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)]]

Given

P= 0.0127 torr

P'=0.0888 torr

T= 50oC= 50+273=323 K

T'= 80 oC= 80+273= 353 K

Plug the values we get L= 61.4*10^3 J = 61.4 kJ

Explanation / Answer

C EN1345 in ela reoamercury so. 127torr at constantinthis temperature range. (14pt October 27, 2017 2. The vapor pressure of mercury is 0.0127 torr at 50.0°C and 0.0888 torr at 80.0°C. Calculate the enthalpy of vaporization of mercury assume it is constant in this temperature range. (14 pts)

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