1 Chapter 11 Gases 11.8 The Ideal Gas Law Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings.

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Chapter 11 Gases

11.8 The Ideal Gas Law

Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings

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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

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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

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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 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

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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 2.86 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

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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

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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

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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

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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)(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

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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) mol•K = 0.00703 mol

RT (0.0821 L • atm)(303K)

2. Set up the molar mass relationship.

Molar mass = g = 0.250 g = 35.6 g/mol mol 0.00703 mol

Molar Mass of a Gas