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If the vapor pressure of ethanol at 34.7°C is 100 mmHg, and its heat of vaporization is 38.6 kJ/mol, at what temperature would ethanol boil at sea level?

Answer :

Answer:

we will use the Clausius-Clapeyron equation to estimate the vapour pressures of the boiling ethanol at sea level pressure of 760mmHg:


ln (P2/P1) = [tex]\frac{ΔvapH}{R}([tex]\frac{1}{T1}[/tex]-[tex]\frac{1}{T2}[/tex])


where

P1 and P2 are the vapour pressures at temperatures T1 and T2


Δ
vapH = the enthalpy of vaporization of the ETHANOL


R = the Universal Gas Constant


In this problem,


P
1
=
100 mmHg

; T
1
=
34.7 °C
=
307.07 K


P
2
=
760mmHg

T
2
=T⁻²=?


Δ
vap
H
=
38.6 kJ/mol


R
=
0.008314 kJ⋅K
-1
mol
-1


ln
(
760/10)=(0.00325 - T⁻²) (38.6kJ⋅mol-1
/0.008314
)

0.0004368=(0.00325 - T⁻²)

T⁻²=0.002813

T² = 355.47K

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