Question

For an ideal gas, starting from state point 1, two different processes take place. The corresponding final states in these two processes are 2 and 3, lying on same isotherm. If \(P\) and \(h\) represent pressure and enthalpy, respectively, then which one of the following options is correct?

• A. \(h_2 = h_3\)
• B. \(h_2 > h_3\)
• C. \(P_2 h_3 = P_3 h_2\)
• D. \(P_3 h_3 = P_2 h_2\)

Hint:

Recall that enthalpy of an ideal gas depends only on temperature, not on pressure.

Answer 1

The Correct Option is A

The key here is remembering that enthalpy of an ideal gas ties only to temperature, so any statement mixing in pressure needs to be checked against that fact.

1. $h_2 = h_3$: correct. States 2 and 3 sit on the same isotherm, so $T_2 = T_3$. Since $h = c_pT$ for an ideal gas, equal temperature means equal enthalpy no matter what the two processes or pressures were.
2. $h_2 > h_3$: wrong, this assumes a temperature difference between states 2 and 3, but they share the same isotherm, so there is none.
3. $P_2h_3 = P_3h_2$: wrong, this drags pressure into an enthalpy relation that for an ideal gas depends on temperature only.
4. $P_3h_3 = P_2h_2$: wrong, for the same reason, pressure has no role in fixing enthalpy equality here.

The direct consequence of sharing an isotherm is $h_2 = h_3$, option A.