DISCLAIMER: these notes are provided to assist you in mastering the course material but they are not intended as a replacement of the
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conjunction with your own notes.
Chapter 4: The Gaseous State



3 physical states of matter under normal conditions: solid, liquid, gas
former 2 are condensed states
gas state unique:
 no interactions between molecules under normal conditions
 single Equation of State for any gas regardless of which

3 universal properties of a gas:
 volume
 pressure
 temperature
Pressure



early work with Torricelli’s barometer, height of liquid in a sealed, inverted tube varies with
atmospheric pressure (Fig. 4.2)
pressure is defined as force per unit area: P = F/A
for units, recall (Newton’s Second Law): F = ma
 m is mass in kg
 a is acceleration in m s-2
 hence, units of force: kg m s-2 = Newton = N

units of pressure:
F
kg m s -2
P 

 N m -2  Pa  pascal
A
m2
 this is the SI unit of pressure (note: weather report in kPa)

see Table 4.2 for other units of pressure
 based on atmospheric pressure, taking a standard atmosphere as that supporting a
column of mercury (Torricelli barometer) of 760 mm at 0oC
 1 atm = 101,325 Pa
 derived unit: 1 bar = 100,000 Pa, defined as Standard Pressure
Boyle’s Law, Variation of Volume with Pressure

Boyle’s J-tube expt, Fig. 4.3, closed system at constant temperature
 P  1/V, or PV = c, (a constant) Boyle’s Law




3 graphical presentations, Fig. 4.4
independent of gas used as long as temperature not too low, pressure not too high
note: at 0oC for 1 mol of gas, pressure in atm: PV = 22.4 L atm
useful relationship: P1V1 = P2V2
(Example 4.1)
Chem 59-110 (’02)
Charles’ Law, Variation of Volume with Temperature

two scales of temperature:
 a relative scale, eg. based on the liquid range of water at standard pressure, Celsius scale
(and Fahrenheit), negative values possible
 an absolute scale with a true zero point:
 Charles: all gases, at low pressures, expand by same relative amount when heated
through the same temperature range
 Converse expt Fig. 4.6, extrapolate to zero volume at –273.15 oC = absolute zero
 Kelvin scale defined: T (Kelvin) = t (Celsius) + 273.15
 Example 4.2
The Ideal Gas Law





Combine Boyle’s and Charles’ Laws, both for fixed amount of gas
 V  1/P (at fixed T)
 V  T (at fixed P),
Also, V  n, the number of moles of gas (at fixed T, P)
Overall, V  nT/P
 Proportionality constant, R: PV = nRT
Ideal Gas Law
 R = universal gas constant = 8.31451 J mol-1 K-1 (in SI, P in Pa, V in m3)
= 0.082058 L atm mol-1 K-1 (when P in atm, V in L)
 (Caution with SI: P in kPa, compensate with V in L)
Useful relationship:
P1 V1
PV
(examples 4.3 and 4.4)
 2 2
n 1T1
n 2 T2
Chemical calculations with gases – worked example 4.5:
Concentrated nitric acid acts on copper as follows:
Cu (s)  4 H  (aq)  2 NO3- (aq)  2 NO 2 (g)  Cu 2 (aq)  2 H 2 O (l)
If 6.80 g of copper is consumed and the nitrogen dioxide is collected at a pressure of 0.970
atm and a temp of 45oC, what volume of NO2 is produced?



6.80 g Cu
 0.107 mol Cu
63.55 g mol -1
calculate moles of NO2 produced from stoichiometry:
 2 mol NO 2 
0.107 mol Cu x 
  0.214 mol NO 2
 1 mol Cu 
use ideal gas law (caution: temp scale, pressure units, etc.)
nRT
(0.214 mol)(0.082 06 L atm mol -1 K -1 ) (273  45 K)
V 

 5.76 L
P
0.970 atm
Convert copper to moles:
Chem 59-110 (’02), ch. 4, The Gaseous State
Mixtures of Gases – Dalton’s Law of Partial Pressures

gas law independent of composition, therefore holds for mixtures:
RT
Psystem  n total
V
but n total  n 1  n 2  n 3  ... n i
and
consequent ly

Pi
Psystem

ni
n total
ni
Psystem
n total
the ratio, ni/ntotal = the mole fraction (symbol, X)
thus, Pi = Xi Psystem
and

RT
V
RT
RT
RT
RT
 n1
 n2
 n3
 ... n i
V
V
V
V
 P1  P2  P3  ... Pi
Psystem  (n 1  n 2  n 3  ... n i )
Pi 
Example 4.6 and related
The Kinetic Theory of Gases



back to the microscopic – attempt to explain generalizations of ideal gas law from particle
dynamics; connect temp to distribution of molecular speeds
assumptions:
 distances between molecules large compared to size
 gas molecules in constant random motion with distribution of speeds
 molecules exert no forces on each other between collisions, when they move in straight
lines at constant velocities
 collisions with walls are elastic
molecule in a box (Fig. 4.10)
 speed, u, related to velocities by: u2 = vx2 + vy2 +vz2 ;
 focus on x direction, momentum mvx
 after collision with wall, momentum -mvx ; change, px = 2mvx
 frequency of collisions: 2l/vx
 rate of momentum transfer to wall: px/t = 2mvx/(2l/vx) = mvx2/l
 from Newton’s second law, force on the wall:
Chem 59-110 (’02), ch. 4, The Gaseous State

mv 2x
p
 f 
, per molecule
dt

N mv 2
or, per mole F  o x , where v x now average velocity

pressure, P = F/A and vx2 is 1/3 of u2 , hence:
1 N mu 2
o
1
P  3
x
; but  x A  V

A
PV  1 N o mu 2 ;
3
from gas law, PV  nRT ; here n  1 and N o m  molar mass, M (in kg)
3RT
3RT
; u 
, root - mean - square speed
M
M
the higher the temp and the smaller the molecules, the higher the speed
Example 4.7
In reality, a distribution of speeds, Fig 4.13, which is also an energy distribution:
 Mu2 = 3RT, from above
 ½ Mu2 = kinetic energy = 3/2 RT (note: independent of which gas!)
sample of gas is an energy reservoir, where different energy content expressed in
speed of molecules
u2 




Some Gas Law Problems
1. Calculate the rms speed of nitrogen at 100oC
(later, do fluorine at 25oC)
R = 8.314 J K-1 mol-1
T = 100.0 + 273.2 = 373.2 K
M = 2 x 14.007 x 10-3 = 0.02801 kg mol-1
u 
3RT

M
3 x 8.314 x 373.2
 576.5 m s -1
0.02801
2. Calculate the density of air at 25oC and 100 kPa. Assume the composition is 20 mol %
oxygen and 80 mol % nitrogen.
Chem 59-110 (’02), ch. 4, The Gaseous State
Ptotal  PO2  PN 2
PO 2
Ptotal
 0.20 ;
PN 2
Ptotal
 0.80
Assume 1 liter, use PV  nRT
PO2 V
0.20 x 100 x 1
 8.067 x 10 -3 mol
RT
8.314 x 298.2
mass of O 2  n O2 x M  8.067 x 10 -3 x 31.98  0.2580 g
n O2 

similarly, calculate mass of N 2  0.9041 g
 density  0.258  0.904  1.162 g L-1
3. Equal amounts of fluorine and bromine pentafluoride are mixed. Determine the ratio of the
rates of effusion of the two gases through a very small; opening in their container.
Rate (F2 )

Rate (BrF5 )
3RT
M F2
3RT
M BrF5

M BrF5
M F2

174.89 x 10 -3
 2.145
38.00 x 10 -3
Real Gases: Intermolecular Forces

Sections 4.6 and 4.7, read for interest only; note deviations from ideality in Fig. 4.19
Suggested Problems
5 – 43, odd.
Chem 59-110 (’02), ch. 4, The Gaseous State