Lecture 6
Gases
Textbook: Chapter 4
Characteristics of Gases
Often referred to as vapors

expand to fill container

highly compressible

form homogeneous mixtures
Pressure
P = F = m·g = dhA·g = dhg
A
A
A
(d = m/V, V=hA)
Chapter 10 Gases 1
1 atm = 760 mm Hg = 760 torr = 1.01325 x 105 Pa = 101.325 kPa
1 bar = 105 Pa
1 bar ≈ 1 atm
Barometer
760 mm Hg
Manometers
a) Open
b) Closed
The Gas Laws
Boyle’s Law – the volume of a fixed quantity of gas maintained at constant temperature is
inversely proportional to the pressure.
V k 
V
1
P
graph is a hyperbola
1
P
k is a constant
Chapter 10 Gases 2
or
PV = k
Charles' Law – the volume of a fixed amount of gas maintained at constant pressure is directly
proportional to its absolute temperature.
VT
graph is linear
V k  T
or
V
= k
T
When the lines for experiments at different (but constant) pressures are plotted, they all
extrapolate to –273.15oC.
It is only when an absolute temperature is defined (K = oC + 273.15) that the linear relation
“doubling the temperature doubles the volume”, holds.
Combined Boyle’s – Charles’ Law
Since V = k /P and V = k’T, then P = k”T
Chapter 10 Gases 3
Avogadro’s hypothesis: Equal volumes of gases at the same temperature and pressure contain
equal numbers of molecules.
V k  n
(n = number of moles of any gas)
Corollary: Since V = k/P, P = k · n
The Ideal-Gas Equation
V
nT
P
nT 
V  R 
 P 
PV  nRT
Ideal Gas Law
All other gas laws can be derived from the ideal-gas law!
The Gas Constant:
R  0.08206
L  atm
mol  K
Beware of units, R = 8.314 J/(mol·K) = 8.314 J·mol-1·K-1
Toolbox 4.1
Standard Temperature and Pressure (STP)


L-atm 273.15 K
1.00 mol  0.08206 mol-K


molar V  nRT  
 22.41 L
P
STP: T = 273.15 K
1.000 atm
P = 1 atm
molar volume of an ideal gas at STP: 22.41 L
Standard Ambient Temperature and Pressure (SATP)
SATP: T = 298.15 K (25oC, 77oF),
P = 1 bar, Vm = 24.79 L
Chapter 10 Gases 4
You should be able to write the ideal gas law in terms of all variables. (P, V, T, and n)
PV  nRT = constant
P1 V1  P2 V2
If P increases V must decrease and vice-versa.
If the number of moles doesn't change, then the ideal-gas law is
PV
k
T
P1 V1
PV
 2 2
T1
T2
Boyle's Law
V  P1
V  k 1P
at constant T,n
PV  k
P1V1  k  P2 V2
A balloon contains 14.0 L of air at a pressure of 760 torr. What is the volume if a balloon is
taken to the bottom of a 10 ft pool where the pressure is 981 torr?
P1 = 760 torr
V1 = 14.0 L
P2 = 981 torr
V2 = ?
P V 760 torr 14.0 L
V2  1 1 
 10.8 L
P2
981 torr 
Chapter 10 Gases 5
Charles' Law
V T
V  kT
V T k
V1
V
k 2
T1
T2
at constant P,n
A sample of CO2 gas occupies 300 mL at 10°C and 750 torr. What will the volume be at 30°C
and 750 torr?
V1 = 300 mL
T1 = 10°C = 10+273.15=283 K
V2 = ?
T2 = 30°C = 303 K
V1 V2

T1 T2
V2 
V1 T2 300 mL303 K

 321 mL
T1
283 K
Ideal-Gas Law
Calculate the volume occupied by 0.54 mol of N2 at 15°C and 0.976 atm.
If given moles, use the ideal-gas law.
n = 0.54 mol
T = 15 + 273.15 = 288 K
P = 0.967 atm
R = 0.08206 L-atm/mol-K
V=?
Chapter 10 Gases 6
PV  nRT
V
nRT 0.54 mol0.08206 Latm
molK 288 K

P
0.967 atm 
 13 L
The average human male consumes 200 mL of O2 per hour at 25°C and 1.0 atm for each
kilogram of body weight. How many moles of O2 are consumed by a 70-kg male for 1 hour.
n=?
V = 70 kg x 200 mL/kg = 14000 ml or 14 L
T = 298 K
P = 1 atm
R = 0.08206 L-atm/mol-K
PV  nRT
n
PV
1.0 atm 14 L

 0.57 mol O 2
Latm
RT 0.08206 mol
K 298 K 
What is the pressure in atmospheres in a 35.0 L balloon at 25°C filled with dried hydrogen gas
produced by the reaction of 39.8 g of NaH with water?
We need to find the number of moles of hydrogen from the balanced equation.
NaH  H2 O  NaOH + H 2
39.8g NaH 
1 mol NaH
1 mol H 2

 1.62 mol H 2
24.00 g NaH 1 mol NaH
Chapter 10 Gases 7
n = 1.62 mol
V = 35.0 L
T = 298 K
PV  nRT
L-atm
nRT 1.62 mol0.08206 mol-K
298 K 
P

V
35.0 L
= 1.13 atm
What is the density (in grams per liter) of ammonia at STP if the gas in a 1.000 L bulb weighs
0.672 g at 25°C and 733.4 mm Hg pressure?
Mass remains constant, so moles remain constant.
Need to find volume, V, at STP.
Use:
P1V1 P2 V2

T1
T2
V2 
P1 = 733.4 mm Hg
P2 = 760 mm Hg
T1 = 298 K
T2 = 273.15 K
V1 = 1.000 L
V2 = ?
V2 
P1V1T2
P2 T1
P1 V1 T2 733.4 mm1.000 L273.15 K

P2 T1
298 K 760 mm
 0.884 L
d
0.672 g
 0.760 g L
0.884 L
Chapter 10 Gases 8
Further Applications of the Ideal-Gas Equation
n
P

V RT
nM PM
(M = molecular mass)

V
RT
nM moles grams grams



V
liter mole
liter
PM
density of a gas is measured in g/L
d
RT
M
dRT
P
If a 2.00 L flask contains 3.11 g of cyclopropane gas at 684 torr and 23°C, what is the molecular
weight of cyclopropane ? (R = 0.08206 L-atm/mol-K)
Use the ideal-gas law.
P  684 torr 
1 atm
 0.900 atm
760 torr
T  23  273.15  296 K
PV  nRT
n
PV
0.900.atm2.00 L

 0.0741 mol
Latm
RT 0.08206 mol
K 296 K 
M =
3.11 g
 42.0 g mol
0.0741 mol
A sample of gas was collected and found to have a density of
5.380 g/L at 15°C and 736 mm of Hg pressure. What is the molar mass of the gas?
Chapter 10 Gases 9
Assume we have 1.000 L of the gas, which weighs 5.380 g.
P  736 mm 
1 atm
 0.968 atm
760 mm
PV  nRT
n
PV 0.968 atm 1.000 L

 0.0409 mol
Latm
RT 0.08206 mol
K 288 K 
M
5.380 g
 131 g mol
0.0409 mol
Gas Mixtures and Partial Pressures
Dalton’s Law of Partial Pressures – the total pressure of a mixture of gases equals the sum of the
pressures that each would exert if it were present alone.
Ptotal  P1  P2  P3 
Ptotal  n1  n2  n3 

RT 
RT
 ntotal  
 V 
V
P1
n RT V
n
 1
 1
Ptotal ntotal RT V ntotal
The mole fraction of gas1:
X1 
n1
ntotal
Chapter 10 Gases 10
 n 
P1   1 Ptotal  X1 Ptotal
ntotal 
What is the mole fraction of each component in a mixture of 12.45 g of H2, 60.67 g of N2, and
2.38 g of NH3?
n = mass/M
12.45 g H2, / 2.016 g·mol-1H2 = 6.176 mol H2
60.67 g N2 / 28.02 g·mol-1N2 = 2.165 mol N2
2.38 g NH3 / 17.0 g·mol-1 NH3 = 0.140 mol NH3
6.176 + 2.165 + 0.140 = 8.481
p·H2 = 6.176 mol H2 / 8.481 mol = 0.728
p·N2 = 2.165 mol N2 / 8.481 mol = 0.255
p·NH3 = 0.140 mol NH3 / 8.481 mol = 0.0165
Question:What is the total pressure (in atmospheres) and what is the partial pressure of each
component if the gas mixture in a 10.00 L steel container at 90°C?
Ptotal = ntotal (RT/V)
= 8.481mol· 0.08206 (L·atm·mol-1·K-1) · (90. +273.)K / 10.00 L = 25. atm
pi = Xi . Ptotal
p·H2 = 0.728 · 25. atm = 18. atm
p·N2 = 0.255 · 25. atm = 6.4. atm
p· NH3 = 0.728 · 25. atm = 0.41 atm
Note. Since n=mass/MW, the partial pressures can also be calculated with the formula
Pi = (mi/Mi)·RT/V
But this involves many more calculator manipulations.
Chapter 10 Gases 11
Kinetic-Molecular Theory
1. Gases consist of large numbers of molecules that are in continuous motion, random motion.
2. The volume of all molecules the gas is negligible compared to the total volume in which the
gas is contained.
3. Attractive and repulsive forces between gas molecules are negligible.
4. Energy can be transferred between molecules during collisions, but the average kinetic
energy of the molecules does not change with time, as long as the temperature of the gas
remains constant.
5. The average kinetic energy of the molecules is proportional to the absolute temperature. At
any given temperature the molecules of all gases have the same kinetic energy.
Molecular Effusion (through a small hole) and Diffusion (movement through open space)
3RT
u
M
The average speed, u of a gas particle is proportional to the square root of the absolute
temperature, T, and the reciprocal of the square root of its gram molecular weight, M.
Chapter 10 Gases 12
Graham’s Law of Effusion
By the Kinetic-Molecular Theory, at one given T, the average kinetic energy of one gas is equal
to the average kinetic energy of another gas.
KEHe = KEAr
½ MHe·u2He = ½ MAr·u2Ar
u2He / u2Ar = MAr / MHe
uHe / uAr = (MAr / MHe)1/2
Proven by experiment
The rate of effusion, r of a gas depends on the speed of the gas particle arriving at the point of
effusion and is found to be proportional to the reciprocal of the square root of its gram molecular
weight, M.
r1
M2

r2
M1
Self-test 4.15B
Maxwell-Boltzmann Distribution of Speeds
Temperature is a measure of the average KE (1/2 mv2) of gas particles. Although at one T there
is one average speed, each individual particle can have any particular speed. Which particles are
flying at which speeds? In the mid eighteen-hundreds, James Clerk Maxwell first derived an
equation based on the Kinetic-Molecular Theory that showed what fraction of the the molecules,
f(u), had which particular speed, u, when the gas was at a particular temperature, T.
f(u) = k·(M/T)3/2 · u2 · e(-Mu2/k’T) (k,k’constants, u2 in exponent)
At the end of that century, Ludwig Boltzmann derived the same equation using statistics and the
gas laws. He found the relation of the constants, k to the gas constant, R.
f(u) = 4·(M/[2·RT)3/2 · u2 · e(-Mu2/2RT) (u2 in exponent)
Chapter 10 Gases 13
Experimental verification came with the invention of the molecular beam apparatus, cf. Box 4.1
in textbook. We will use this equation when we study topics such as evaporation and the rate of
chemical reactions.
When you are confronted with the equations of functions, it is good to pare them down to the
essentials. When you see a “function of u”, f(u), isolate all the u on the right side of the equation.
All the other parameters such as M, T, and R can be set to 1. Change u to x if you want, and f(u)
to y and you can graph the function on a graphing calculator. The above equation reduces to
y = f(x) = x2 · exp(-x2)
The function “e to the minus x squared” or exp(-x2) is called a Gaussian function in Statistics.
(Plot it for your self.) Multiplying the gaussian by x2 starts the function at the origin with y=zero
and “skews” or biases the function.
The distribution curves vary in a definite way when the T is increased:
1. The maximum fraction is at higher speed as T increases
2. For any two T, after both maximum fractions, the fraction at Thigher is higher than the
fraction at Tlower.
The distribution curves vary the same way with reciprocal of the molecular weight, M.
Chapter 10 Gases 14
Assume that you have a sample of hydrogen gas containing H2, HD, and D2 that you want to
separate into pure components. What are the relative rates of diffusion of the three molecules
according to Graham's Law?
Rate of HD diffusion
mass of D 2

Rate of D2 diffusion
mass of HD
Real Gases: Deviations from Ideal Behavior
P
nRT
V
Ideal Gas Law
nRT
n2a
P

V  nb V 2
or
 n 2 a 

P  V 2 
V  nb  nRT


van der Waals equation
a: correction factor for intermolecular forces of attraction
b: correction factor for excluded volume
Chapter 10 Gases 15