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. $\endgroup$ – Gaurang Tandon 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. $\endgroup$ – Chet Miller Mar 10 '18 at 12:16

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