The normal boiling point of water is 100 Celcius and

anniferathonniz8km 2022-04-12 Answered
The normal boiling point of water is 100 Celcius and the molar enthalpy of vaporization of water is 40.7 KJ/mol. What is the boiling point of water under a pressure of 0.500 atm?
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Answers (1)

empatteMattmkezo
Answered 2022-04-13 Author has 22 answers
Given,
The normal boiling point of water ( T 1 ) = 100 C = ( 273.15 + 100 K ) = 373.15 K
Note: pressure at normal boiling point of water ( P 1 ) = 1   a t m
Boiling point of water under a pressure of 0.500 atm ( T 2 ) = ?
Pressure ( P 2 ) = 0.500
Molar enthalpy of vaporization ( H vap ) = 40.7   k J / m o l = 40.7 × 1000   J / m o l = 40700   J / m o l
Note: universal gas constant ( R ) = 8.314   J / m o l . K
Calculating the boiling point of water under a pressure of 0.500 atm ( T 2 ):
ln ( P 2 P 1 ) = H v a p R ( 1 T 2 1 T 1 ) ln ( 0.500   a t m 1.0   a t m ) = 40700 J m o l 8.314 J m o l K ( 1 T 2 1 373.15 K ) 0.693 = 4895.357 K ( 1 T 2 0.0268 K ) 1 T 2 0.00268 K = 0.0001415267 K 1 T 2 = 0.0028215627 K T 2 = 1 0.0028215627 K T 2 = 354.41 K T 2 = ( 354.41 273.15 ) C T 2 = 81.26 C
Conclusion,
So, the boiling point of water under a pressure of 0.500 atm is 81.26 C
Answer: 81.26 C

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