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The boiling point of ethanol is 73.4 °c at 1.00 atm. calculate the vapor pressure (in atm) of ethanol at 10 °c.

User Mfluehr
by
7.5k points

1 Answer

1 vote

T
1
=
(50.0+ 273.15) K = 323.15 K
;
P
1
=
?

T
2
=
(78.4 + 273.15) K = 351.15 K
;
P
2
=
760 Torr
ln
(
P
2
P
1
)
=
Δ
H
vap
R
(
1
T
1

1
T
2
)
ln
(
760 Torr
P
1
)
=
(
38
560
J⋅mol

1
8.314
J⋅K

1
mol

1
)
(
1
323.15
K

1
351.55
K
)
ln
(
760 Torr
P
1
)
=
4638
×
2.500
×
10

4
=
1.159
760 Torr
P
1
=
e
1.159
=
3.188
P
1
=
760 Torr
3.188
=
238.3 Torr
User Sooriya Dasanayake
by
7.0k points