1. Temperature and Pressure (Gay Lussac’s Law)
Chapter 6 Gases
6.5
Temperature and Pressure (Gay Lussac’s Law)
Copyright © by Pearson Education, Inc.
Publishing as Benjamin Cummings

2. 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

3. Learning Check
Solve Gay-Lussac’s Law for P2.
P1 = P2
T1 T2

4. Solution Solve Gay-Lussac’s Law for P2. P1 = P2 T1 T2
Multiply both sides by T2 and cancel
P1 x T2 = P2 x T1
T T1
P2 = P1 x T2 T1

5. Calculation with Gay-Lussac’s Law
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 =
T1 = 18°C T2 = 62°C + 273
= 291 K = K ?

6. Calculation with Gay-Lussac’s Law (continued)
2. Solve Gay-Lussac’s Law for P2:
P1 = P2
T T2
P2 = P1 x T2 T1
P2 = 2.0 atm x 335 K = atm
291 K
Temperature ratio
increases pressure

7. Learning Check A gas has a pressure of 645 torr at 128°C. What is the
temperature in Celsius if the pressure increases to
824 torr (n and V remain constant)?

8. Solution A gas has a pressure of 645 torr at 128°C. What is the
temperature in Celsius if the pressure increases to
1.50 atm (n and V remain constant)?
1. Set up a data table:
Conditions Conditions 2
P1 = 645 torr P2 = 1.50 atm x 760 torr = 1140 torr
1 atm
T1 = 128°C T2 = K – 273 = ?°C
= 401 K

9. Solution 2. Solve Gay-Lussac’s Law for T2: P1 = P2 T1 T2 T2 = T1 x P2
T2 = 401 K x torr = K = 436°C
645 torr Pressure ratio
increases temperature