Chapter 2 Gases
1.Give two reasons why the van der Waals equation gives more accurate prediction of gas
pressures, especially when the pressure is high.
Molecules have volume and molecules interact.
2. Analysis of a newly discovered gaseous compound containing only Si and F shows that it
contains 33.01% Si by mass. At 27oC, 2.60 g of this compound exerts a pressure of 1.50 atm in
a 0.250 L vessel. What is the molecular formula of the gas? Assume ideal gas behaviour.
n

PV
RT
150
. atm  0.250 L
0.082 Latm K 1mol 1  (27  273) K
 0.0152 mol
Thus, the molecular weight is MW 
2.60 g
 170.6 g mol 1
0.0152 mol
Assume a 100 g sample. It therefore contains 33.01 g Si and (100.00 – 33.01) = 66.99 g F
Moles Si = 33.01g / 28.09 g mol-1 = 1.18 mol Si
Moles F = 66.99 g / 18.99 g mol-1 = 3.53 mol F
The empirical formula is therefore SiF3. This compound would have a molecular weight of
28.09 + 3(18.99) = 85.1 g mol-1. The molecular formula, which must have a molecular weight of
170.6 g mol-1, must therefore be Si2F6.
3. Chlorine gas, Cl2(g), is produced from seawater via the “chlor-alkali” process. The gas is
stored in containers to prevent unwanted and explosive reactions. If a 15.0 L container holds
0.580 kg Cl2(g) at 200oC:
(a) Calculate the pressure assuming ideal behaviour.
n
580g
 817
. mol
710
. g mol 1
nRT 817
. mol  0.082 L atm K 1mol 1  (200  273) K
P

V
15.0 L
 211
. atm
(b) Calculate the pressure using the van der Waals equation. For Cl2(g), a = 6.49 atm L2 mol-2
and b = 0.0562 L mol-1.
nRT
 n
P
 a 
V  nb  V 
2
817
.  0.082  (200  273)
. 
 818

 6.49


15.0  817
.  0.0562
15.0 
 2179
.  193
. atm
2
 19.9 atm
(c) Why are the two calculated pressures so different?
The van der Waals equation takes account of the fact that the chlorine molecules have
volume and interact.
4. Natural gas is a mixture of two components: methane (CH4(g)) and ethane (C2H6(g)). A typical
mixture might have XCH4 = 0.915 and XC2H6 = 0.085. Assume you have a 15.50 g sample of natural
gas in a volume of 15.00 L at 20.0°C.
(a) How many total moles of gas are there in the sample?
ntotal = nCH4 + nC2H6
and nCH4 / nC2H6 = 0.915 / 0.085 = 10.76
or, nCH4 = 10.76 nC2H6
masstotal = massCH4 + massC2H6
= nCH4 x MWCH4 + nC2H6 x MWC2H6
= (10.76 nC2H6) x MWCH4 + nC2H6 x MWC2H6
= nC2H6 (10.76 MWCH4 + MWC2H6) = 15.50 g
thus, nC2H6 = 15.50 g / (10.76 (16.0 g mol-1) + (30.0 g mol-1)) = 0.0767 mol C2H6
0.0767 mol C2H6 = 0.0767 mol x 30.0 g mol-1 = 2.30 g C2H6
Thus, mass CH4 = 15.50 – 2.30 = 13.20 g CH4
Total moles gas = (2.30 g / 30.0 g mol-1) + (13.2 g / 16.0 g mol-1) = 0.902 mol
(b) What is the pressure of the sample (in atm)?
P = nRT/V = 0.902 mol (0.082 L atm K-1 mol-1)(293 K) / 15.00 L = 1.44 atm
(c) What is the partial pressure of each component (in atm)?
PCH4 = PtotXCH4 = 1.44 atm (0.915) = 1.32 atm
PC2H6 = PtotXC2H6 = 1.44 atm (0.085) = 0.12 atm
(d) What are the average speeds of the two components?
v
5 (a)
3RT
MW
v CH 4 
3(8.314 J K 1mol 1 )(293 K)
 676 ms1
1
0.016 kg mol
v C2 H 6 
3(8.314 J K 1mol 1 )(293 K)
 494 ms1
1
0.030 kg mol
SiCl4(g) is a starting material in the electronics industry. Calculate the density of this gas
(in g/L) at 85C and 705 Torr.
P(MW)
RT
 705Torr 
1

 (28.1  4(35.4))g mol
760
Torr
/
atm


0.082 L atm K 1mol 1 (85  273)K

 5.4 g L1
(b)
A sample of oxygen gas is collected over water at 25C and a total pressure of 745 Torr.
The volume of gas collected is 15.0 mL. What mass of oxygen is collected? (The vapor
pressure of water at 25C is 23.8 Torr.)
pO2  p total  p H2O
 745  23.8Torr
 721.2 Torr
n O2
 721.2 Torr 

 (0.015 L)
p O2 V
760 Torr / atm 



RT
0.082 L atm K 1mol 1 (25  273)K
 0.000583mol
 32 g O 2 mol1  0.0187 g O 2
(c)
Calculate the pressure exerted by 1.00 mol of H2O(g) in 0.500 L at 500C.
(i) Use the ideal gas law.
p
nRT 1.00 mol(0.082 L atm K 1mol 1 )(500  273)K

 126.8atm
V
0.5L
(ii) Use the van der Waals equation. For water, a = 5.46 atm L2 mol-2 and b = 0.0305 L
mol-1
p
nRT
n
a 
V  nb
V
2
1.00 mol(0.082 L atm K 1mol1 )(500  273)K
atm L2  1.00 mol 


5.46


0.5 L  1.00 mol(0.0305 L mol 1 )
mol 2  0.5 L 
2
 135.0 atm  21.8atm
 113.2 atm
6 (a) Calculate which has the higher average speed: He(g) at 25C or F2(g) at 500C.
1/ 2
v He
 3RT 


 MW 
1/ 2
 3RT 
v F2  

 MW 
1/ 2
 3(8.314 J K 1mol 1 )(25  273)K 


0.004 kg mol 1


 1363m s 1
1/ 2
 3(8.314 J K 1mol 1 )(500  273)K 


0.038 kg mol 1


 722 m s 1
Thus, the He(g) has a higher average speed.
(b) 1.00 mol of Cl2(g) diffuses through a porous barrier in 8.0 minutes. Hwo long (in minutes) will
it take for 1.00 mol of SiH4(g) to diffuse under the same conditions?
timeSiH4
timeCl2

rateCl2
rateSiH4

MWSiH4
MWCl2
32.12 g mol1

 0.673
70.9g mol 1
timeSiH4  0.673  timeCl2  0.673  8.0 min  5.38 min
(c) A gas sample contains N2(g), O2(g) and H2(g). The total pressure is 4.5 atm. The number of moles
of each gas is: n(N2(g)) = 5 mol, n(O2(g)) = 1 mol and n(H2(g)) = 0.2 mol. Calculate the
partial pressure of O2(g) in this sample.
pO2  X O2 p total
X O2 
n O2
n N 2  n O2  n H 2

1
 0.161
5  1  0.2
 pO2  0.161(4.5atm)  0.726 atm
7.
At a particular temperature, Kp = 2.5 for the reaction SO2(g) + NO2(g)  SO3(g) + NO(g). If
a container initially contains only SO2(g) and NO2(g), each with a partial pressure of 0.1
atm, find the equilibrium partial pressures of each gas.

SO2(g)
NO2(g)
SO3(g)
NO(g)
Initial, atm
0.1
0.1
0
0
Change, atm
-x
-x
+x
+x
Equilibrium, atm
0.1-x
0.1-x
x
x
x(x)
 2.5
(0.1  x)(0.1  x)
This can solved using the quadratic formula, but note that it is a perfect square!
x
 (2.5)1/ 2  1.58
0.1  x
x  0.158  1.58x

2.58x  0.158
x  0.061
Thus, at equilibrium,
pSO2  p NO2  0.1  x  0.1  0.061  0.039 atm
pSO3  p NO  x  0.061atm