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H20(l) ➡ H2O(g),. ∆H(vap)= +40.79kJ/mol Calculate the temperature at which water and water vapour are at equilibrium at 1 atm pressure.

H20(l) ➡ H2O(g),.    ∆H(vap)= +40.79kJ/mol
Calculate the temperature at which water and water vapour are at equilibrium at 1 atm pressure.

Grade:12

1 Answers

Vikas TU
14149 Points
7 years ago
The temperature at which water and water vapour are at equilibrium at 1 atm pressure,
∆H(vap)= +40.79kJ/mol = > ngRT
where ng = 1-0 = 1
Therefore,40.79 = 1*R*T
T = 40.79*1000/8.341 = >  4890.30 K
 

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