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

Step 1: Recall the key property of enthalpy for an ideal gas.
For an ideal gas, enthalpy is a function of temperature alone, \(h = h(T)\), and does not depend on pressure or volume. Internal energy \(u\) is a function of \(T\) only, and \(h = u + Pv = u + RT\), both terms depending only on \(T\).

Step 2: Use the condition that states 2 and 3 lie on the same isotherm.
"Lying on the same isotherm" means states 2 and 3 have the same temperature, \(T_2 = T_3\), even though they were reached by two different processes starting from state 1.

Step 3: Connect temperature equality to enthalpy equality.
Since \(h\) depends only on \(T\) for an ideal gas, and \(T_2 = T_3\), it follows that \(h_2 = h_3\), regardless of what the pressures \(P_2\) and \(P_3\) turn out to be.

Final Answer:
Equal temperature on the same isotherm forces equal enthalpy for an ideal gas. \[ \boxed{h_2 = h_3} \]