Question

The gas inside of a container exerts `6 Pa` of pressure and is at a temperature of `360 ^o K`. If the pressure in the container changes to `32 Pa` with no change in the container's volume, what is the new temperature of the gas?

Answer 1

`T_2=1920" K"`

Explanation:

`P_1/T_1=P_2/T_2`
`cancel(6)/cancel(360)=32/T_2`
`1/60=32/T_2`
`T_2=32*60`
`T_2=1920" K"`