Chapter 8 Gases

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Chapter 8 Gases

8.1 Gases and Kinetic Theory 8.2 Gas Pressure8.8 Ideal Gas Law

* You do not need to know Boyle’s (8.3), Charles’ (8.4), Gay-Lussac’s (8.5), Avogadro’s (8.7) or the Combined gas (8.6) laws. They are all contained within the ideal gas law. We will not cover Dalton’s law (8.9).

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Particles of a gas Move rapidly in straight lines. Have kinetic energy that increases with an

increase in temperature. Are very far apart. Have essentially no attractive (or repulsive)

forces. Have very small volumes compared to the

volume of the container they occupy.

8.1 Kinetic Theory of Gases

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Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n).

Properties of Gases

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A barometer measures the pressure exerted by the gases in the atmosphere.

The atmospheric pressure is measured as the height in mm of the mercury column.

8.2 Barometer

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A. The downward pressure of the Hg in a barometer is _____ than/as the weight of the atmosphere.

1) greater 2) less 3) the same

B. A water barometer is 13.6 times taller than a Hg barometer (DHg = 13.6 g/mL) because

1) H2O is less dense

2) H2O is heavier

3) air is more dense than H2O

Learning Check

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A.The downward pressure of the Hg in a barometer is 3) the same as the weight of the atmosphere.

B. A water barometer is 13.6 times taller than a Hg barometer (DHg = 13.6 g/mL) because

1) H2O is less dense

Solution

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A gas exerts pressure, which is defined as a force acting on a specific area.

Pressure (P) = Force Area

One atmosphere (1 atm) is 760 mm Hg. 1 mm Hg = 1 torr

1.00 atm = 760 mm Hg = 760 torr

Pressure

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In science, pressure is stated in atmospheres (atm), millimeters of mercury (mm Hg), and Pascals (Pa).

Units of Pressure

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A. What is 475 mm Hg expressed in atm?1) 475 atm2) 0.638 atm3) 3.61 x 105 atm

B. The pressure in a tire is 2.00 atm. What is this pressure in mm Hg?1) 2.00 mm Hg2) 1520 mm Hg3) 22,300 mm Hg

Learning Check

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A. What is 475 mm Hg expressed in atm?2) 0.638 atm485 mm Hg x 1 atm = 0.638 atm

760 mm HgB. The pressure of a tire is measured as 2.00 atm.

What is this pressure in mm Hg?2) 1520 mm Hg2.00 atm x 760 mm Hg = 1520 mm Hg

1 atm

Solution

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The relationship between the four properties (P, V, n, and T) of gases can be written equal to a constant R.

PV = RnT

Rearranging this expression gives the expression called the ideal gas law.

PV = nRT

8.8 Ideal Gas Law

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The volumes of gases can be compared when they have the same conditions of temperature and pressure (STP).Standard temperature (T)

273 K (0°C )

Standard pressure (P)

1 atm (760 mm Hg)

STP

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At STP, 1 mole of a gas occupies a volume of 22.4 L.

The volume of one mole of a gas is called the molar volume.

Molar Volume

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The molar volume at STP can be used to form conversion factors.

22.4 L and 1 mole 1 mole 22.4 L

Molar Volume as a Conversion Factor

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A. What is the volume at STP of 4.00 g of CH4?

1) 5.60 L 2) 11.2 L 3) 44.8 L

B. How many grams of He are present in 8.00 L

of gas at STP?

1) 25.6 g 2) 0.357 g 3) 1.43 g

Learning Check

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A. 1) 5.60 L4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L 16.0 g CH4 1 mole CH4

B. 3) 1.43 g8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 L 1 mole He

Solution

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The universal gas constant, R, can be calculated using the molar volume of a gas at STP.

At STP (273 K and 1.00 atm), 1 mole of a gas occupies 22.4 L.

P V R = PV = (1.00 atm)(22.4 L)

nT (1 mole) (273K) n T

= 0.0821 L atm mole K

Note there are four units associated with R.

Universal Gas Constant, R

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Another value for the universal gas constant is obtained using mm Hg for the STP pressure. What is the value of R when a pressure of 760 mm Hg is placed in the R value expression?

Learning Check

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What is the value of R when the STP value for P is 760 mmHg?

R = PV = (760 mm Hg) (22.4 L)

nT (1 mole) (273K)

= 62.4 L mm Hg mole K

Solution

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Dinitrogen oxide (N2O), laughing gas, is used by dentists as an anesthetic. If a 20.0 L tank of laughing gas contains 2.8 moles N2O at 23°C, what is the pressure (mm Hg) in the tank?

Learning Check

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1. Adjust the units of the given properties to match the units of R.

V = 20.0 L, T = 296 K, n = 2.8 moles, P = ? 2. Rearrange the ideal gas law for P.

P = nRT VP = (2.8 moles)(62.4 L mm Hg)(296 K)

(20.0 L) (mole K)

= 2.6 x 103 mm Hg

Solution

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A cylinder contains 5.0 L of O2 at 20.0°C and 0.85 atm. How many grams of oxygen are in the cylinder?

Learning Check

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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)(mole K) = 0.18 mole O2

(0.0821atm L)(293 K) 3. Convert moles to grams using molar mass.

= 0. 18 mole O2 x 32.0 g O2 = 5.8 g O2

1 mole O2

Solution

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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)

RT (0.0821 L atm/mole K)(303K) = 0.00703 mole

2. Set up the molar mass relationship.

Molar mass = g = 0.250 g = 35.6 g/mole

mole 0.00703 mole

Molar Mass of a Gas

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Gases in Equations

The amounts of gases reacted or produced in a chemical reaction can be calculated using the ideal gas law and mole factors.

Problem:

What volume (L) of Cl2 gas at 1.2 atm and 27°C is needed to completely react with 1.5 g of aluminum?

2Al(s) + 3Cl2 (g) 2AlCl3(s)

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Gases in Equations (continued)

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 mole Al x 3 moles Cl2 = 0.083 mole Cl2

27.0 g Al 2 moles Al

2. Place the moles Cl2 in the ideal gas equation.V = nRT = (0.083 mole Cl2)(0.0821 Latm/moleK)(300K) P 1.2 atm

= 1.7 L Cl2

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What volume (L) of O2 at 24°C and 0.950 atm are needed to react with 28.0 g NH3?

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

Learning Check

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1. Calculate the moles of O2 needed.

28.0 g NH3 x 1 mole NH3 x 5 mole O2

17.0 g NH3 4 mole NH3

= 2.06 mole O2

2. Place the moles O2 in the ideal gas equation.

V = nRT = (2.06 moles)(0.0821 L atm/moleK)(297K)P 0.950 atm

= 52.9 L O2

Solution

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Solution

V1 = V2

T1 T2

Cross multiply to give V1T2 = V2T1

Isolate T2 by dividing through by V1

V1T2 = V2T1 T2 = V2T1

V1 V1 V1