Renata A.

asked • 01/31/22

A certain substance has a heat of vaporization of 65.01 kJ/mol. At what Kelvin temperature will the vapor pressure be 8.00 times higher than it was at 303 K?

T = _____ K



1 Expert Answer

By:

Ajinkya J. answered • 01/31/22

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The vapour pressure will be 8.00 times higher at 303 K.


We use the Clausius-Clapeyron equation to determine the vapour pressure at different temperatures:


ln(p2​/p1​​) = Δvap​H/R​(1/T1​​−1/T2​)​ 


where 

p1​ and p2​ are the vapour pressures at temperatures T1​ and T2

Δvap​H = the enthalpy of vaporization of the liquid

R = the Universal Gas Constant


In the given problem problem,

p1 = p1​; T1​ = 303 K 

p2 ​= 8.00 p1​; T2 =? 

Δvap​H = 65.01 kJ⋅mol 

R = 8.314 J⋅K−1 mol−1


ln(​8.00p1​​/p1) = 65010 J⋅mol/8.314 J⋅K-1 mol-1​(1/303 K−1/T2​) 


2.08 = 7819 K×(1/303 K−1/T2​)


Solving for T2, we get:


T2 = 330 K


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