CHAPTER 20
THE THERMAL BEHAVIOR OF
MATTER
ActivPhysics
can help with these problems:
Activities 8.1–8.4
Section 201:
Gases
Problem
1. At Mars’s surface, the planet’s atmosphere has a
pressure only 0.0070 times that of Earth, and an
average temperature of 218 K. What is the volume
of 1 mole of the Martian atmosphere?
Solution
The molar volume of an ideal gas at STP for the
surface of Mars can be calculated as in Example 201.
However, expressing the ideal gas law for 1 mole of
gas at the surfaces of Mars and Earth as a ratio,
P
M
V
M
/T
M
=
P
E
V
E
/T
E
,
and using the previous
numerical result, we find
V
M
= (
P
E
/P
M
)(
T
M
/T
E
)
×
V
E
= (1
/
0
.
0070)(218
/
273)(22
.
4
×
10

3
m
3
) = 2
.
56 m
3
.
Problem
2. How many molecules are in an ideal gas sample at
350 K that occupies 8.5 L when the pressure is
180 kPa?
Solution
From Equation 201,
N
=
PV/kT
=
(180 kPa)(8
.
5
×
10

3
m
3
)
/
(1
.
38
×
10

23
J
/
K)(350 K) =
3
.
17
×
10
23
.
Problem
3. What is the pressure of an ideal gas if 3.5 moles
occupy 2.0 L at a temperature of

150
◦
C?
Solution
The ideal gas law in terms of the gas constant
per mole, Equation 202, gives
P
=
nRT/V
=
(3
.
5 mol)(8
.
314 J
/
K
·
mol)(123 K)
/
(0
.
002 m
3
) =
1
.
79
×
10
6
Pa
.
(The absolute temperature must be
used, but any convenient units for the gas constant
can be used, e.g.,
R
’
0
.
0821 L
·
atm
/
K
·
mol
.
Then
P
= (3
.
5 mol)(0
.
0821 L
·
atm
/
K
·
mol)(123 K)
/
(2 L) =
17
.
7 atm
.
)
Problem
4. An ideal gas occupies a volume
V
at 100
◦
C
.
If the
gas pressure is held constant, by what factor does
the volume change (a) if the Celsius temperature is
doubled and (b) if the kelvin temperature is
doubled?
Solution
To compare different states of an ideal gas, it is often
convenient to express Equation 201 as a ratio:
(
P
1
V
1
/P
2
V
2
) = (
N
1
T
1
/N
2
T
2
)
.
In this problem, the
pressure is constant, and if no gas escapes or enters, so
is
N
, therefore
V
1
/V
2
=
T
1
/T
2
.
(a)
T
is the absolute
temperature. If
T
2
= 100
◦
C = 373 K
,
and
T
1
=
200
◦
C = 473 K
,
then
V
1
= (473
/
373)
V
2
= 1
.
27
V
2
.
(b) If
T
1
= 2
T
2
,
then
V
1
= 2
V
2
.
(The fact that
V
∼
T
for a given mass of ideal gas at constant pressure is
known as the law of Charles and GayLussac.)
Problem
5. If 2.0 mol of an ideal gas are at an initial
temperature of 250 K and pressure of 1.5 atm,
(a) what is the gas volume? (b) The pressure is now
increased to 4.0 atm, and the gas volume drops to
half its initial value. What is the new temperature?
Solution
(a) From Equation 202:
V
=
nRT
P
=
(2 mol)(8
.
314 J
/
mol
·
K)(250 K)
(1
.
5 atm)(1
.
013
×
10
5
Pa
/
atm)
= 2
.
74
×
10

2
m
3
= 27
.
4 L
.
(b) The ideal gas law in ratio form (for a fixed
quantity of gas,
N
1
=
N
2
) gives:
T
2
T
1
=
P
2
V
2
P
1
V
1
,
or
T
2
=
4
.
0 atm
1
.
5 atm
¶
0
.
5
V
1
V
1
¶
(250 K) = 333 K
.
Problem
6. The solar corona is an extended atmosphere of hot
(2
×
10
6
K) gas surrounding the cooler visible
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CHAPTER 20
surface of the Sun. The gas pressure in the solar
corona is about 0.03 Pa. What is the coronal
density in particles per cubic meter? Compare with
Earth’s atmosphere.
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 Spring '08
 WILKE
 Thermodynamics, Energy

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