115℃ water vapor pressure 3.50 atm temp

115℃ water vapor pressure 3.50 atm temp

 

 

The temperature inside a pressure cooker is 115°C.

Calculate the vapor pressure of water inside the pressure cooker.

What would be the temperature inside the pressure cooker

if the vapor pressure of water was 3.50 atm?

 

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▶ 참고: Clausius-Clapeyron 식

[ https://ywpop.tistory.com/3133 ]

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> P1 = 1 atm, T1 = 273 + 100 = 373 K

---> 물의 정상 끓는점

 

> P2 = ? atm, T2 = 273 + 115 = 388 K

 

 

 

> 물의 증발 엔탈피, ΔH_vap = 40660 J/mol

( 인터넷에서 검색하여 찾은 값. 문제에서 주어져야 할 값. )

 

 

 

ln(P2/P1) = –ΔH/R (1/T2 – 1/T1)

 

ln(P2/1) = –(40660/8.314) (1/388 – 1/373)

 

P2 = e^[–(40660/8.314) (1/388 – 1/373)] = 1.66 atm

 

 

 

 

> P1 = 1 atm, T1 = 273 + 100 = 373 K

 

> P2 = 3.50 atm, T2 = ? K

 

 

 

ln(P2/P1) = –ΔH/R (1/T2 – 1/T1)

 

ln(3.50/1) = –(40660/8.314) (1/T2 – 1/373)

 

1/T2 – 1/373 = –ln(3.50/1) / (40660/8.314)

 

1/T2 = –ln(3.50/1) / (40660/8.314) + 1/373

 

T2 = 1 / [–ln(3.50/1) / (40660/8.314) + 1/373] = 412 K

 

412 – 273 = 139℃

 

 

 

 

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