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
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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
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 Spring '08
 HANSON
 Chemistry, Thermodynamics

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