Physical Chemistry II
Chem 402
Spring 2012
Chapter 1 (3, 10, 13, 20, 24, 37)
P1.3) Calculate the pressure exerted by Ar for a molar volume of 1.04 L mol-1 at 475 K using the van der
Waals equation of state. The van der Waals parameters a and b for Ar are 1.355 bar dm6 mol–2 and 0.0320
dm3 mol–1, respectively. Is the attractive or repulsive portion of the potential dominant under these
conditions?
P

RT
a
 2
Vm  b Vm
8.314×102 bar dm3 mol1K 1 ×475 K 1.355 bar dm 6 mol2

 37.93 bar
3
1 2
1.04 dm3 mol1  0.0321 dm3 mol1
1.04
dm
mol


Pideal 
RT 8.3145  102 ×L bar mol1 K 1 ×300. K

 38.0 bar
Vm
1.04 L mol1
P1.10) A typical diver inhales 0.500 liters of air per breath and carries a 20. L breathing tank containing
air at a pressure of 300. bar. As she dives deeper, the pressure increases by 1 bar for every 10.08 m. How
many breaths can the diver take from this tank at a depth of 25 m? Assume that the temperature remains
constant.
V25 m 
Vsurface Psurface
P25 m
number breaths =

20. L  300. bar
 1.72  103 L
25 m
1
10.08 bar m 1
V25 m
 3.5  103
Vbreath
P1.20) In the absence of turbulent mixing, the partial pressure of each constituent of air would fall off
with height above sea level in the Earth’s atmosphere as Pi  Pi 0 e  M i gz
RT
where Pi is the partial pressure at
the height z, Pi 0 is the partial pressure of component i at sea level, g is the acceleration of gravity, R is the
gas constant, T is the absolute temperature, and Mi is the molecular mass of the gas. As a result of
turbulent mixing, the composition of the Earth’s atmosphere is constant below an altitude of 100 km, but
the total pressure decreases with altitude as
air. At sea level,
P  P0eMave gz RT
xN 2  0.78084 and xHe  0.00000524
where Mave is the mean molecular weight of
and T = 300 K.
a) Calculate the total pressure at 10.0 km assuming a mean molecular mass of 28.9 g mol-1 and that T =
300 K throughout this altitude range
b) Calculate the value that xN 2 xHe would have at 10.0 km in the absence of turbulent mixing. Compare
your answer with the correct value.
Ptot  PN02 e

M i gz
RT
 28.9  103 kg  9.81 m s 2 10.0 103 m 
 1.0125 105 Pa exp  

8.314 J mol1 K 1  300. K


 3.25 104 Pa

PN2  P e
0
N2
M i gz
RT

 28.02  103 kg  9.81 m s 2 10.0 103 m 
0.78084  1.0125  105 Pa exp  

8.314 J mol1 K 1  300. K


4
 2.63 10 Pa
PHe  PHe0 e

M i gz
RT

 4.003 103 kg  9.81 m s 2 10.0 103 m 
5.24  10  1.0125  10 Pa exp  

8.314 J mol1 K 1  300. K


 0.453 Pa
6
PN2
PHe
PN2
PHe
5
 5.79  104 without mixing

xN 2
xHe

0.78084
 1.49  105 with mixing
6
5.24 10
P1.24) When Julius Caesar expired, his last exhalation had a volume of 500. cm3 and contained 1.00 mole
percent argon. Assume that T = 300. K and P = 1.00 atm at the location of his demise. Assume further that
T has the same value throughout the Earth’s atmosphere. If all of his exhaled Ar atoms are now uniformly
distributed throughout the atmosphere, how many inhalations of 500. cm3 must we make to inhale one of
the Ar atoms exhaled in Caesar’s last breath? Assume the radius of the Earth to be 6.37 × 106 m. [Hint:
Calculate the number of Ar atoms in the atmosphere in the simplified geometry of a plane of area equal to
that of the Earth’s surface. See Problem 1.20 for the dependence of the barometric pressure on the height
above the Earth’s surface.]



is the number of Ar atoms
The total number of Ar atoms in the atmosphere is N Ar   N Ar
Adz where N Ar
0

per m3 at the surface of the earth. N Ar
is given by
N P
6.023  1023  0.0100  1 105 Pa

N Ar
 A Ar 
 2.411023 m 3
RT
8.314 J K 1 mol1  300 K
The total number of Ar atoms in the atmosphere is
N Ar


0
0


  N Ar Adz  N Ar
e
M Ar gz
RT
RT

Adz  N Ar
A
M Ar g
2.411023 m 3  4   6.37 106 m   8.314 J mol 1 K 1  300 K
2

39.9 103 kg  9.81 m s 2
 7.85  1041
The fraction of these atoms that came from Caesar’s last breath, f, is given by
f 

N Ar
V 2.411023 m 3  0.500  103 m3

 1.53 1022
7.85 1041
N Ar
The number of Ar atoms that we inhale with each breath is
N  NA
PV
102  1 105 Pa  0.500  103 m3
 6.023 1023 mol1 
 1.211020
1
1
RT
8.314 J mol K  300 K
The number of these that came from Caesar’s last breath is fN fN  1.53 1022  1.211020  1.85 102
The reciprocal of this result, or 54, is the number of breaths needed to inhale one Ar atom that Caesar
exhaled in his last breath.
 df  x  dx 
P1.37) Use L’Hôpital’s rule, lim  f  x  g  x   x 0  lim 
 to show that the expression
dg
x
dx



 x 0
derived for Pf in part b of Example Problem 1.1 has the correct limit as  → 0.




Pf Vi 1    Pf  Pi 
PV
i i

;
Ti
Tf


PT
i f  Pf Ti 1    Pf  Pi 
Pf2Ti  Pf Ti 1  Pi   PT
i f 0
Ti 1  Pi   Ti 2 1  Pi   4TT
i f  Pi
2
Pf 
2Ti

2
i f  Pi
df    d  d Ti 1  Pi   Ti 1  Pi   4TT
lim

  0 dg    d 
d  2Ti  d 
Ti Pi 
 lim
2Ti 2 1  Pi   Pi   4TT
i fP
2 Ti 2 1  Pi   4TT
i f  Pi
Ti Pi   PT
i i  2i T f Pi
2Ti
Ti Pi 
2
2Ti
 0

2

Tf P
Ti

 d
2Ti 2   Pi   4TT
i fP
2Ti
2Ti