Chapter 1
Gases and the Zeroth Law of Thermodynamics
1.2.
1.4.
A system is any part of the universe under observation. The “surroundings” includes
everything else in the universe. Consider a solution calorimeter in which two aqueous
solutions are mixed and temperature changes are recorded. In this case the solutes
(reactants and products) would be considered the “system”. The water, calorimeter, and
rest of the lab and world would be the “surroundings”.
1000 mL 1 cm 3

 12,560 cm 3  1.256  10 4 cm 3
(a) 12.56 L 
1L
1 mL
(b) 45ºC + 273.15 = 318 K
(c) 1.055 atm 
1.01325 bar 100,000 Pa

 1.069  105 Pa
1 atm
1 bar
(d) 1233 mmHg 
1 torr
1 atm 1.01325 bar


 1.644 bar
1 mmHg 760 torr
1 atm
1 cm 3
 125 cm 3
(e) 125 mL 
1 mL
(f) 4.2 K – 273.15 = –269.0°C
(g) 25750 Pa 
1.6.
1 bar
 0.2575 bar
100,000 Pa
patm = pmouth + hg, where hg correspond to the pressure exerted by the liquid
patm - pmounth = hg = (1.0x103 kg/m3)(0.23m)(9.80m/s2) = 2254 N/m2 = 2.3 x103 Pa
1.8.
In terms of the zeroth law of thermodynamics, heat will flow from the (hot) burner or
flame on the stove into the (cold) water, which gets hotter. Then heat will move from the
hot water into the (colder) egg.
1.10.
For this sample of gas under these conditions, F(T) = 2.97 L  0.0553 atm = 0.164 Latm.
If the pressure were increased to 1.00 atm: 0.164 Latm = (1.00 atm)  V; therefore V =
0.164 L.
1.12.
V  R
pV
nT
, which rearranges to R 
.
nT
p
Therefore: R 
(2.66 bar)(27.5 L)
L  bar
 0.0830
(1.887 mol)(466.9 K)
mol  K
1
Instructor’s Manual
1.14.
V1=67 L, p1=1.04 atm.
x atm
1 atm
; x = 6.34 atm. Therefore p2 = 1.04 atm + 6.34 atm = 7.38 atm.

64.0 m 10.1 m
p1V1 = p2V2; so V2 =
1.16.
1.82  1013 g SO 2 
nRT

p
V
(1.04 atm)(67 L)
 9 .4 L
(7.38 atm)
1 moleSO 2
 2.84  1011 molesSO 2
64.07 g
L  atm
)(273.15  17.0K )
mol  K
 7  10 10 atm
21
8  10 L
(2.84  1011 molesSO 2 )(0.0821
L  atm
L  atm 5 / 9 K
 
 0.0456
mol  K 1 R
mol  R
1.18.
0.0821
1.20.
Calculations using STP and SATP use different numerical values of R because the sets of
conditions are defined using different units. It’s still the same R, but it’s expressed in
different units of pressure, atm for STP and bar for SATP.
1.22.
The partial pressure of N2 = 0.80 
14.7 lb
lb
 11.8 2
2
in.
in.
The partial pressure of O2 = 0.20 
14.7 lb
lb
 2.9 2
2
in.
in.
1.24.
4.50torr(44.01g / mole)  0.01706 g / L  1.71 10 5 g / cm3
m pM
m
pV   RT;


L  torr
V RT
M
(62.36
)(273.15  87K)
mol  K
1.26.
Using the ideal gas law, the number of moles of CO2 =
(.965 atm)(1.56 L)
= 0.0621 moles
L·atm
(0.0821
)(273.15K+22.0K)
K·mol
1 mole C6 H12 O6
180.16 g
×
= 5.60 g C6 H12 O6
0.0622 moles CO2 ×
1 mole C6 H12 O6
2 moles CO2
2
Chapter 1
1.28. Following the normal rules of derivation: (a) 3 y 2 
3w 2 z 3 xy 2 z 3
y2 z3

(b)
(c)
32 y
w
w2
w 3 z 3 2 xyz 3
3y 2 z 2
3y 2 z 2
6 xy 

(d) Using the answer from part a, we get 
(e) 
(f)
w
w
w
32 y 2
Using the answer from part e, we get
1.30
3y 2 z 2
.
w2
p
 n 
(a) 
 
 V  T, P R  T
 T 
V
(b)   
 p  V,n n  R
- pV
 n 
(c)   
2
 T  P,V RT
RT
 p 
(d)   
V
 n  T ,V
1.32.
(a) V(T, p) 
nRT
p
 V 
 V 
(b) dV  
 dp
 dT  
 T  p
 p  T
(c)



Latm  
Latm 


 1mole 0.0821
 1 mole 0.0821
350K  

  nRT 
 nR 
Kmole  
Kmole 


0.10atm 
dp  
dT  

dV  
10.0K   
2
2




p
1
.
08
atm
p
1.08atm 












= -1.70 L
 V 
 nRT   2 V 
2nRT
 
1.34. (a) 
;  2  
2
p
p3
 p  n ,T
 p  n ,T
nR   2 p 
 p 
 0
(b)   
;
V  T 2  n , V
 T  n , V
1.36.
(a) Z=1 for an ideal gas. (b) If the gas truly follows the ideal gas law, Z will always be 1
regardless of the pressure, temperature, or molar volume.
3
Instructor’s Manual
1.38.
Using equation 1.23, TB 
a
, and using data from Table 1.6, we have:
bR
L2 atm
mol 2
for CO2: TB 
 1026 K
L atm 

0.04267 L/mol  0.08205

mol K 

3.592
L2 atm
mol 2
for O2: TB 
 521 K
L atm 

0.03183 L/mol  0.08205

mol K 

1.360
L2 atm
mol 2
for N2: TB 
 433 K
L atm 

0.03913 L/mol  0.08205

mol K 

1.390
1.40.
1.42.
C
. In order for the term to be unitless, C should have units of
V2
(volume)2/(moles)2, or L2/mol2. The C’ term is C’p2, and in order for this term to have the
L·mol
. (The unit
same units as p (which would be Latm/mol), C’ would need units of
atm
bar could also be substituted for atm if bar units are used for pressure.)
The C term is
Gases that have lower Boyle temperatures will be most ideal (at least at high
temperatures). Therefore, they should be ordered as He, H2, Ne, N2, O2, Ar, CH4, and
CO2.
1.44. (a) He: b=
m3
0.0237L 1 m 3

 2.37  10 5
mole
mole
1000 L
3
m3
1mole
 29 m

 3.94 10
2.37 10
atom
mole 6.02 10 23 atoms
5
V
4
4 3
m3
r  3.94  10  29
; r  2.11  10 10 m or 2.11 Å
atom
3
(b) H2O: b=
0.03049L
; using a similar method to part a, r = 2.30 Å
mole
(c) C2H6: b=
0.0638L
; using a similar method to part a, r = 2.94 Å
mole
Chapter 1
1.46.
Let us assume standard conditions of temperature and pressure, so T = 273.15 K and p =
1.00 atm. Also, let us assume a molar volume of 22.412 L = 2.2412  104 cm3. Second
virial coefficient terms can be calculated using values in Table 1.6.
Therefore, we have for hydrogen:
pV
B
15.7cm 3 / mol
 1  1
 1.00070 , which is a 0.070% increase in the
V
RT
2.2412 10 4 cm 3 / mol
compressibility.
For H2O, we have:
pV
B
213.3cm3 / mol
 1  1
 0.9905 , which is a 0.95% decrease in the
V
RT
2.2412 10 4 cm 3 / mol
compressibility with respect to an ideal gas.
1.48.
By comparing the two expressions from the text
2
B C
a 1 b

Z  1 b 
      and Z  1   2  
V V
RT  V  V 

it seems straightforward to suggest that, at the first approximation, C = b2. Additional
terms involving V 2 may occur in later terms of the first expression, necessitating
additional corrections to this approximation for C.
1.50. The Redlich-Kwong equation of state: p 
RT
a

Vb
T V ( V  b)
 RT
a
a
 p 



 
2
2
2
T V ( V  b)
  V  T ,n ( V  b)
T V ( V  b)
an 2
1.52. The van der Waals equation of state: (p  2 )(V  nb)  nRT ; As V approaches ∞, the
V
an2/V2 term goes to 0 and the nb term becomes negligible. The equation then reduces to
pV=nRT.
1.54. The Redlich-Kwong equation of state: p 
RT
a

Vb
T V ( V  b)
As T approaches infinity, the second term on the right side goes to 0. The molar volume
of a gas at high temperature will generally be high so the correction factor, b, is
negligible. The equation reduces to p V  RT
5
Instructor’s Manual
RT
1.56. Using the ideal gas law, p 
,p 
V
(0.0821
L  atm
)(273.15K )
K  mol
 1.00atm
22.41L
Using the Dieterici equation of state:
p  (0.0821
L  atm
e
)(273.15K )
K  mol
Latm


 (10.91atmL2 ) /  ( 22.41L )( 0.0821
)( 273.15 K ) 
K mol


(22.41L  0.0401L)
 0.981atm ; It varies
by ~2%.
1.58. In terms of p, V, and T, we can also write the following two expressions using the cyclic
rule:
 T 
 p 


 
 V 
 V 
 T V
 p V
 
and 
. There are other constructions possible that

 
p  T
 p 
 T 
 T  p





 V  T
 V  p
would be reciprocals of these relationships or the one given in Figure 1.11.
1.60. Since the expansion coefficient is defined as
1  V 

 ,  will have units of
V  T  p
1
volume
1


, so it will have units of K-1. Similarly, the
volume temperature temperature
1  V 
isothermal compressibility is defined as  
 , so  will have units of
V  p  T
1
volume
1


, or atm-1 or bar-1.
volume pressure pressure
1.62.

1 1
 1  V 
 1   nRT  1  nRT  1  1

  
   V   

 For an ideal gas,  
2
V  P  T
V  p  V  p  p  V
p p
This is equal to 1 bar-1 at STP and SATP
nRT
 1 nRT
 V , this last
. Since
 
2
p
 V p
T
1V 1

for an ideal gas. The expression  is evaluated as
expression becomes
V p p
p
1.64. For an ideal gas,   
1  V

V  p

1   nRT
  

V p  p
T
T
T 1  V 
T   nRT 
T nR
. For an ideal gas, the ideal gas law can be

 


 
p
p V  T  p pV T  p  pV p
nR V
 , so we substitute to get that this last expression is
rearranged to give
p T
6
Chapter 1
T V
1
, which  . Thus, the two sides of the equation ultimately yield the same
pV T
p
expression and so are equal.
RT
. Therefore, the expression for density becomes, substituting
p
M
pM

. The derivative of this expression with respect
for the molar volume, d 
RT / p RT
pM
 d 
to temperature is 
. Using the definition of V , this can be rewritten as
 
RT 2
 T  p.n
1.66. For an ideal gas, V 
M
 d 
.

 
VT
 T  p.n
1.68.
 g  m 
 m 

Mgh  mol  s 2 
pe

; If we convert g to kg and recognize that a J=
J
RT


K 

 K  mole 
2
 kg  m 


 2 
kg  m 2 Mgh  mol  s 

,
. All of the units cancel and the exponent is unitless.
RT
s2
 J 

K 
 K  mol 
Mgh

RT
1.70. If we assume that the average molecular weight of air is 28.967 g/mole, p  e

Mgh
RT
=e-X
m
J
kg 



X
X   0.028967
 9.8 2  432m    8.314
273.15  39.0K   0.0473; e  1.05atm
K

mole
mole
s





4
3
 5   10
7
7
 7.14
1.72. N=7 score   
4 3
  
7 7
1.74.
The probability that the particle is in the higher state = e  E / RT .
(a) At 200K, probability = e
(b) At 500K, probability = e
J



1000 J /   8.314
  200 K  
K mole 


J



1000 J /   8.314
 500 K  
K mole 


(c) At 1000K, probability = e
=0.548
=0.786
J



1000 J /   8.314
 100 K  
K mole 


=0.887
7
Instructor’s Manual
probability
. For the three temperatures
1  probability
these ratios equal: 1.21, 3.67, and 7.85; as the temperature goes up.
To calculate ratios we can use the equation:
1.76.
8
3
RT; Erot  RT
2
3
3
(b) H2O is a non-linear molecule. Etrans  RT; Erot  RT
2
2
3
(c) Kr is an atom. Etrans  RT; Erot  0
2
3
3
(d) C6H6 is a non-linear molecule. Etrans  RT; Erot  RT
2
2
(a) (CN)2 is a linear molecule. Etrans 