Question

The gas inside of a container exerts `15 Pa` of pressure and is at a temperature of `60 ^o K`. If the pressure in the container changes to `36 Pa` with no change in the container's volume, what is the new temperature of the gas?

Answer 1

`144K`

Explanation:

`PV=nr∆T`
`V=(nr∆T)/P`

`P_1=15Pa`
`T_1=60K`
`P_2=36Pa`

`V_1=V_2`
`T_2=?`

And, `n" & " r` will remain the same. So,

`(cancel(nr)60K)/(15Pa)=(cancel(nr)T_2)/(36Pa)`

`=>T_2=(36cancel(Pa)*60K)/(15cancel(Pa))`

`=>T_2=144K`