### Chapter 5 Gases

```Chapter 5
Gases
A Gas• Uniformly fills any container
• Mixes completely with any other gas
• Can easily be compressed
• Exerts pressure on its surroundings
The properties of a gas depends upon four variables• Pressure (P)
– Is equal to force/unit area
– Measured by a barometer
– SI unit = Newtons/meter2 = 1 Pascal (Pa)
– 1 standard atmosphere = 101,325 Pa
– 1 atm = 760 mm Hg = 760 torr
• Volume (V) of the gas = volume of the container
• Temperature (T) – measured in Kelvin
• Number of moles (n)
A Torricellian Barometer
A Simple Manometer
The Gas Laws
• Boyle’s Law: At constant temperature, the
volume of a gas is inversely proportional to the
pressure
V  1/P  V = k x 1/P  PV = k (where k is
constant)
P1V1 = P2V2 = constant (k) = PV (1 and 2
represents two states)
Pressure x Volume = constant (T = constant)
(Holds precisely only at very low pressures)
A J-Tube Similar to the One Used by Boyle
Plotting Data for Boyle’s Law
Example: A sample of CO2 gas is placed in a 125
mL flask where it has a pressure of 67.5 mm Hg.
What pressure will it has when it is transferred to a
500 mL flask at the same temperature?
P1 = 67.5 mm Hg
P2 = ?
V1 = 125 mL
V2 = 500 mL
P1V 1 = P2V2
P2 = (P1V 1 )/V2
= (67.5 mm Hg x 125 mL)/500 mL
= 16.9 mm Hg
Charle’s Law: At constant pressure (P), the volume
of a gas is directly proportional to temperature of the
gas and extrapolates to zero at zero Kelvin.
V  T  V = bT
Where,
b = proportionality constant
T = Kelvin temperature,
V/T = b  V1/T1 = V2/T2 = b
Plots of V Versus T (ºC) for Several Gases
Example: A sample of gas at 15oC and 1 atm has
a volume of 2.58 L. What volume will this gas
occupy at 38oC and 1 atm?
T1 = 15oC + 273 = 288 K,
V1 = 2.58 L ,
T2 = 38oC + 273 = 311 K,
V2 =?
V1/T1 = V2/T2
Solving for V2 = V1 T2 / T1
= (2.58 L x 311 K)/288 K
= 2.79 L
• Avogadro’s Law: Equal volume of gases at the
same temperature and pressure contain the same
number of particles. At constant temperature and
pressure, the volume is directly proportional to the
number of moles of gas (at low pressures).
V = an
Where, V = the volume of the gas, n = the number
of moles of gas particle, and a = proportionality
constant
V1/n1 = V2/n2 = a = constant
At constant temperature and pressure, the volume
of a gas is directly proportional to the number of
moles of gas.
Balloons Holding 1.0 L of Gas at 25º C and 1 atm
Example: 12.2 L sample containing 0.50 mol
oxygen gas at a pressure of 1 atm and a temperature
of 25oC. If all this O2 were converted to ozone at the
same temp. and pressure, what would be the volume
of ozone?
3O2  2O3(g)
0.50 mol O2 x (2 mol O3 )/(3 mol O2 ) = 0.33 mol O3
V1= 12.2 L,
V2 = ?
n1 = 0.50 mol, n2 = 0.33 mol
V1/n1 = V2/n2
Solving for V2 = V1n2 / n1
= (12.2 L x 0.33 mol)/(0.50 mol) = 8.1 L
The Ideal Gas Law: (An equation of state for a gas)
Boyle’s law: V = k/P (at constant T and n)
Charles law: V = bT (at constant P and n)
Avogadro’s law: V = an (at constant T and P)
The volume of gas depends on pressure, temperature,
and number of moles of gas present, can be combined
as follows:
V = R(Tn/P); where, R = combined proportionality
constant known as universal gas constant.
PV = nRT (Holds closely at P < 1 atm)
P = Pressure in atm
V = Volume in liters
n = number of moles T = Temperature in Kelvins
R = Universal gas constant = 0.08206 L.atm K-1 mol-1
Example: Nitrogen gas in an air bag at 850 mm
Hg and 25oC has a volume of 3.5 x 104 mL. How
many moles of gas are in there?
P = 850 mm Hg x 1 atm / 760 mm Hg = 1.118 atm
T = 25oC + 273 = 298 K
V = 3.5 x 104 mL x 1L / 1000 mL = 35 L
PV = nRT
n = PV/RT
=(1.118 atm x 35 L)/(0.0821 L.Atm K-1mol-1 x
298K) = 1.6 moles N2
Example: A gas containing 0.35 mol argon at a temperature
of 13oC and a pressure of 568 torr is heated to 56oC and
pressure of 897 torr. Calculate the change in volume that
occurs.
P1 = 568 torr x 1 atm/760 torr P2 = 897 torr x 1 atm/760 torr
= 0.747 atm
= 1.18 atm
V1 = ?
V2 = ?
T1 = 13oC + 273 = 286 K
T2 = 56oC + 273 = 329 K
n1 = 0.35 mol
n2 = 0.35 mol
P1V1 = n1RT1
P1V1 = n1RT1
V1 = n1RT1 /P1 = (0.35 mol x 0.08206 L.atm.mol-1.K-1 x
286K) / (0.747 atm) = 11 L
V2 = n2RT2 /P2 = (0.35 mol x 0.08206 L.atm.mol-1.K-1 x
329K) / (1.18 atm) = 8.0 L
V = V2 - V1 = 8.00 L – 11.0 L = -3 L (Volume decreases)
Gas Stoichiometry
The conditions 0oC and 1 atm called Standard
Temperature and Pressure (STP)
Suppose we have 1 mol of an ideal gas at 0oC
and 1 atm, the volume of the gas is given by
ideal gas law,
V = nRT / P
= (1 mol x 0.08206 L.atm.mol-1.K-1 x
273K) / (1 atm) = 22.42 L
 Molar Volume of an ideal gas at STP
A Mole of Any Gas Occupies a Volume
of Approximately 22.4 L at STP
Example: A sample of N2 gas has a volume of
1.75 L at STP. How many moles of N2 are
present?
1.75 L N2 x (1 mol N2)/22.42 L N2 = 7.81 x 10-2
mol N2
OR
PV = nRT
n = PV/RT = (1 atm x 1.75 L)/(0.08206
L.atm.mol-1.K-1 x 273 K)
= 7.81 x 10-2 mol N2
Molar Mass
By using ideal gas law we can calculate the molar
mass (molecular weight) of a gas from its measured
density.
PV = nRT
n = grams of gas/molar mass = m/molar mass
PV = (m/molar mass)RT  P = mRT/V(molar
mass)
m/V is the gas density “d” in units of grams per liter,
Thus, P = dRT/molar mass
molar mass = dRT/P
If density of a gas at a given temperature and
pressure is known, its molar mass can be calculated.
Example: The density of a gas was measured at
1.50 atm and 27oC and found to be 1.95 g/L.
Calculate the molar mass of the gas.
P = 1.50 atm
T = 27oC + 273 = 300K
d = 1.95 g/L
Molar mass = dRT/P
1.95
=
g
L
 0.08206
= 32.0 g/mol
L . atm
K . m ol
1.50 atm
 300 K
Dalton’s Law of Partial Pressures
• For a mixture of gases in a container, the total pressure
exerted is the sum of the pressure that each gas would
exert if it were alone.
Ptotal = P1+ P2 + P3 + --- (P1, P2 , P3 --- represent partial P)
Partial pressure of each gas can be calculated from ideal
gas law,
P1 = n1RT/V, P2 = n2RT/V, P3 = n3RT/V
Ptotal = P1+ P2 + P3 + ---- = n1RT/V + n2RT/V + n3RT/V
= (n1+n2+n3+---)RT/V = ntotal (RT/V)
• For a mixture of ideal gas, the total number of moles of
particles is important, not the identity or composition of
the involved gas.
Partial Pressure of Each Gas in a Mixture
Partial Pressure
Example: 46 L He at 25oC and 1.0 atm and 12 L O2 at
25oC and 1.0 atm were pumped into a tank with a
volume of 5.0 L. Calculate the partial pressure of each
gas and the total pressure in the tank at 25oC.
PV = nRT
n = PV/RT
1.0 atm x 46 L
nHe 
 1.9 m ol
0.08206 L.atm.mol-1.K-1 x (25+273) K
1.0 atm x 12 L
 0.49 m ol
no 
-1
-1
0.08206 L.atm.mol .K x 298K
2
…Partial Pressure (example) continued…
Again, PV = nRT  P = nRT /V
P He
1.9 mol x 0.08206 L.atm.mol-1.K-1 x 298K
 9 .3atm

5.0 L
PO 2
0.49 mol x 0.08206 L.atm.mol-1.K-1 x 298K
 2 .4 atm

5.0 L
Ptotal = PHe + PO2 = (9.3 + 2.4) atm = 11.7 atm
Mole Fraction
• The ratio of the number of moles of a given component
in a mixture to the total number of moles in the mixture.
Mole fraction = 1 
n1
n Total

n1
n 1  n 2  n 3  ....
n = P(V/RT) [# of moles proportion to pressure]
 V 
n 1  P 1

 RT 
 V 
n 2  P 2

 RT 
…Mole Fraction continued…
1  n 1
n T otal



1 
2 
n1
n Total
n2
n Total


P1
P Total
P2
P Total
P 1(V / R T )
P 1(V / R T )  P 2 (V / R T )  P 3 (V / R T )  ...
(V / R T ) P 1
(V / R T )( P 1  P 2  P 3  ... )
P1
P 1  P 2  P 3  ...
 P 1  1 x PTotal
 P 2  2 x PTotal

P1
P T otal
Example: The partial pressure of oxygen was
observed to be 156 torr in air with a total
atomospheric pressure of 743 torr. Calculate
the mole fraction of O2 present.
Mole fraction of O2,
O2 = PO2/PTotal = 156 torr/743 torr = 0.210
Example: The mole fraction of N2 in the air
is 0.7808. Calculate the partial pressure of
N2 in air when the atomospheric pressure is
760 torr.
The partial pressure of N2,
PN2 = N2 x PTotal = 0.7808 x 760 torr
= 593 torr
The Kinetic Molecular Theory of Gases
• Postulates for an Ideal Gas:
1. The particles are so small that the volume of the
particles can be assumed to be negligible (zero).
2. The particles are in constant motion. The collision of
the particles with the walls of the container are the
cause of the pressure exerted by the gas.
3. The particles are assumed to exert no forces on each
other, they are assumed neither to attract nor to repel
each other.
4. The average kinetic energy of a collection of gas
particles is assumed to be directly proportional to the
Kelvin temperature of the gas.
The Effects of Decreasing the Volume of
a Sample of Gas at Constant Temperature
The Effects of Increasing the Temperature
of a Sample of Gas at Constant Volume
The Effects of Increasing the Temperature
of a Sample of Gas at Constant Pressure
The Effects of Increasing the Number of Moles of
Gas Particles at Constant Temperature and Pressure
Effusion And Diffusion
• Effusion: Flow of gas particles through tiny pores
or pinholes due to pressure differences. Describes
the passage of gas into an evacuated chamber.
• Diffusion: Movement of gas particles through
space, from a region of high concentration to one
of low concentration. Describes the mixing of
gases. The rate of diffusion is the rate of gas
mixing.
Graham’s law of effusion: The rate of effusion of
a gas is inversely proportional to the square root of
the mass of its particle. Relative rates of effusion
of two gases at the same temperature and pressure
are given by the inverse ratio of the square roots of
the masses of the gas particle:
M2
Rate of effusion for gas 1

Rate of effusion for gas 2
M1
M1 and M2 represents the molar masses of the
gases.
Diffusion:
Distance traveled by gas 1
M2

Distance traveled by gas 2
M1
The Effusion of a Gas into
an Evacuated Chamber
HCI(g) and NH3(g) Meet in a Tube
Real Gases
• An ideal gas is a hypothetical concept. No gas exactly
follows the ideal gas law. Real gas typically exhibit
behavior that is closest to ideal behavior at low pressure
and high temperature. Must correct ideal gas behavior
when at high pressure (smaller volume) and low
temperature (attractive forces become important)
P obs
 n 2

 a 
V 
V  nb
nRT
Rearrangement gives vander Waals equation:

 n
 P obs  a     V  nb   nRT
V 

2
Plots of PV/nRT Versus P for Several Gases (200 K)
Plots of PV/nRT Versus P for Nitrogen Gas at Three Temperatures
Chemistry in the Atmosphere
• The principal components are N2 and O2. Other
important gases are H2O, CO2, Ar, Ne, He, CH4,
Kr, H2, NO, Xe etc.
• Because of gravitational effects, the composition
of the earth’s atmosphere is not constant.
• Heavier molecules tend to be near the earth’s
surface and light molecules tend to migrate to
higher altitudes.
• Troposphere is the lowest layer of the atmosphere.
…Chemistry in the Atmosphere continued…
• Ozone in the upper atmosphere helps prevent
high energy ultraviolet radiation from penetrating
to the earth.
• Two main sources of pollution are–
transportation and the production of electricity.
• The combustion of petroleum produces CO, CO2,
NO, and NO2, along with unburned molecules
from petroleum.
Reactions Occurring in the Atmosphere
NO2(g)
energy
NO(g) + O(g)
O(g) + O2(g)
Light
O3
O* + H2O
OH + NO2
S(in coal) + O2(g)
2SO2(g ) + O2(g)
SO3(g ) + H2O(l)
O3(g)
O2* + O*
2OH
HNO3
SO2(g)
2SO3(g )
H2SO4(aq) [corrosive]
acid rain
Summary
• Gas properties depends on four variables –
pressure, volume, temperature, and amount of
gas.
• Boyle’s law: PV = constant
• Charle’s law: V/T = constant
• Avogadro’s law: V/n = constant
• Ideal gas law: PV = nRT
• Molar mass = dRT/P
• Dalton’s law of partial pressures:
Ptotal = P1+ P2 + P3 + ---
…Summary continued…
• Mole fraction: 1 = n1/ntotal = P1/Ptotal
• Kinetic molecular theory of gasses: Postulates
• Grahams law of effusion:
M2
Rate of effusion for gas 1

Rate of effusion for gas 2
M1
• Chemistry in atmosphere
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