클라우지우스-클라페이롱. ΔH_vap 35.5 kJ/mol 1 atm 122℃ 113℃

클라우지우스-클라페이롱. ΔH_vap 35.5 kJ/mol 1 atm 122℃ 113℃

 

 

A liquid has a ΔH_vap of 35.5 kJ/mol

and a boiling point of 122℃ at 1.00 atm.

What is its vapor pressure at 113℃?

 

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> P1 = 1.00 atm, T1 = 273.15+122 = 395.15 K

> P2 = ? atm, T2 = 273.15+113 = 386.15 K

 

 

 

Clausius-Clapeyron 식

ln(P1/P2) = ΔH_vap/R (1/T2 – 1/T1)

( 참고 https://ywpop.tistory.com/3133 )

 

 

 

ln(1.00 / ?) = (35500 / 8.314) (1/386.15 – 1/395.15)

1.00 / ? = e^[(35500 / 8.314) (1/386.15 – 1/395.15)]

 

? = 1.00 / e^[(35500 / 8.314) (1/386.15 – 1/395.15)]

= 0.77736 atm

 

 

 

답: 0.777 atm