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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.

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  • $\begingroup$ 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, 2018 at 11:59
  • $\begingroup$ @GaurangTandon Yes, answer is (2). $\endgroup$
    – Amit Kumar
    Mar 10, 2018 at 12:11
  • $\begingroup$ Your assessment using PV=nRT is spot on. The question is bogus. $\endgroup$ Mar 10, 2018 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, 2021 at 10:40
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    $\begingroup$ The question most probably meant constant external pressure. $\endgroup$ Aug 20, 2021 at 6:30

3 Answers 3

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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.

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$$\Delta H = \Delta U + \Delta PV$$
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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I will give an answer which explains what happens at the molecular level which necessitates a change in gas pressure during isothermal expansion on a purely theoretical basis as the other answers explains them precisely with formulas.

In a gas the gas pressure is a result of two components.

  1. The average kinetic energy of the gas molecules - The faster a molecule collides (elastically) with the wall the greater is the pressure exerted by it.

  2. The number of molecules colliding a given area of the wall at a given time - With more number of molecules colliding the wall at a given time we have more force exerted on the wall (a simple summation of force exerted by all molecules) and greater is the pressure.

With this two things in mind, as we perform an isothermal heat addition (isothermal expansion) the heat added to the gas molecule is such that it causes an infinitesimal increase in the temperature of the kinetic energy of the gas molecules. Thus now the molecules collide with higher velocity. Since the walls are movable the collisions of these higher velocity molecules will displace the walls by doing some amount of work. And subsequently the kinetic energy of molecules will drop back to the initial velocity of the gas. Thus the condition of the constant temperature remains satisfied since the temperature changes are infinitesimally small and temperature is back to initial value.

Now since the walls are displaced by the expanding gas, it means that the gas can now occupy more volume. Now remember the way gas pressure is manifested (mentioned above). Here there is no change in kinetic energy of the molecules effectively, so on this basis the gas pressure should be constant. But the second factor i.e the number of molecules colliding with the wall at a given area at a given time will now be lesser. This is because as the same amount of gas now occupies more volume, the mean free path will increase and thus there will be less number of molecular collisions with wall (and in between the molecules). So this second factor is responsible for the reduction in pressure of the gas in the isothermal expansion process.

Thus the pressure has to change in the isothermal expansion process.

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