Ideal Gas Enthalpy Property

For any ideal gas, enthalpy ($h$) is exclusively a function of temperature ($T$). This means $h$ changes only when $T$ changes, and is independent of pressure ($P$) or volume ($V$). We can represent this as $h = f(T)$.

Isotherm Condition Analysis

The question specifies that the final states, point 2 and point 3, lie on the same isotherm.

• An isotherm represents a process where the temperature remains constant.
• Therefore, the temperature at state 2 ($T_2$) must be equal to the temperature at state 3 ($T_3$). Mathematically, $T_2 = T_3$.

Deriving Enthalpy Relationship

Based on the properties of an ideal gas:

• Since $h = f(T)$ for an ideal gas, and we have established that $T_2 = T_3$, the enthalpy values at these two states must be identical.
• Therefore, $h_2 = h_3$.

This relationship holds true regardless of the pressures ($P_2$ and $P_3$) or volumes at states 2 and 3, as long as they reside on the same isotherm.

The correct option is the one stating $h_2 = h_3$.