Chem142 – Introduction to Physical Chemistry
Homework #1
September 4, 2014
due September 10, 2014
0.10.
a) extensive
b) intensive
c) intensive
d) intensive
e) extensive
Discussion 1.7.
The van der Waals equation accounts for repulsive interactions between molecules by supposing that they cause the
molecules to behave as small but impenetrable spheres. The nonzero volumes of the molecules implies that instead
of moving in a volume V, they are restricted to a smaller volume V-nb, where b is proportional to the volume of a
mole of molecules. This changes the ideal gas law to P = nRT / (V-nb).
Attractive forces reduce the pressure of the gas and are proportional to the concentration of the gas squared (squared
for pairs of molecules interacting). If the reduction in pressure is written as –a(n/V)2, where a is a positive constant
nRT
an 2
 2
characteristic of each gas, the complete van der Waals equation is: P 
V  nb V
1-1.
N2 3.055g V=3.00 dm3=3.00 L T=32°C

1mol 
 3.055 g *
  0.08206 L * atm / mol * K  305 K 
28.014 g 
nRT 

 0.910atm
P
V
3.00 L
1-3.
V=300 cm3
n
PV

RT
T=14.5°C
P=34.5 kPa
3

1atm


3  1dm 
 34.5kPa *
  300.0cm * 
 
101.325kPa  

 10cm  
 0.08206 L * atm / mol * K  287.5K 
1-6.
T1 = 18 °C P1 = 125 kPa
T2 = 700 °C P2 = ?
PV
PV
1 1
 2 2
T1
T2
P1 P2

T1 T2
(V is not changing)
P
125kPa
 2
291K
973K
P2  418kPa
 0.00433mol NO
1-9.
T1 = 315 K
increase V by 25%  V2 = 1.25 * V1
PV
PV
1 1
 2 2
T1
T2
V1 V2

T1 T2
(P is not changing)
1.25V1
V1

315 K
T2
T2  394 K
1-15.
320 mg CH4
175 mg Ar
225 mg N2 PN2 = 15.2 kPa
T=300 K
1mol
 0.0199mol
16.043 g
1mol
 0.175 g *
 0.00438mol
39.95 g
1mol
 0.225 g *
 0.00803mol
28.014 g
 0.0199mol  0.00438mol  0.00803mol  0.0323mol
nCH 4  0.320 g *
nAr
nN 2
ntotal
PN2  xN2 Ptotal
 0.00803mol 
15.2kPa  
 Ptotal
 0.0323mol 
Ptotal  61.1kPa
PtotalV  ntotal RT
V
ntotal RT  0.0323mol  0.08206 L * atm / mol * K  300 K 

 1.32 L
1atm
Ptotal
61.1kPa *
101.325kPa
1-17.
d=1.23 g/dm3
T=330 K
P=25.5 kPa
PM  dRT
M 
dRT 1.23 g / L  0.08206 L * atm / mol * K  330 K 

 132.3g / mol
1atm
P
25.5kPa *
101.325kPa
1-36.
1.0 mol C2H6
V=22.414 dm3
i) T = 273.15 K
ideal gas
P
nRT 1.0mol  0.08206 L * atm / mol * K  273.15 K 

 1.00atm
V
22.414 L
VDW gas
2
2
nRT
an 2 1.0mol  0.08206 L * atm / mol * K  273.15 K   5.507atm * L / mol  1.0mol 
P
 2 

 0.992atm
2
V  nb V
22.414 L  1.0mol   6.51*102 L / mol 
 22.414 L 
2
3
i) T = 1000 K
ideal gas
P
 1dm 
3
V= 100cm3 * 
  0.100dm
10
cm


nRT 1.0mol  0.08206 L * atm / mol * K 1000 K 

 821atm
V
0.100 L
VDW gas
2
2
nRT
an 2 1.0mol  0.08206 L * atm / mol * K 1000 K   5.507atm * L / mol  1.0mol 
P



 1801atm
2
V  nb V 2
0.100 L  1.0mol   6.51*102 L / mol 
 0.100 L 
2
1-37.
10.00 g CO2 V=100cm3
ideal gas
T=25.0°C

1mol 
10.00 g *
  0.08206 L * atm / mol * K  298K 
44.009
g
nRT 
P

 55.6atm
V
0.100 L
VDW gas
2
2
nRT
an 2  0.227 mol  0.08206 L * atm / mol * K  298 K   3.610atm * L / mol   0.227 mol 
 2 

 42.9atm
2
V  nb V
0.100 L   0.227 mol   4.29 *10 2 L / mol 
 0.100 L 
2
P
difference = 55.6 atm – 42.9 atm = 12.7 atm
12.7atm
*100%  29.6%
% difference =
42.9atm
Pvdw < Pideal implies that attractive molecular forces are significant at these conditions.
Discussion 2.6.
The general expression for work is w    Pext dV
When the expansion work is against a constant external pressure, the expression becomes w   Pext V
When the expansion work is reversible, P = Pext. In the case of a reversible isothermal expansion of an ideal gas, the
Vf
expression becomes w  nRT ln
.
Vi
These examples demonstrate that work depends upon the particulars (i.e., the path) of the process. Work is not a
state function. Care must be taken to select or derive the work relation that is applicable to a particular process.
2-1.
a) expand 1.0 cm3 against P = 100 kPa
w   Pext V
3

 
3  1m
  100 *10 Pa  1.0cm * 
 

 100cm  

 0.1J
To compress back, w = + 0.1 J
3
b) expand 1.0 dm3 against P = 100 kPa
w   Pext V
3

 1m  
  100*103 Pa   1.0dm3 * 
 

 10dm  

 100 J
To compress back, w = + 100 J
2-3.
4.50 g CH4
V=12.7 dm3
T=310 K
a) isothermal expansion against constant external P = 30.0 kPa
V  3.3dm3
3

 1m  
w   Pext V    30.0 *103 Pa   3.3dm3 * 
  99 J


 10dm  

b) isothermal and reversible
Vf
w  nRT ln
Vi

1mol 
12.7 dm3  3.3dm3
   4.50 g *
  8.314 J / K * mol  310 K  ln
16.043g 
12.7 dm3

 167 J
2-7.
10.0 g C12H22O11
T=20.0 °C Pext = 1.20 atm
a) C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(l)
Δng = 0 for the reaction
 ΔV = 0  w = -PextΔV = 0
b) C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(g)
10.0 g C12 H 22 O11 *
1mol C12 H 22 O11
11mol H 2 O
*
 0.321mol H 2 O
342.29 g C12 H 22 O11 1mol C12 H 22 O11
 change in moles of gas = +0.321 mol
nRT  0.321mol  0.08206 L * atm / mol * K  293.0 K 

 6.44 L
P
1.20atm
w   Pext V   1.20atm  6.44 L   7.73L * atm  783 J
V 
2-15.
n=1.00 mol T = 300 K
expand reversibly, isothermally
Vf = 30.0 dm3
Vi = 22 dm3
w   nRT ln
Vf
Vi
  1.00mol  8.314 J / K * mol  300 K  ln
 774 J
U  0 since T  0
U  q  w  q  w  0  q   w  774 J
30.0dm3
22dm3
4.
The following diagram represents the P-V changes of a gas. Write an expression for the total work done.
P1
P2
P
P3
V1
V2
V
wtotal = -P2(V2-V1) –P3(V3-V2)
V3

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