Question

The vapor pressure of the dietary ether is 100 Torr at -115 ° C and 400 torr at 17.9 ° C

a)Calculate the heat of vaporization and the normal boiling point in a city where the barometric pressure is 620 torr

b) Calculate the entropy of vaporization at the boiling point

c) ΔG of vaporization at 25 ° C

Answer 1

We have, Clausius-Clapeyron Equation

ln(P₂/P₁) = -(H^o_vap/R)[(1/T₂)-(1/T₁)]

ln(400/100) = -(H^o_vap/8.314)[(1/291)-(1/158)]

-(H^o_vap/8.314) = -479.24

H^o_vap = 3984 J/mol = 3.98 kJ/mol

Normal boiling point in a city where the barometric pressure is 620 torr

ln(620/400) = -(3984/8.314)[(1/T₂)-(1/291)]

(1/T₂) = 0.00252

T₂ = 396.5 K = 123.3 ^oC

Entropy of vaporization at the boiling point

We have,

S^o_vap = H^o_vap/Tb

S^o_vap = 3984 J/396.5 K = 17.61 J/K

G of vaporization at 25 ^oC

We have,

G^o_vap = H^o_vap –TS^o_vap

             = 3984 – (298 x 17.61) = -1263 J/mol = -1.26 kJ/mol

G^o_vap = -1.26 kJ/mol