The vapour pressure of benzene is `1.53 xx 10^(4) Nm^(-2)` at 303 K and `5.2 xx 10^(4) Nm^(-2)` at 333 K. Calculate the mean latent heat of evaportion of benzene over this temprature range

Answer

31.1 kj

Explanation

Using clausinus-Clapeyron equation,
2.303 log `(P_(2))/(P_(1)) = (Delta H_(v))/(R) ([T_(2) - T_(1)])/(T_(1)T_(2))`
`Delta H_(v) = (2.303 RT_(1) T_(2))/((T_(2) - T_(1))) log. (P_(2))/(P_(1))`
We have
`Delta H_(V) = (2.303 xx 8.314 xx 303 xx 333)/((333 - 300)) log_("to"). (5.2 xx 10^(4))/(1.53 xx 10^(4))`
`=31.1 xx 10^(3) J = 31.1 kJ`