Question

In a rigid container if the temperature of a gas is -131 oC and the pressure is 5.0 atm what is the pressure of the gas when it is warmed to 110 oC

Answer 1

Answer

Pressure 2 = 13.49 atm

Explanation

Given:

Temperature 1 = -131 °C = 142 K

Temperature 2 = 110 °C = 383 K

Pressure 1 = 5.0 atm

Solution

To solve this problem we will use Gay-Lussac's Law

P1/T1 = P2/T2

P2 = P1T2/T1

P2 = (5.0 atm x 383 K)/142 K

P2 = 13.49 atm