Answer:
H vaporization = 100.0788 kJ/mol
Explanation:
Use clausius clapyron's adaptation for the calculation of Hvap as:
[tex]ln\frac {P_2}{P_1}=\frac {H_{vap}}{R}(\frac {1}{T_1}-\frac {1}{T_2})[/tex]
Where,
P₂ and P₁ are the pressure at Temperature T₂ and T₁ respectively.
R is the gas constant.
T₂ = 823°C
T₁ = 633°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So, the temperature,
T₂ = (823 + 273.15) K = 1096.15 K
T₁ = (633 + 273.15) K = 906.15 K
P₂ = 400.0 torr , P₁ = 40.0 torr
R = 8.314 J/K.mol
Applying in the formula to calculate heat of vaporization as:
[tex]ln \frac {400}{40}=\frac {H_{vap}}{R} (\frac {1}{906.15}-\frac {1}{1096.15})[/tex]
Solving for heat of vaporization, we get:
H vaporization = 100078.823 J/mol
Also, 1 J = 10⁻³ kJ
So,
H vaporization = 100.0788 kJ/mol
