17.1
THE THERMAL BEHAVIOR OF MATTER
EXERCISES
Section 17.1 Gases
18.
The molar volume of an ideal gas at STP for the surface of Mars can be calculated as in Example 17.1. However,
expressing the ideal gas law for 1 mole of gas at the surfaces of Mars and Earth as a ratio,
PV T
MM M
EE E
//
,
=
and using the previous numerical result, we find
VP
P
T
T
V
ME
M
M
EE
=×
=
(
/
)(
/
)
( / .
/
10 0070 218 273 22.4
×
10
2 56
3
−
=
mm
33
).
.
19.
INTERPRET
We are dealing with an ideal gas. We are given the pressure, temperature, and volume, and want to
find the number of gas molecules.
DEVELOP
We shall use the ideal-gas law,
pV
NkT
=
, given in Equation 17.1, to find the number of molecules.
EVALUATE
From Equation 17.1, we have
N
PV
kT
==
×
×
−
−
()
(
.
)
(.
180
8 5 10
138 10
3
23
kPa
m
J/
3
K
K
)
.
350
317 10
23
ASSESS
One mole has
N
A
602 10
23
.
molecules. Thus, we have about 0.53 mole of molecules in the system.
20.
The ideal gas law in terms of the gas constant per mole, Equation 17.2, gives
Pn
R
T
V
/(
.
3 5 mol)
(8.314 J/K mol)(123 K)/
m
3
⋅=
×
)
.
0 002
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)/
L
atm.)
() .
21
7
7
=
21.
INTERPRET
We are dealing with an ideal gas. We want to know how volume changes with temperature.
DEVELOP
To compare different states of an ideal gas, it is often convenient to express Equation 17.1 as a ratio:
NT
11
22
=
In this problem, the pressure is constant, and if no gas escapes or enters, then
N
is also constant. Therefore, the
above equation becomes
V
V
T
T
1
2
1
2
=
where
T
is in the Kelvin scale.
EVALUATE
(a)
If
T
2
100
=°
=
C
373 K and
T
1
200
=
C
473 K then
VV
V
12
2
473
373
127
=
=
K
K
.
(b)
If
TT
2
=
, then
2
=
.
ASSESS
The fact that
VT
~
for a given mass of ideal gas at constant pressure is known as the law of Charles and
Gay-Lussac.
22.
(a)
From Equation 17.2:
V
nRT
P
⋅
(
.
)
(
)
2
8 314
250
15
mol
J/mol K
K
at
m
P
a
/
a
t
m
mL
3
)( .
)
..
1 013 10
274 10
274
5
2
×
=
−
17