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
