Chapter 11 - Gases
Properties of Gases
Gas Pressure
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 2 of 77
Gases We Breathe
The air we breathe • Is a gas mixture.• Contains mostly
N2 and O2 and small amounts of other gases.
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 3 of 77
Kinetic Theory of Gases
Gases consist of _____ particles that
• Move rapidly in ___________ lines.
• Have essentially no attractive (or repulsive) ___________.
• Are very ______ ____________.
• Have very small volumes compared to the volume of the ____________ they occupy.
• Have kinetic energies that increase with an __________ in temperature.
Chapter 11 Slide 4 of 77
Properties of Gases• Gases are described in terms of four properties:
pressure (P), volume (V), temperature (T), and amount (n, number of moles).
Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings
Chapter 11 Slide 5 of 77
Units of PressureGas pressure• Is described as a force acting on a specific area. • Pressure (P) = Force Area
Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings
Chapter 11 Slide 6 of 77
Atmospheric Pressure
The atmospheric pressure• Is the pressure exerted
by a column of air from the top of the atmosphere to the surface of the Earth.
• Is about 1 atmosphere or a little less at sea level.
Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings
Chapter 11 Slide 7 of 77
Altitude and Atmospheric Pressure
Atmospheric pressure• Depends on the
altitude and the weather.
• Is lower at high altitudes where the density of air is less.
• Is higher on a rainy day than on a sunny day.
Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings
Chapter 11 Slide 8 of 77
What is 475 mm Hg expressed in atm?
475 mm Hg x 1 atm = 0.625 atm 760 mm Hg
The pressure of a tire is measured as 2.00 atm. What is this pressure in mm Hg?
2.00 atm x 760 mm Hg = 1520 mm Hg 1 atm
Learning Check
Chapter 11 Slide 9 of 77
Barometer
A barometer• Measures the
pressure exerted by the gases in the atmosphere.
• Indicates atmospheric pressure as the height in mm of the mercury column.
Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings
Chapter 11 Slide 10 of 77
Chapter 11 Gases
Pressure and Volume Boyle’s Law
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 11 of 77
Boyle’s Law
Boyle’s Law states that
• The pressure of a gas is ___________ related to its volume when T and n are constant.
• If volume decreases, the pressure ______________.
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 12 of 77
In Boyle’s Law• The product P x V is constant as long as T and
n do not change.
P1V1 = 8.0 atm x 2.0 L = 16 atm L
P2V2 = 4.0 atm x 4.0 L = 16 atm L
P3V3 = 2.0 atm x 8.0 L = 16 atm L
• Boyle’s Law can be stated as
P1V1 = P2V2 (T, n constant)
PV Constant in Boyle’s Law
Chapter 11 Slide 13 of 77
Solving for a Gas Law Factor
The equation for Boyle’s Law can berearranged to solve for any factor.
P1V1 = P2V2 Boyle’s Law
To obtain V2 , divide both sides by P2.
P1V1 = P2V2
P2 P2
P1V1 = V2
P2
Chapter 11 Slide 14 of 77
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 15 of 77
Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of a 8.0 L sample of Freon gas initially at 550 mm Hg after its pressure is changed to 2200 mm Hg at constant T?
1. Set up a data tableConditions 1 Conditions 2 P1 = 550 mm Hg P2 = 2200 mm Hg
V1 = 8.0 L V2 =
(predict smaller
V2)
Calculation with Boyle’s Law
?
Chapter 11 Slide 16 of 77
2. Because pressure increases, we predict that the volume will decrease.
Solve Boyle’s Law for V2:
P1V1 = P2V2
V2 = V1P1
P2
V2 = 8.0 L x 550 mm Hg = 2.0 L
2200 mm Hg pressure ratio
decreases volume
Calculation with Boyle’s Law (Continued)
Chapter 11 Slide 17 of 77
Learning Check
For a cylinder containing helium indicate if cylinder A or cylinder B represents the new volume for the following changes (n and T are constant):
1) Pressure decreases
2) Pressure increases
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 18 of 77
Learning Check
If the helium in a cylinder has a volume of 120mL and a pressure of 850 mm Hg, what is the newvolume if the pressure is changed to 425 mm Hginside the cylinder? A) 60 mL B) 120 mL C) 240 mL
P1 = 850 mm Hg P2 = 425 mm Hg
V1 = 120 mL V2 = ??
V2 = P1V1 = 120 mL x 850 mm Hg = 240 mL P2 425 mm Hg
Chapter 11 Slide 19 of 77
Learning Check A sample of helium gas in a balloon has a volume
of 6.4 L at a pressure of 0.70 atm. At 1.40 atm (T constant), is the new volume represented by A, B, or C?
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 20 of 77
If the sample of helium gas has a volume of 6.4 L at a pressure of 0.70 atm, what is the new volume when the pressure is increased to 1.40 atm (T constant)?
A) 3.2 L B) 6.4 L C) 12.8 L
V2 = V1P1
P2
V2 = 6.4 L x 0.70 atm = 3.2 L
1.40 atmVolume decreases when there is an increase in the pressure (Temperature is constant.)
Learning Check
Chapter 11 Slide 21 of 77
A sample of oxygen gas has a volume of 12.0 L at 600. mm Hg. What is the new pressure when the volume changes to 36.0 L? (T and n constant).
1) 200. mm Hg
2) 400. mm Hg
3) 1200 mm Hg
Learning Check Practice At Home
Chapter 11 Slide 22 of 77
Chapter 11 Gases
Temperature and Volume (Charles’ Law)
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 23 of 77
Charles’ Law
In Charles’ Law• The Kelvin
temperature of a gas is directly related to the volume.
• P and n are constant.• When the
temperature of a gas increases, its volume increases.
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 24 of 77
• For two conditions, Charles’ Law is writtenV1 = V2
(P and n constant)
T1 T2
• Rearranging Charles’ Law to solve for V2
T2 x V1 = V2 x T2
T1 T2
V2 = V1T2
T1
Charles’ Law: V and T
Chapter 11 Slide 25 of 77
Learning CheckSolve Charles’ Law expression for T2.
V1 = V2
T1 T2
Cross multiply to give V1T2 = V2T1
Isolate T2 by dividing through by V1V1T2 = V2T1
V1 V1
T2 = V2T1
V1
T2
Chapter 11 Slide 26 of 77
A balloon has a volume of 785 mL at 21°C. If the
temperature drops to 0°C, what is the new volume of
the balloon (P constant)?
1. Set up data table:Conditions 1 Conditions 2V1 = 785 mL V2 = ? (decrease)
T1 = 21°C = 294 K T2 = 0°C = 273 K
Be sure to use the Kelvin (K) temperature ingas calculations.
Calculations Using Charles’ Law
Chapter 11 Slide 27 of 77
Calculations Using Charles’ Law (continued)
2. Solve Charles’ law for V2
V1 = V2
T1 T2
V2 = V1 T2
T1
V2 = 785 mL x 273 K = 729 mL 294 K
When temperature decreased, the volume decreased as predicted.
Chapter 11 Slide 28 of 77
A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. At what temperature (in °C) will the volume of the oxygen be 640 mL (P and n constant)?
A) 443°C B) 170°C C) - 82°C
T2 = T1V2
V1
T2 = 291 K x 640 mL = 443 K
420 mL
= 443 K - 273 K = 170°C
Learning Check
18o C + 273 = 291 K
Chapter 11 Slide 29 of 77
Chapter 11 Gases
Temperature and Pressure (Gay-Lussac’s Law)
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 30 of 77
Gay-Lussac’s Law: P and T
In Gay-Lussac’s Law• The pressure exerted by
a gas is directly related to the Kelvin temperature.
• V and n are constant. P1 = P2
T1 T2
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 31 of 77
A gas has a pressure at 2.0 atm at 18°C. What
is the new pressure when the temperature is
62°C? (V and n constant)
1. Set up a data table.
Conditions 1 Conditions 2
P1 = 2.0 atm P2 = (increase)
T1 = 18°C + 273 T2 = 62°C + 273
= 291 K = 335 K
Calculation with Gay-Lussac’s Law
?
Chapter 11 Slide 32 of 77
Calculation with Gay-Lussac’s Law (continued)
2. Solve Gay-Lussac’s Law for P2
P1 = P2
T1 T2
P2 = P1 T2
T1
P2 = 2.0 atm x 335 K = 2.3 atm
291 K
Chapter 11 Slide 33 of 77
Chapter 11 Gases
The Combined Gas Law
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 34 of 77
• The combined gas law uses Boyle’s Law, Charles’ Law, and Gay-Lussac’s Law (n is constant).
P1 V1 = P2 V2
T1 T2
Combined Gas Law
Chapter 11 Slide 35 of 77
A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm (n constant)?
1. Set up Data Table
Conditions 1 Conditions 2
P1 = 0.800 atm P2 = 3.20 atm
V1 = 0.180 L (180 mL) V2 = 90.0 mL
T1 = 29°C + 273 = 302 K T2 = ??
Combined Gas Law Calculation
Chapter 11 Slide 36 of 77
2. Solve for T2 P1 V1 = P2 V2
T1 T2
T2 = T1 P2V2
P1V1
T2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180.0 mL
T2 = 604 K - 273 = 331 °C
Combined Gas Law Calculation (continued)
Chapter 11 Slide 37 of 77
A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the volume(mL) of the gas at -95°C and a pressure of 802 mm Hg (n constant)?
Learning Check
Chapter 11 Slide 38 of 77
1. Data TableConditions 1 Conditions 2
T1 = 308 K T2 = -95°C + 273 = 178K
V1 = 675 mL V2 = ???
P1 = 646 mm Hg P2 = 802 mm Hg
2. Solve for V2
V2 = V1 P1 T2
P2T1
V2 = 675 mL x 646 mm Hg x 178K = 314 mL 802 mm Hg x 308K
A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the volume (mL) of the gas at -95°C and a pressure of 802 mm Hg (n constant)?
P1 V1 = P2 V2
T1 T2
Chapter 11 Slide 39 of 77
Chapter 11 Gases
Volume and Moles (Avogadro’s Law)
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 40 of 77
Avogadro's Law: Volume and Moles
In Avogadro’s Law
• The volume of a gas is directly related to the number of _____ of gas.
• T and P are constant.
V1 = V2
n1 n2
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 41 of 77
Learning Check
If 0.75 mol of helium gas occupies a volume of 1.5 L, what volume will 1.2 mol of helium occupy at the same temperature and pressure?
1) 0.94 L
2) 1.8 L
3) 2.4 L Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 42 of 77
If 0.75 mol of helium gas occupies a volume of 1.5 L, what volume will 1.2 mol of helium occupy at the same temperature and pressure? 1) 0.94 L 2) 1.8 L 3) 2.4 L
Conditions 1 Conditions 2V1 = 1.5 L V2 = ???n1 = 0.75 mol He n2 = 1.2 mol HeV2 = V1n2
n1
V2 = 1.5 L x 1.2 mol He = 2.4 L 0.75 mol He
V1 = V2 n1 n2
Chapter 11 Slide 43 of 77
• The volumes of gases can be compared when they have the same conditions of temperature and pressure (STP, Standard Temperature and Pressure).
Standard temperature (T)
0°C or 273 K
Standard pressure (P)
1 atm (760 mm Hg)
STP
Chapter 11 Slide 44 of 77
Molar VolumeAt standard temperature and pressure (STP).• 1 mol of a gas occupies a volume of 22.4 L,
which is called its molar volume.
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 45 of 77
The molar volume • At STP • Can be used to form conversion factors.
22.4 L and 1 mol
1 mol 22.4 L
Molar Volume as a Conversion Factor
Chapter 11 Slide 46 of 77
Guide to Using Molar Volume
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 47 of 77
1. What is the volume at STP of 4.00 g CH4?
A) 5.59 L B) 11.2 L C) 44.8 L
4.00 g CH4 x 1 mol CH4 x 22.4 L (STP) = 16.04 g CH4 1 mol CH4
Learning Check
5.59 L
Chapter 11 Slide 48 of 77
2. How many grams of He are present in 8.00 L at STP?
A) 25.6 g B) 0.357 g C) 1.43 g
8.00 L x 1 mol He x 4.003 g He = 22.4 L 1 mol He
Learning Check
1.43 g He
Chapter 11 Slide 49 of 77
AT STP, the density of gas can be calculated using the mass of the gas and its volume.
Density = Molar mass Molar volume
Density of a Gas at STP
Chapter 11 Slide 50 of 77
Density of a Gas
Calculate the density in g/L of O2 gas at STP. At STP, we know that 1 mol O2 (32.00 g) occupies a volume of 22.4 L.
Density O2 at STP = 32.00 g O2 = 1.43 g/L
22.4 L (STP)
Chapter 11 Slide 51 of 77
Chapter 11 Gases
The Ideal Gas Law
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 52 of 77
The ideal gas law• Provides a relationship between the four
properties (P, V, n, and T) of gases that can be written equal to a constant R.
PV = RnT
• Rearranges these properties to give the ideal gas law expression.
PV = nRT
Ideal Gas Law
Chapter 11 Slide 53 of 77
The universal gas constant, R• Can be calculated using the molar volume of a
gas at STP. • Calculated at STP uses 273 K,1.00 atm, 1 mol
of a gas, and a molar volume of 22.4 L.
P V
R = PV = (1.00 atm)(22.4 L)
nT (1 mol) (273K) n T = 0.0821 L • atm
mol • K
Universal Gas Constant, R
Chapter 11 Slide 54 of 77
Another value for the universal gas constant is obtained using mm Hg for the STP pressure.
The units of R are dependent on the units of the parameters that are used to calculate R.
What is the value of R when a pressure of 760 mm Hg is placed in the R value expression?
Learning Check
Chapter 11 Slide 55 of 77
What is the value of R when a pressure of 760 mm Hg is placed in the R value expression?
R = PV = (760 mm Hg) (22.4 L)
nT (1 mol) (273 K)
= 62.4 L • mm Hg mol • K
Solution
Chapter 11 Slide 56 of 77
Dinitrogen oxide (N2O), laughing gas, is used by dentists as an anesthetic. If a 20.0 L tank of laughing gas contains 2.86 mol N2O at 23°C, what is the pressure (mm Hg) in the tank?
Learning Check
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 57 of 77
1. Adjust the units of the given properties to match the units of R.
V = 20.0 L 20.0 L
T = 23°C + 273 296 K
n = 2.86 mol 2.86 mol
P = ? ?
Solution
Chapter 11 Slide 58 of 77
2. Rearrange the ideal gas law for P.P = nRT V
3. Substitute quantities and solve.
P = (2.86 mol)(62.4 L • mm Hg)(296 K) (20.0 L) (mol • K)
= 2.64 x 103 mm Hg
Solution
Chapter 11 Slide 59 of 77
Learning Check
A cylinder contains 5.0 L O2 at 20.0°C and 0.85 atm. How many grams of oxygen are in the cylinder?
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 60 of 77
1. Determine the given properties.P = 0.85 atm, V = 5.0 L, T = 293 K, n ( or g =?)
2. Rearrange the ideal gas law for n (moles).n = PV RT = (0.85 atm)(5.0 L)(mol • K) = 0.18 mol O2
(0.0821atm • L)(293 K) 3. Convert moles to grams using molar mass.
= 0. 18 mol O2 x 32.0 g O2 = 5.8 g O2
1 mol O2
Solution
Chapter 11 Slide 61 of 77
What is the molar mass of a gas if 0.250 g of the gasoccupy 215 mL at 0.813 atm and 30.0°C?
1. Solve for the moles (n) of gas.
n = PV = (0.813 atm) (0.215 L) mol•K = 0.00703 mol
RT (0.0821 L • atm)(303K)
2. Set up the molar mass relationship.
Molar mass = g = 0.250 g mol 0.00703 mol
= 35.6 g/mol
Molar Mass of a Gas
Chapter 11 Slide 62 of 77
Chapter 11 Gases
Gas Laws and Chemical Reactions
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 63 of 77
Gases in Equations
The volume or amount of a gas in a chemical
reaction can be calculated from
• STP conditions or the ideal gas law.
• Mole factors from the balanced equation.
Chapter 11 Slide 64 of 77
STP and Gas Equations
What volume (L) of O2 gas is needed to completelyreact with 15.0 g of aluminum at STP?
4Al(s) + 3O2 (g) 2Al2O3(s)
Plan: g Al mol Al mol O2 L O2 (STP)
15.0 g Al x 1 mol Al x 3 mol O2 x 22.4 L (STP)
26.98 g Al 4 mol Al 1 mol O2
= 9.34 L O2 at STP
Chapter 11 Slide 65 of 77
Ideal Gas Equation and Reactions
What volume (L) of Cl2 gas at 1.2 atm and 27°C is needed to completely react with 1.5 g aluminum?
2Al(s) + 3Cl2(g) 2AlCl3(s)
Plan?
g Al moles Al moles Cl2 L Cl2
Chapter 11 Slide 66 of 77
Ideal Gas Equation and Reactions
2Al(s) + 3Cl2 (g) 2AlCl3(s) 1.5 g ? L 1.2 atm, 300K
1. Calculate the moles of Cl2 needed.
1.5 g Al x 1 mol Al x 3 mol Cl2 = 0.083 mol Cl2 26.98 g Al 2 mol Al
2. Place the moles Cl2 in the ideal gas equation.
V = nRT =(0.083 mol Cl2)(0.0821 L• atm)(300 K) P 1.2 atm mol K
= 1.7 L Cl2
Chapter 11 Slide 67 of 77
What volume (L) of O2 at 24°C and 0.950 atm is needed to react with 28.0 g NH3?
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
Learning Check
Chapter 11 Slide 68 of 77
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
1. Calculate the moles of O2 needed.
28.0 g NH3 x 1 mol NH3 x 5 mol O2
17.03 g NH3 4 mol NH3
= 2.06 mol O2
2. Place the moles of O2 in the ideal gas equation.
V = nRT =(2.06 mol)(0.0821 L• atm/mol K)(297 K)
P 0.950 atm
= 52.9 L O2
Solution
Chapter 11 Slide 69 of 77
Chapter 11 Gases
Partial Pressure (Dalton’s Law)
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 70 of 77
The partial pressure of a gas• Is the pressure of each gas in a mixture.• Is the pressure that gas would exert if it were
by itself in the container.
Partial Pressure
Chapter 11 Slide 71 of 77
Dalton’s Law of Partial Pressures statesthat the total pressure• Depends on the total number of gas
particles, not on the types of particles.• Exerted by a gas mixture is the sum of
the partial pressures of those gases.•
PT = P1 + P2 + .....
Dalton’s Law of Partial Pressures
Chapter 11 Slide 72 of 77
Illustrating Partial Pressures
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 73 of 77
At STP, • One mole of a pure gas in a volume of 22.4 L will
exert the same pressure as one mole of a gas mixture in 22.4 L.
• VSTP = 22.4 L Gas mixtures
Total Pressure
0.5 mol O2
0.3 mol He0.2 mol Ar1.0 mol
1.0 mol N2
0.4 mol O2
0.6 mol He1.0 mol
1.0 atm 1.0 atm 1.0 atm
Chapter 11 Slide 74 of 77
Learning Check
A scuba tank contains O2
with a pressure of 0.450 atm and He at 855 mm Hg.
What is the total pressure in mm Hg in the tank?
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 11 Slide 75 of 77
1. Convert the pressure in atm to mm Hg 0.450 atm x 760 mm Hg = 342 mm Hg = PO2
1 atm
2. Calculate the sum of the partial pressures.
Ptotal = PO2 + PHe
Ptotal = 342 mm Hg + 855 mm Hg
= 1197 mm Hg
Solution
Chapter 11 Slide 76 of 77
For a deep dive, some scuba divers are using a mixture of helium and oxygen gases with a pressure of 8.00 atm.
If the oxygen has a partial pressure of 1280 mm Hg, what is the partial pressure of the helium?
A) 520 mm Hg B) 2040 mm Hg C) 4800 mm Hg
Learning Check
Chapter 11 Slide 77 of 77
A) 520 mm Hg B) 2040 mm Hg C) 4800 mm Hg
PTotal = 8.00 atm x 760 mm Hg = 6080 mm Hg 1 atm
PTotal = PO + PHe 2
PHe = PTotal - PO2
PHe = 6080 mm Hg - 1280 mm Hg = 4800 mm Hg
Solution
Top Related