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For water at 25°C, dps /dTs = 0.189 kPa/K (ps is the saturation pressure in kPa and Ts 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
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- 2445.25
- 2553.25
- 2543.25
- 2443.25
Correct Option: D
| = 0.189 kPa/K | dTs |
Tsat =273 + 25 = 298 K
v g = 43.38 m3 /kg
v f = 0
v fg = v g – v f
= 43.38 – 0= 43.38 m3 /kg
| = | ||||
| dTs | Tsat × Vfg |
| = | 298 × 43.38 |
hfg = 2443.25 kJ/kg
