Question

Using Clausius-Claperyon’s equation, estimate the enthalpy of vapourisation. The following data is given : 

At 200°C : v_g = 0.1274 m³/kg ; v_f = 0.001157 m³/kg ; \(\left(\cfrac{dp}{dT}\right)\) = 32 kPa/K.

Answer 1

Using the equation \(\left(\cfrac{dp}{dT}\right)\) = \(\cfrac{h_{fg}}{T_s(v_g-v_f)}\) 

where, h_fg = Enthalpy of vapourisation. 

Substituting the various values, we get

32 × 10³ = \(\cfrac{h_{fg}}{(200+273)(0.1274-0.001157)}\) 

h_fg = 32 × 10³ (200 + 273)(0.1274 – 0.001157) J 

= 1910.8 × 10³ J/kg = 1910.8 kJ/kg.