The gas inside of a container exerts 6 Pa of pressure and is at a temperature of 120 ^o K. If the pressure in the container changes to 49 Pa with no change in the container's volume, what is the new temperature of the gas?

1 Answer
Mar 21, 2018

The new temperature is =980K

Explanation:

Apply Gay Lussac's Law

"Pressure"/"Temperature"= "Constant", at "Constant Volume"

P_1/T_1=P_2/T_2

The initial pressure is P_1=6Pa

The initial temperature is T_1=120K

The final pressure is P_2=49Pa

The final temperature is

T_2=P_2/P_1*T_1

=49/6*120=980K