Gases
I. Factors that affect gas pressure
1. Temperature
2. Volume
3. Amount of gas
II. Gas Laws
A. Boyle’s Law: Pressure – Volume Relationship
Constant Temperature
P1V1 = P2V2
P1 = initial pressure P2 = new pressure
V1 = initial volume V2 = new volume
A helium-filled balloon contains 125 mL of gas at a
pressure of 0.974 atm. What volume will the gas occupy at
standard pressure?
Write down given:
V1 = 125 mL P1 = 0.974 atm V2 = ? P2 = 1 atm
P1V1 = P2V2
Solve for V2
V2 = P1V1 = (0.974 atm)(125 mL) = 122 mL
P2 (1 atm)
Problem: A gas has a pressure of 1.26 atm and occupies a
volume of 7.40 L. If the gas is compressed to a volume of
2.93 L, what will its pressure be if temperature is held
constant?
B. Charles Law: Volume-Temp. Relationship
Pressure is constant
Temperature must be in Kelvin
Absolute zero = 0 K or -273 ºC
0 ºC = 273 K
Kelvin (K) = 273 + ºC
Charles Law: V1 = V2
T1 T2
V1 = initial volume T1 = initial temperature
V2 = new volume T2 = new temperature
A balloon filled with 02 (g) occupies a volume of 5.5 L at 25 ºC.
What volume will the gas occupy at 100 ºC?
Write down given & convert temps to Kelvin
V1 = 5.5 L T1 = 25 ºC = 298 K
V2 = ? T2 = 100 ºC = 373 K
V1 = V2
T1 T2
Solve for V2 V2 = V1T2 = (5.5 L) ( 373 K) = 6.9 L
T1 (298 K)
Problem: At 0.0 ºC, the volume of a gas is 375 mL. To what
temperature (in ºC) must the gas be heated to occupy a
volume of 500mL?
C. Gay-Lussac’s Law: Pressure-Temperature Relationship
Constant Volume
Temperature must be in Kelvin
P1 = P2
T1 T2
P1 = initial pressure T1 = initial temperature
P2 = new pressure T2 = new temperature
At 120 ºC, the pressure of a sample of nitrogen is 1.07 atm.
What will the temperature (in ºC) be at 1.30 atm, assuming
constant volume?
Write down given & convert temps to Kelvin
T1 = 120 ºC = 393 K P1 = 1.07 atm
T2 = ? P2 = 1.30 atm
P1 = P2
T1 T2
Solve for T2
T2 = P2T1 = (1.30 atm)(393 K) = 477.5 K = 205 ºC
P1 (1.07 atm)
D. Combined Gas Law
P1V1 = P2V2
T1 T2
P1 = initial pressure T1 = initial temperature (K)
V1 = initial volume
P2 = new pressure T2 = initial temperature (K)
V2 = new volume
A gas occupying 75 mL at standard conditions is heated to
17 ºC while the pressure is reduced to 0.97 atm. What is the
new volume occupied by the gas?
Write down given & convert temps to Kelvin
V1= 75 mL P1 = 1 atm T1 = 0 ºC = 273 K
V2 = ? P2 = 0.97 atm T2 = 17 ºC = 290 K
P1V1 = P2V2
T1 T2
Solve for V2
V2 = P1V1T2 = (1 atm)(75 mL)(290 K) = 82 mL
T1P2 (273 K)(0.97 atm)
Problem: What pressure is required to reduce 60.0 mL of a gas
at standard conditions to 10.0 mL at a temperature of 25.0 ºC?
E. Dalton’s Law of Partial Pressures
Total pressure = sum of partial pressures of component gases
PT = P1 + P2 + P3 + ….
What is the pressure of a mixture of He, N2, and O2 if their
partial pressures are 600. mm Hg, 150. mm Hg, and 102 mm
Hg?
1. Gases Collected by Water Displacement (Gases collected
over water)
Pgas = PT – PH20
Pgas = partial pressure of a gas
PT = Total pressure (atmospheric pressure)
PH20= vapor pressure of water
Vapor pressure of water varies with temperature.
Must look it up
A sample of nitrogen gas is collected over water at a
temperature of 23.0oC. What is the pressure of the nitrogen
gas if the atmospheric pressure is 785 mm Hg?
PN2 = ? Patm = 785 mm Hg PH20 = 21.1 mm Hg
PN2 = Patm – PH20
= 785 mm Hg – 21.1 mm Hg = 764 mm Hg
III. Kinetic Molecular Theory and Gases
A. 5 assumptions
Ideal Gas (does not really exist)
1. Consists of large number of tiny particles that are far
apart
2. Collisions between gas particles & between gas particles
and container walls are elastic
3. Gas particles are in constant random motion
4. No forces of attraction/repulsion between gas particles
5. Average kinetic energy depends on temperature
B. Deviations of real gases from ideal behavior
London Dispersion Forces (aka Van der Waals Forces) –
gas particles exert attractive forces on each other
Real gases behave near ideal if pressure is low and
temperature is high
Noble gases are close to ideal
Nonpolar, diatomic molecules are the closest to ideal
under low pressure & high temp.
The more polar a molecule, the more attractive forces
act upon it and the more deviation from ideal gas
behavior
IV. Volume – Mass Relationship of Gases
A. Measuring & Comparing the volumes of gases
Gay- Lussac’s law of combining volumes of gases
2H2 + O2 2H2O
Mol Ratio 2 mol H2 : 1 mol O2 : 2 mol H2O
2 molecules H2 : 1 molecules O2 : 2 molecules H2O
2 L H2 : 1 L O2 : 2 L H2O
B. Avogadro’s Law
Equal volumes of gases at the same temperature and
pressure contain equal number of molecules
C. Molar Volume of gases
Standard molar volume of a gas (volume occupied by 1 mol
of a gas at STP) = 22.4 L
22.4 L /mol for any gas at STP
You need 0.0580 mol NO for an experiment. What volume
of NO gas would you need at STP?
0.0580 mol NO 22.4 L NO = 1.30 L NO
1 mol NO
V. Ideal Gas Law
Relationship of pressure, volume, temp, and number of
mols of a gas
PV = nRT
P = pressure n = number of mols
V = volume R = ideal gas constant
T = temperature in Kelvin
Values of R (look at pressure units to determine which
one to use)
0.0821 L atm
mol K
62.4 L mm Hg 8.314 L kPa
mol K mol K
A 2.07 L cylinder contains 2.88 mol of helium gas at 22oC.
What is the pressure in atmospheres of the gas in the cylinder?
Write down given and change temp to Kelvin
Choose appropriate value for R:
V = 2.07 L n = 2.88 mol
o
T = 22 C = 295 K R = 0.0821 L atm
P=? mol K
PV = nRT Solve for P
L atm
P = nRT = (2.88 mol)(0.0821 mol K )(295 K) = 33.7 atm
V (2.07 L)
Problem: A reaction yields 0.00856 mol of O2 gas. What
volume in mL will the gas occupy if it is collected at 43 oC and
0.926 atm pressure?
B. Other ways to express the Ideal gas law
Mass
m =mass M=Molar mass n=m/M
Ideal gas equation becomes: PV = mRT
M
Density
D = density D=m/V
Ideal gas equation becomes: D = MP
RT
VI. Stoichiometry of Gases
A. Volume – Volume Calculations
Use volume ratio – can only be used to convert between
gases under the same conditions
If 708 L of NO2 gas react with water, what volume of NO
gas will be produced? Assume the gases are measured
under the same conditions.
3NO2 (g) + H2O (l) 2HNO3 (l) + NO (g)
708 L NO2 1 L NO = 236 L NO
3 L NO2
B. Volume – Mass and Mass – Volume Calculations
Use mol ratio and ideal gas law
Problem 1: What mass (in grams) of Aluminum would be
needed to produce 4.00 L of hydrogen gas at STP?
2NaOH (aq) + 2Al (s) + 6H2O (l) 2NaAl(OH)4 (aq) + 3H2 (g)
?g 4.00 L
1. Balance Equation
2. Find mass of Al --- at STP all gases have 22.4 L/mol
4.00 L H2 1mol H2 2 mol Al 26.98 g Al = 3.21 g Al
22.4 L H2 3 mol H2 1 mol Al
Problem 2: What volume (in Liters) of N2 gas, measured at
1.30 atm and 87 oC, would be produced by the reaction of
70.0 g of NaN3?
2NaN3 (s) 3 N2 (g) + 2 Na (s)
70.0 g ?L
1. Balance equation
2. Need mols of N2
70.0 g NaN3 1 mol NaN3 3 mol N2__ = 1.61 mol N2
65.02 g NaN3 2 mol NaN3
3. Find volume of N2
PV = nRT
P = 1.30 atm n = 1.61 mol R = 0.0821 L atm
o
T = 87 C = 360 K mol K
L atm
V = nRT = (1.61 mol)(.0821 mol K )(360K) = 36.6 L N2
P (1.30 atm)
Problem: What volume of chlorine gas at 38oC and 1.63 atm is
needed to react with 10.4 g of sodium?
Na + Cl2 NaCl
VII. Effusion and Diffusion
Graham’s law of effusion: rate of effusion is inversely
proportional to the molar mass
v1 = mw2
v2 mw1
mw = molecular weight = Molar Mass
Problem 1: An unknown gas passed through a tiny hole at
a speed of 75 m/s. Oxygen gas passed through the hole at
a speed of 30. m/s. What is the molar mass of the
unknown gas?
v1 = mw2 comp’d #1 = O2
v2 mw1 comp’d #2 = unknown gas
v1 = 30 s mw1 = 32.00 g/mol
v2 = 75 s mw2 = ?
mw2 = v1 (mw1) = 30 s (32.00 g/mol)
v2 75 s
mw2 = 2.2627 Square both sides
mw2 = 5.1 g/mol
Problem 2: Estimate the molar mass of a gas that effuses
at 1.6 times the effusion rate of CO2.
v1 = mw2 comp’d #1 = CO2
v2 mw1 comp’d #2 = unknown gas
v1 = rate of CO2 mw1 = 44.01 g/mol
v2 = 1.6(rate of CO2) mw2 = ?
rate of CO2___= mw2_____
1.6(rate of CO2) 44.01 g/mol
“Rate of CO2“ cancels out:
rate of CO2___= mw2____
1.6(rate of CO2) 44.01 g/mol
1_= mw2_____
1.6 44.01 g/mol
1 = mw2____
1.6 6.63
6.63( 1 ) = mw2
1.6
4.14 = mw2
Square both sides of the equation
mw2 = 17.17 g/mol