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Alexxx [7]
4 years ago
10

Assume that the ΔHo and ΔSo of vaporization do not change significantly with temperature. Calculate the vapor pressure of CH3OH

at 42 oC (in atm).
CH3OH (l) ↔ CH3OH(g) . . . ΔHo = 38.0 kJ and ΔSo = 112.9 J/K
Chemistry
1 answer:
expeople1 [14]4 years ago
7 0
We are given with
<span>ΔHo = 38.0 kJ
ΔSo = 112.9 
for these standard conditions:
To = 25 C
Po = 1 atm

Using the Clapeyron Equation
ln (P/Po) = (</span><span>ΔHo / R)(1/T1  - 1/T2)
Substitute the values and solve for P (Temperature should be in Kelvin)</span>
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