. During the process:

  1. enthalpy increases but entropy decreases.
  2. enthalpy remains constant but entropy increases.
  3. enthalpy decreases but entropy increases.
  4. Both enthalpy and entropy remain constant.

I applied this formula. $\Delta H = \Delta U + p\Delta V$ (pressure is constant)
$\Delta U = 0$ (since, process is isothermal)
$\Delta V >0$ (gas is expanding)
So $\Delta H > 0$
So enthalpy must be increasing $\left(\Delta S =\frac{\Delta Q_{\text{rev}}}{T}\right)$

Since heat is supplied to expand gas $\Delta S$ must be positive. And entropy must increase. But none of the option is matching with my answer.

Also I think question is wrong since $PV= nRT$ and, at constant temperature (isothermal) and constant pressure, volume can not change.

  • $\begingroup$ Is the answer (2)? It appears like they've assumed $\Delta H=nC_\mathrm{p}\Delta T=0$ (for isothermal process). And entropy is increasing because expansion (=> disorder of gas molecules) is there. Though I agree with your points. Perhaps, gas moles are being externally fed into the system? But then it is an open system instead, so I am not sure. $\endgroup$ Mar 10 '18 at 11:59
  • $\begingroup$ @GaurangTandon Yes, answer is (2). $\endgroup$
    – Amit Kumar
    Mar 10 '18 at 12:11
  • $\begingroup$ Your assessment using PV=nRT is spot on. The question is bogus. $\endgroup$ Mar 10 '18 at 12:16
  • $\begingroup$ Late now, but questions like this should generally be brought to the instructor's attention for not making sense. In this case, yes, an ideal gas with constant-pressure and constant-temperature doesn't expand unless added to, but that can't be the case since both enthalpy and entropy would increase, which isn't a choice. $\endgroup$
    – Nat
    Apr 21 '21 at 10:40
  • $\begingroup$ The question most probably meant constant external pressure. $\endgroup$ Aug 20 '21 at 6:30

The first step in answering the multiple-choice question lies in remembering that there is no change in the internal energy and enthalpy of an ideal gas undergoing an isothermal expansion. A proof involves the second law of thermodynamics.

That leaves options 2 and 4 as the only possibilities.

Next note again that $\Delta U = 0 $ so that $w=-q$. Since the work is negative (a spontaneous expansion with $w=-p_\mathrm{ext}\Delta V$) the heat is positive, which means the entropy of the surroundings must have decreased (at constant T, $\Delta S_\mathrm{surroundings} = -q/T$). For the process to be spontaneous, however, the overall change in entropy, $\Delta S_\mathrm{universe}=\Delta S_\mathrm{system}+\Delta S_\mathrm{surroundings}$, must be positive. This implies that $\Delta S_\mathrm{system}>0$ and option 2 is the correct choice.

Regarding the question in the title

Can pressure remain constant in isothermal expansion?

the pressure of the surroundings can certainly remain constant during an isothermal expansion. It is worth remembering that such expansion work is defined in terms of the pressure of the surroundings (not of the system) as $w=-\int_{V_\mathrm{ini}} ^{V_\mathrm{fin}} p_\mathrm{ext} \mathrm d V=-p_\mathrm{ext} \Delta V$.

However, for an ideal gas of constant composition, $$p=nRT/V=\mathrm{constant}/V$$ implies that a change in volume must be accompanied by a change in the internal pressure of the system.


$$\Delta H = \Delta U + \Delta PV$$
Here you were wrong you wrote that $$\Delta H = \Delta U + P\Delta V,$$ thus the product of $PV$ remains constant and the $\Delta H$ is zero, so that the enthalpy is constant. And you are wrote that entropy will increase, hence the answer is B.


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