Question

Calculate the latent heat of vaporization for water at normal boiling point and at 250oC using Riedel and Watson equations. Compare with values from steam table.

Answer 1

Calculation of heat of vapourization of water at normal boiling point :

Using Riedel's equation

H₁ = 1.093×RT₁ ( lnP_c - 1.013 ) / { 0.930 - (T₁/T_c) }

where, H₁ = heat of vapourization of water at normal boiling point

R = gas constant

T₁ = normal boiling point of water

T_c = critical temperature of water

P_c = critical pressure of water

Using the standard values of

R = 8.314 J/ K.mol ,

T₁ = 373 K,

T_c = 647 K

and P_c = 221 bar , in the Riedel's equation

H₁ = 1.093×8.314×373{ln221 - 1.013} / {0.930 - (373/647)}

H₁ = 3389.526 {5.398 - 1.013} / {0.930 - 0.576}

H₁ = 3389.526× 4.385 / 0.354₂₂

H₁ = 41986 J/mol

Let H₂ = heat of vapourization of water at 250°C

T₂ = 250°C = 523 K

Using watson equation,

H₂ / H₁ = [ (T_c - T₂) / (T_c - T₁ ) ]^0.38

H₂ / 41986 = [ (647 - 523) / (647 - 373) ]^0.38

H₂ / 41986 = [ 0.45255 ]^0.38

H₂ = 0.73987 × 41986

H₂ = 31064.18 J/mol