1
1
When water vaporizes at 100
o
C and 1 atm
pressure, 2.26 kJ/g of energy are transferred from
the surroundings to the water.
At 100
o
C the
density of liquid water is 0.958 g/mL, and the
density of water vapor (steam) is 5.98x10
4
g/mL.
What is the enthalpy change,
H
vap
,
when 1 mole of water vaporizes?
(A) 2.26 kJ
(B) 40.7 kJ
(C) 3.04 kJ
H =
E + P
V
101.3 J = 1 L atm
H
2
O = 18.0 g/mol
2
When
1 mol
water vaporizes at 100
o
C and 1 atm
pressure,
40.7 kJ
of energy are transferred from the
surroundings to the water.
At 100
o
C the density of liquid water is 0.958 g/mL,
and the density of water vapor (steam) is
5.98x10
4
g/mL.
How much work in joules is done in pushing back
the atmosphere when mole of liquid water at 100
o
C
is turned into steam at 100
o
C?
H =
E + P
V
101.3 J = 1 L atm
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document2
3
Asking for work done, I know work w =  P
V
And
V = V
final
– V
initial
, and P = 1 atm is given.
V
initial
= volume of 1 mol liquid water at 100
o
C
18.0 g / (0.958 g/mL) = 18.79 mL = 18.7 x 10
3
L
V
final
= volume of 1 mol of steam at 100
o
C
18.0 g /(5.98x10
4
g/mL) = 3.010 x 10
4
mL
3.010 x 10
4
x 10
3
L = 30.10 L
So w =  1 atm (30.10 L – 0.0187 L) =
 30.08 L atm
Asks for work in joules so
 30.08 L atm (101.3 J / L atm) =  3.05 kJ
4
When
1 mol
water vaporizes at 100
o
C and 1 atm
pressure,
40.7 kJ
of energy are transferred from the
surroundings to the water and
– 3.05 kJ
of work is
done in pushing back the atmosphere.
The negative sign means work is done by the
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 HANSON
 Chemistry, Thermodynamics

Click to edit the document details