Question

For water at 25°C, dp_s/dT_s = 0.189 kPa/K (p_s is the saturation pressure in kPa and T_s is the saturation temperature in K) and the specific volume of dry saturated vapour is 43.38 m³/kg. Assume that the specific volume of liquid is negligible in comparison with that of vapour. Using the Clausius-Clapeyron equation, an estimate of the enthalpy of evaporation of water at 25°C (in kJ/kg) is _________.

Answer 1

Concept:

According to Clausius Clapeyron equation

\(\frac{{dP}}{{dT}} = \frac{{Latent\;heat \times {P_{sat}}}}{{T_{sat}^2 \times R}}\left[ {When\;{V_g} \gg > {V_f}} \right]\) 

\(\frac{{dP}}{{dT}} = \frac{{Latent\;heat}}{{{T_{sat}}\left( {{V_g} - {V_f}} \right)}}\)

Calculation:

\(0.189 = \frac{{LH}}{{298\left( {43.38 - 0} \right)}}\)

LH = 2443.24 kJ/kg