Benzaldehyde, C6H5CHO, has a normal boiling point of 179.0 ∘C and a critical point at 422 ∘C and 45.9 atm.

1 answer:

5 0

Ln(P2/P1) = (L/R).(1/T1 - 1/T2)

with P1 = 1 atm, P2 = 45.9 atm, R = 8.314 J/mol.K, T1 = 452.15 K, T2 = 695.15 K. This gave the latent heat L as 41.2 kJ/mol.K. 

Using this value, and calculating the vapour pressure at 120°C = 393.15 K 

ln(P2) = (41.2 x 10^3/8.314).(1/452.15 - 1/393.15) 

which gives the vapour pressure P2 as 0.1931 atm = 146.7 torr.

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