Question

The gas inside of a container exerts `5 Pa` of pressure and is at a temperature of `120 ^o C`. If the temperature of the gas changes to `180 ^oK` with no change in the container's volume, what is the new pressure of the gas?

Answer 1

`P_2~=2,29 Pa`

Explanation:

`"we can use the Gay-Lussac's law"`
`P_1/T_1=P_'/T_2`
`"change "120 ^o C " to " ^o K" 120+273=393 ^o K`
`5/393=P_2/180`
`P_2=(180*5)/393`
`P_2~=2,29 Pa`