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 `60 Pa` with no change in the container's volume, what is the new temperature of the gas?

Answer 1

The new temperature is `525 K`

Explanation:

The answer can be determined by using Gay-Lussac's law:

`P_1/T_1 = P_2/T_2`

The number 1 represents the initial conditions and the number 2 represents the final conditions.

Identify your known and unknown variables:

`color(indigo)("Knowns:"`
`P_1`= 24 Pa
`P_2`= 60 Pa
`T_1`= 210K

`color(gold) ("Unknowns:"`
`T_2`

Rearrange the equation to solve `T_2`

`T_2=(P_2xxT_1)/P_1`

Plug in your given values to obtain the final temperature:

`T_2=(60cancel"Pa"xx210K)/(24\cancel"Pa")` = `525K`

I should also mention that when you are dealing with the Kelvin scale, don't write the degrees symbol. Just write the letter K.