# Can pressure remain constant in isothermal expansion?

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.

• 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. – Gaurang Tandon Mar 10 '18 at 11:59
• @GaurangTandon Yes, answer is (2). – Amit Kumar Mar 10 '18 at 12:11
• Your assessment using PV=nRT is spot on. The question is bogus. – Chet Miller Mar 10 '18 at 12:16
• 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. – Nat Apr 21 at 10:40

$$\Delta H = \Delta U + \Delta PV$$
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.

A recurrent issue with solving expansion problems in thermodynamics is the question of what values of the pressure describe the system and the expansion during a process, but in this case you can circumvent that worry altogether.

Starting from the first law, the definition of the enthalpy, and assuming the gas is ideal and n=constant (closed system of constant composition),

$$\Delta U = 0 = \Delta H - \Delta (pV) = \Delta H - \Delta (nRT) = \Delta H$$

in other words the enthalpy change is zero (something worth memorizing, it is true for isothermal processes involving ideal gases at constant composition and closed systems with only pV work).

The entropy change is equal to the same change as for the reversible process between the same states. Since this is a case of a spontaneous process (isothermal expansion of a gas under these constraints of constant composition, closed system) then the entropy change is positive.

You can compute the entropy change explicitly for an isothermal volume change of an ideal gas (constant composition, closed system) as

$$\Delta S = R \log\left( \frac{V_{\text{fin}}}{ V_{\text{ini}}} \right)$$

which you can convince yourself is positive for an expansion.