If we next assume that the gas is ideal (so that pV ϭ nRT ), then because
there is no change in temperature, there can be no change in the product pV.
Thus, instead of Eq. 19-53 a free expansion involves the relation
p i V i ϭ p f V f (free expansion).
(19-63)
574
CHAPTE R 19 TH E KI N ETIC TH EORY OF GASES
expression into Eq. 19-64 and integrating lead us to
(19-66)
Before we substitute in given data, we must determine the
ratio g of molar specific heats for a gas of diatomic molecules
with rotation but no oscillation. From Table 19-3 we find
g
(19-67)
We can now write the work done by the gas as the following
(with volume in cubic meters and pressure in pascals):
ϫ [(8.00 ϫ 10
Ϫ6
)
Ϫ1.4ϩ1 Ϫ (4.00 ϫ 10
Ϫ6
)
Ϫ1.4ϩ1
]
ϭ 0.48 J.
(Answer)
The first law of thermodynamics (Eq. 18-26) tells us that
⌬E int ϭ Q Ϫ W. Because Q ϭ 0 in the adiabatic expansion,
we see that
⌬E int ϭ Ϫ0.48 J.
(Answer)
With this decrease in internal energy, the gas temperature
must also decrease because of the expansion.
W ϭ
1
Ϫ1.4 ϩ 1
(2.00 ϫ 10
5
)(4.00 ϫ 10
Ϫ6
)
1.4
ϭ
C p
C V
ϭ
7
2 R
5
2 R
ϭ 1.4.
ϭ
1
Ϫg ϩ 1
p i V
g
i [V
Ϫgϩ1
f
Ϫ V
Ϫgϩ1
i
] .
ϭ p i V
g
i ͵
V f
V i
V
Ϫ g dV ϭ
1
Ϫg ϩ 1
p i V
g
i [V
Ϫgϩ1
]
V f
V i
W ϭ ͵
V f
V i
p dV ϭ ͵
V f
V i
V
Ϫ g
p i V
g
i dV
Sample Problem 19.09 Work done by a gas in an adiabatic expansion
Initially an ideal diatomic gas has pressure p i 2.00 10
5
Pa
and volume V i ϭ 4.00 ϫ 10
Ϫ6
m
3
. How much work W does
it do, and what is the change ⌬E int in its internal energy if
it expands adiabatically to volume V f ϭ 8.00 ϫ 10
Ϫ6
m
3
?
Throughout the process, the molecules have rotation but
not oscillation.
KEY IDEA
(1) In an adiabatic expansion, no heat is exchanged between the gas and its environment, and the energy for the
work done by the gas comes from the internal energy.
(2) The final pressure and volume are related to the initial
pressure and volume by Eq. 19-54
. (3) The
work done by a gas in any process can be calculated by
integrating the pressure with respect to the volume (the
work is due to the gas pushing the walls of its container
outward).
Calculations: We want to calculate the work by filling out
this integration,
(19-64)
but we first need an expression for the pressure as a function of volume (so that we integrate the expression with
respect to volume). So, let’s rewrite Eq. 19-54 with indefinite
symbols (dropping the subscripts f ) as
(19-65)
The initial quantities are given constants but the pressure p
is a function of the variable volume V. Substituting this
p ϭ
1
V
g
p i V
g
i ϭ V
Ϫg p i V
g
i .
W ϭ ͵
V f
V i
p dV,
(p i V
g
i ϭ p f V
g
f )
ϫ
ϭ
2. When the process is adiabatic (no energy is transferred as
heat), then the energy required for the work can come only
from the internal energy of the gas.
3. Because the internal energy decreases, the temperature
T must also decrease.
Calculations: We can relate the initial and final temperatures and volumes with Eq. 19-56:
(19-68)
Because the molecules are diatomic and have rotation but
not oscillation, we can take the molar specific heats from
T i V
gϪ1
i
ϭ T f V
gϪ1
f
.
Sample Problem 19.10 Adiabatic expansion, free expansion
Initially, 1 mol of oxygen (assumed to be an ideal gas) has
temperature 310 K and volume 12 L. We will allow it to
expand to volume 19 L.
(a) What would be the final temperature if the gas expands adiabatically? Oxygen (O 2 ) is diatomic and here has rotation but
not oscillation.
KEY IDEAS
1. When a gas expands against the pressure of its environment, it must do work.
there is no change in temperature, there can be no change in the product pV.
Thus, instead of Eq. 19-53 a free expansion involves the relation
p i V i ϭ p f V f (free expansion).
(19-63)
574
CHAPTE R 19 TH E KI N ETIC TH EORY OF GASES
expression into Eq. 19-64 and integrating lead us to
(19-66)
Before we substitute in given data, we must determine the
ratio g of molar specific heats for a gas of diatomic molecules
with rotation but no oscillation. From Table 19-3 we find
g
(19-67)
We can now write the work done by the gas as the following
(with volume in cubic meters and pressure in pascals):
ϫ [(8.00 ϫ 10
Ϫ6
)
Ϫ1.4ϩ1 Ϫ (4.00 ϫ 10
Ϫ6
)
Ϫ1.4ϩ1
]
ϭ 0.48 J.
(Answer)
The first law of thermodynamics (Eq. 18-26) tells us that
⌬E int ϭ Q Ϫ W. Because Q ϭ 0 in the adiabatic expansion,
we see that
⌬E int ϭ Ϫ0.48 J.
(Answer)
With this decrease in internal energy, the gas temperature
must also decrease because of the expansion.
W ϭ
1
Ϫ1.4 ϩ 1
(2.00 ϫ 10
5
)(4.00 ϫ 10
Ϫ6
)
1.4
ϭ
C p
C V
ϭ
7
2 R
5
2 R
ϭ 1.4.
ϭ
1
Ϫg ϩ 1
p i V
g
i [V
Ϫgϩ1
f
Ϫ V
Ϫgϩ1
i
] .
ϭ p i V
g
i ͵
V f
V i
V
Ϫ g dV ϭ
1
Ϫg ϩ 1
p i V
g
i [V
Ϫgϩ1
]
V f
V i
W ϭ ͵
V f
V i
p dV ϭ ͵
V f
V i
V
Ϫ g
p i V
g
i dV
Sample Problem 19.09 Work done by a gas in an adiabatic expansion
Initially an ideal diatomic gas has pressure p i 2.00 10
5
Pa
and volume V i ϭ 4.00 ϫ 10
Ϫ6
m
3
. How much work W does
it do, and what is the change ⌬E int in its internal energy if
it expands adiabatically to volume V f ϭ 8.00 ϫ 10
Ϫ6
m
3
?
Throughout the process, the molecules have rotation but
not oscillation.
KEY IDEA
(1) In an adiabatic expansion, no heat is exchanged between the gas and its environment, and the energy for the
work done by the gas comes from the internal energy.
(2) The final pressure and volume are related to the initial
pressure and volume by Eq. 19-54
. (3) The
work done by a gas in any process can be calculated by
integrating the pressure with respect to the volume (the
work is due to the gas pushing the walls of its container
outward).
Calculations: We want to calculate the work by filling out
this integration,
(19-64)
but we first need an expression for the pressure as a function of volume (so that we integrate the expression with
respect to volume). So, let’s rewrite Eq. 19-54 with indefinite
symbols (dropping the subscripts f ) as
(19-65)
The initial quantities are given constants but the pressure p
is a function of the variable volume V. Substituting this
p ϭ
1
V
g
p i V
g
i ϭ V
Ϫg p i V
g
i .
W ϭ ͵
V f
V i
p dV,
(p i V
g
i ϭ p f V
g
f )
ϫ
ϭ
2. When the process is adiabatic (no energy is transferred as
heat), then the energy required for the work can come only
from the internal energy of the gas.
3. Because the internal energy decreases, the temperature
T must also decrease.
Calculations: We can relate the initial and final temperatures and volumes with Eq. 19-56:
(19-68)
Because the molecules are diatomic and have rotation but
not oscillation, we can take the molar specific heats from
T i V
gϪ1
i
ϭ T f V
gϪ1
f
.
Sample Problem 19.10 Adiabatic expansion, free expansion
Initially, 1 mol of oxygen (assumed to be an ideal gas) has
temperature 310 K and volume 12 L. We will allow it to
expand to volume 19 L.
(a) What would be the final temperature if the gas expands adiabatically? Oxygen (O 2 ) is diatomic and here has rotation but
not oscillation.
KEY IDEAS
1. When a gas expands against the pressure of its environment, it must do work.