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S45.
One mole of an ideal gas is allowed to expand reversibly and isothermally (25 °C) at a pressure of 1 atm to a
pressure of 0.1 atm.
a.
What is the change in free energy
Starting with the Gibbs equation for
dG
dG
=
−
SdT
+
VdP
For constant temperature we obtain
dG
=
VdP
Using the ideal gas law and integrating from
P
1
to
P
2
we obtain
dG
=
nRT
dP
P
Δ
G
=
nRT
ln
P
2
P
1
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎟
This is the expression for finding the change in Gibbs Free Energy for an
isothermal reversible expansion of an ideal gas.
Δ
G
=
(1 mole)(8.31451 J/molK)(298 K)ln
0.1 atm
1.0 atm
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎟
=
−
5.7 kJ
b.
What would be the change in free energy if the process occurred irreversible?
Δ
G
=
−
5.7 kJ
It doesn’t matter if the expansion is reversible or not because
G
is a state
function; therefore, it is independent of path.
S46.
Consider the spontaneous process for one mole of water going from H
2
O (
) at 110 °C and 1.00 atm to H
2
O (g) at
110 °C and 1.00 atm.
Calculate
∆
H
, and
∆
S
and show that
∆
S
>
Q
/T.
The enthalpy of vaporization of water at 100
°C is found on p. 746 while the heat capacities of liquid water and steam can be assumed constant at 4.18 J/g K and
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 Fall '08
 Owens,G
 Mole

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