Question

Vapor Pressures of a solid and liquid ammonia near the triple point are given by: log P solid/ torr=10.00-1630 K/T log P liquid / torr=8.46-1330 K/T Calculate the ratio of the slopes of the sublimation and vapor pressure curves at the triple point.

Answer 1

Clausius - Clapeyron equation :
dP/dT = L / TΔV
L is the specific latent heat
Assuming that the ammonia vapor obeys the ideal gas law and the volume of the solid and the liquid is negligibly small compared to the volume of the gas, we can get
ln P = (- L / RT) + constant ------------ (1)
2.303 ln P = 10 - (1630 K / T ) ------------ (2)
compare equations 1 and 2 (the slope part, the one part which is in brackets) ,
as vapor pressure of solid is considered L = L_sub
1630 K / 2.303 = L_sub / R
1630 K R / 2.303 = L_sub
K = 3.12 *10⁴ J/mol for sublimation R = 8.314 J /mol K
707 * 3.12*10⁴ * 8.314 = L_sub
L_sub = 183393537.6 J/ K = 183393.53 kJ/K
ratio of slopes of sublimation and vapor pressure curve of solid ammonia,
( L_sub /R ) / (1630K/2.303) = 0.9989
similarly it can be done for L_vap ,
2.303 ln P = 8.46 - (1330 K / T ) ---------------(3)
ln P = (- L_sub / RT) + constant ------------- (1)
K = 2.55 *10⁴ for equation 3
ratio of the slopes of sublimation and vapor pressure curve of liquid ammonia,
(L_sub / R ) / (1330K/2.303) = 1.49787