Question

The gas inside of a container exerts `27 Pa` of pressure and is at a temperature of `180 ^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

Approximately `267` kelvin.

Explanation:

At a constant number of moles and volume, we can use Gay-Lussac's law, which states that

`PpropT`

or

`P_1/T_1=P_2/T_2`

So, we can plug in our given values into the equation, and get,

`(27 \ "Pa")/(180 \ "K")=(40 \ "Pa")/T_2`

`=>T_2=(180 \ "K")/(27color(red)cancelcolor(black)"Pa")*40color(red)cancelcolor(black)"Pa"`

`=266.bar(6) \ "K"`

`~~267 \ "K"`