클라우지우스-클라페이롱. 760 mmHg 1.75 atm ΔH_vap 40.7 kJ/mol

클라우지우스-클라페이롱. 760 mmHg 1.75 atm ΔH_vap 40.7 kJ/mol

 

 

At sea level, the barometric pressure is 760 mmHg. Calculate the difference in the normal boiling point for water boiled in an open pot and water boiled in a pressure cooker set for 1.75 atm. (ΔH_vap for water = 40.7 kJ/mol)

 

---------------------------------------------------

 

> T1 = 100 + 273.15 = 373.15 K, P1 = 760 mmHg = 1 atm

> T2 = ?, P2 = 1.75 atm

 

 

 

클라우지우스-클라페이롱 식. Clausius-Clapeyron relation

ln(P1/P2) = ΔH_vap/R (1/T2 – 1/T1)

( 참고 https://ywpop.tistory.com/3133 )

 

ln(1 / 1.75) = (40700 / 8.314) (1/T2 – 1/373.15)

 

T2 = 1 / [ln(1 / 1.75) / (40700 / 8.314) + 1/373.15]

= 389.78 K

 

389.78 – 373.15 = 16.63 K

= 16.63℃

 

 

 

답: 16.6℃

 

 

 

 

[키워드] 해수면에서 기압 압력 밥솥 열린 냄비