Chapter 11 - Gases Properties of Gases Gas Pressure Copyright © 2008 by Pearson Education, Inc....

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.



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.


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


Units of PressureGas pressure• Is described as a force acting on a specific area. • Pressure (P) = Force Area



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.



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.



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


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.



Chapter 11 Gases

Pressure and Volume Boyle’s Law



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 ______________.



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


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


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

?


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)


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



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


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?



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


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 Gases

Temperature and Volume (Charles’ Law)



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.



• 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


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


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


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.


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 Gases

Temperature and Pressure (Gay-Lussac’s Law)



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



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

?


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 Gases

The Combined Gas Law



• 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


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


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)


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


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 Gases

Volume and Moles (Avogadro’s Law)



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



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


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


• 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


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.



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


Guide to Using Molar Volume



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


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


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


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 Gases

The Ideal Gas Law



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


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


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


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


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



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


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


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?



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


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 Gases

Gas Laws and Chemical Reactions



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.


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


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


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


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


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 Gases

Partial Pressure (Dalton’s Law)



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


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


Illustrating Partial Pressures



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


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?



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


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


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

Chapter 11 - Gases Properties of Gases Gas Pressure Copyright © 2008 by Pearson Education, Inc....

Documents

Transcript of Chapter 11 - Gases Properties of Gases Gas Pressure Copyright © 2008 by Pearson Education, Inc....