Question

The gas inside of a container exerts `27 Pa` of pressure and is at a temperature of `150 ^o K`. If the pressure in the container changes to `40 Pa` with no change in the container's volume, what is the new temperature of the gas?

Answer 1

The new temperature is `=222.2K`

Explanation:

Apply Gay Lussac's law

`P_1/T_1=P_2/T_2` at constant volume

The initial pressure is `=P_1=27Pa`

The initial temperature is `=T_1=150K`

The final pressure is `P_2=40Pa`

The final temperature is

`T_2=P_2/P_1*T_1=40/27*150=222.2K`

Answer 2

`220^@K`

Explanation:

Using the combined Gas Law, `(P_1V_1)/T_1=(P_2V_2)/T_2`, you can simplify it to `P_1/T_1=P_2/T_2` since volume is constant.

Since to want to find `T_2`(Temperature 2), you rearrange the formula to become `T_2=(T_1*P_2)/P_1`

Then you can plug and chug
`T_2=(150^@K*40Pa)/(27Pa)`
`=222.222222222...^@K`
`=220^@K` (with Significant Figures)