© 2014 Pearson Education, Inc.

Chapter 11 Gases11.4 Temperature and Volume (Charles’s Law)

Learning Goal Use the temperature–volume relationship (Charles’s law) to determine the final temperature or volume when the pressure and amount of gas are constant.

Chapter 11 Lecture

Basic ChemistryFourth Edition

© 2014 Pearson Education, Inc.

Charles’s law states that

• the temperature (K) of a gas is directly related to its volume (V) when P and n are constant

• if the temperature (K) increases, then the volume (V) also increases

Charles’s Law

Figure 11.5 Charles’s law: The Kelvin temperature of a gas is directly related to the volume of the gas when there is no change in the pressure and amount of gas. When the temperature increases, making the molecules move faster, the volume must increase to maintain constant pressure.

© 2014 Pearson Education, Inc.

Charles’s law is a direct relationship between volume and temperature. A direct relationship between variables means the variables increase or decrease together.

All temperatures used in gas law calculations must be converted to Kelvin (K) temperature.

Charles’s Law

No change in number of moles and pressure

© 2014 Pearson Education, Inc.

The equation for Charles’s law can be rearranged to solve for any factor.

To obtain V2, multiply both sides by T2.

Charles’s Law Equation

© 2014 Pearson Education, Inc.

Use the gas laws to complete the following statements with increases or decreases.

A. Temperature _______ when V decreases.

B. When T decreases, V _______.

C. Pressure _______ when V changes from 12 L to 4 L.

D. Volume _______ when T changes from 15 oC to 45 oC.

Learning Check

© 2014 Pearson Education, Inc.

Use the gas laws to complete the following statements with increases or decreases.

A. Temperature decreases when V decreases.

B. When T decreases, V decreases.

C. Temperature decreases when V changes from 12 L to 4 L.

D. Volume increases when T changes from 15 oC to 45 oC.

Solution

© 2014 Pearson Education, Inc.

Solve Charles’s law expression for T2.

Learning Check

© 2014 Pearson Education, Inc.

Solve Charles’s law expression for T2.

Cross multiply to give: V1T2 = V2T1

Solve for T2 by dividing through by V1

Solution

© 2014 Pearson Education, Inc.

Learning Check

A balloon has a volume of 785 mL at 21 oC. If the temperature drops to 0 oC, what is the new volume of the balloon (n, P constant)?

© 2014 Pearson Education, Inc.

A balloon has a volume of 785 mL at 21 oC. If the temperature drops to 0 oC, what is the new volume of the balloon (n, P constant)? Step 1 Organize the data in a table of initial and final conditions. We can predict that as T decreases, V decreases.

Conditions 1 Conditions 2

V1 = 785 mL V2 = ?

T1 = 21 °C = 294 K T2 = 0 °C = 273 K

Learning Check

© 2014 Pearson Education, Inc.

A balloon has a volume of 785 mL at 21 oC. If the temperature drops to 0 oC, what is the new volume of the balloon (n, P constant)?

Step 2 Rearrange the gas law equation to solve for the unknown quantity.

Solve Charles’s law for V2. Decreasing temperature decreases the volume when moles and pressure are constant.

Solution

© 2014 Pearson Education, Inc.

A balloon has a volume of 785 mL at 21 oC. If the temperature drops to 0 oC, what is the new volume of the balloon (n, P constant)? Step 2 Rearrange the gas law equation to solve for the unknown quantity.

Solution

© 2014 Pearson Education, Inc.

A balloon has a volume of 785 mL at 21 oC. If the temperature drops to 0 oC, what is the new volume of the balloon (n, P constant)?

Step 3 Substitute values into the gas law equation and calculate.When temperature decreases, the volume decreases as predicted.

Solution