Temperature and Pressure (Gay Lussac’s Law) Chapter 6 Gases 6.5 Temperature and Pressure (Gay Lussac’s Law) Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

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

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

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 T1 T1 P2 = P1 x T2 T1

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 + 273 T2 = 62°C + 273 = 291 K = 335 K ?

Calculation with Gay-Lussac’s Law (continued) 2. Solve Gay-Lussac’s Law for P2: P1 = P2 T1 T2 P2 = P1 x T2 T1 P2 = 2.0 atm x 335 K = 2.3 atm 291 K Temperature ratio increases pressure

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)?

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 P1 = 645 torr P2 = 1.50 atm x 760 torr = 1140 torr 1 atm T1 = 128°C + 273 T2 = K – 273 = ?°C = 401 K

Solution 2. Solve Gay-Lussac’s Law for T2: P1 = P2 T1 T2 T2 = T1 x P2 T2 = 401 K x 1140 torr = 709 K - 273 = 436°C 645 torr Pressure ratio increases temperature