1 Chapter 6Gases 6.5 Temperature and Pressure (Gay Lussac’s Law) Copyright © 2005 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. P 1 = P 2 T 1 T 2

3 Learning Check Solve Gay-Lussac’s Law for P 2. P 1 = P 2 T 1 T 2

4 Solution Solve Gay-Lussac’s Law for P 2. P 1 = P 2 T 1 T 2 Multiply both sides by T 2 and cancel P 1 x T 2 = P 2 x T 1 T 1 P 2 = P 1 x T 2 T 1

5 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 1Conditions 2 P 1 = 2.0 atmP 2 = T 1 = 18°C T 2 = 62°C = 291 K = 335 K Calculation with Gay-Lussac’s Law ?

6 Calculation with Gay-Lussac’s Law (continued) 2. Solve Gay-Lussac’s Law for P 2 : P 1 = P 2 T 1 T 2 P 2 = P 1 x T 2 T 1 P 2 = 2.0 atm x 335 K = 2.3 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 1.50 atm (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 1 Conditions 2 P 1 = 645 torr P 2 = 1.50 atm x 760 torr = 1140 torr 1 atm T 1 = 128°C T 2 = K – 273 = ?°C = 401 K

9 Solution 2. Solve Gay-Lussac’s Law for T 2 : P 1 = P 2 T 1 T 2 T 2 = T 1 x P 2 P 1 T 2 = 401 K x 1140 torr = 709 K = 436°C 645 torr Pressure ratio increases temperature