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 idealgas 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
GayLussac.
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
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document17.2
Chapter 17
(b)
The ideal gas law in ratio form (for a fixed quantity of gas,
NN
12
=
) gives:
T
T
PV
T
2
1
22
11
2
40
0
==
,
.
.
or
atm
1.5 atm
5
250
333
1
1
V
V
=
()
KK
23.
INTERPRET
We treat air molecules as ideal gas. Given the pressure, temperature, and volume, we want to find
the number of air molecules.
DEVELOP
We shall use the idealgas law,
pV
NkT
=
, given in Equation 17.1, to find the number of molecules.
EVALUATE
The number of air molecules is
N
kT
×
−−
−
(.
10
10
138 10
10
3
23
Pa
m
J/K)(
3
273 K)
2.65 10
7
=×
ASSESS
One mole has
N
A
602 10
23
.
molecules. Thus, we have about44 10
17
.
×
−
mole of molecules in the
system.
24.
The thermal speed (also called the rms, or rootmeansquare speed) is, from Equation 17.4,
vk
T
m
th
=
3/
, where
m
is the mass of a molecule. For molecular hydrogen,
m
<
2u, so
v
th
J/K
K
kg
=
×
×
−
−
3 1 38
10
800
2 1 66
10
23
27
)
(
)
)
.
=
316km/s
25.
INTERPRET
In this problem we want to compare the speeds of two different molecules that are at different
temperatures.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '10
 Pheong
 Thermodynamics, Heat

Click to edit the document details