Question

The gas inside of a container exerts `24 Pa` of pressure and is at a temperature of `210 ^o K`. If the pressure in the container changes to `64 Pa` with no change in the container's volume, what is the new temperature of the gas?

Answer 1

`T_2=560" "^o K`

Explanation:

`P_1=24 " "Pa`
`T_1=210" "^o K`

`P_2=64 " "Pa`
`T_2=?`

`P_1/T_1=P_2/T_2`

`24/210=64/T_2`

`T_2=(210*64)/24`

`T_2=560" "^o K`