According to Clausius-Clayperon equation, ln (P\'/P) = (-L/R) x [(1/T\') - (1/T)

Question

According to Clausius-Clayperon equation,
ln (P'/P) = (-L/R) x [(1/T') - (1/T)]
Where
P' = vapour pressure at 29.8 oC = 427.4 mm Hg
P = vapour pressure at 46.5 oC = 760.0 mm Hg
R = gas constant = 8.314 J/(mol-K)
T' = initial temperature = 29.8 oC = 29.8+273 = 302.8 K
T = final temperature = 46.5 oC = 46.5+273 = 319.5 K
L = heat of vaporization of this substance = ?
Plug the values we get
ln (P'/P) = (-L/R) x [(1/T') - (1/T)]
L = - [ln (P'/P) xR] / [ [(1/T') - (1/T)]]

T=225oC= 225+273=498K

T'= 250oC= 250+273=523K

P= 17.3 mm Hg

P'= 74.4 mm Hg

Plug the values we get L= 126.3 *10^3 J= 126.3kJ

Explanation / Answer

3.) Vapor pressure measurements at several different temperatures are shown below for mercury. Determine the molar heat of vaporization for mercury (Hint graph the In P vrs 1/T or pick 2 points). 250 74.4 330 226.8 340 356.3 350 457.9 225 T°C P (mmHg) 17.3

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