Question

In a liquid-vapour phase change process, $\left(\dfrac{dP}{dT}\right)_{sat}$ at 100$^\circ$C for saturated water is 3750 Pa/K. If the resulting change in specific volume ($v_g - v_f$) is 1.672 m$^3$/kg, the enthalpy of vaporization ($h_{fg}$) will be $____________$ kJ/kg (in integer).

Hint:

For enthalpy of vaporization, always use the Clapeyron equation involving slope of saturation line and specific volume difference.

Answer 1

Correct Answer: 2337

Step 1: Clapeyron equation. \[ \frac{dP}{dT} = \frac{h_{fg}}{T (v_g - v_f)} \]
Step 2: Substitution. At $T = 373 \, K$, $\dfrac{dP}{dT} = 3750 \, Pa/K$, $(v_g - v_f) = 1.672 \, m^3/kg$: \[ h_{fg} = \frac{dP}{dT} . T . (v_g - v_f) \] \[ h_{fg} = (3750)(373)(1.672) = 2.33 \times 10^6 \, J/kg = 2330 \, kJ/kg \] Final Answer: \[ \boxed{2330 \, kJ/kg} \] % Quicktip