Question

The gas inside of a container exerts `24 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

`300"K"`

Explanation:

First the gas to be ideal. Then, from Gay-Lussac's Law, we know that

`T_1/p_1 = T_2/p_2`

So,

`(180"K")/(24"Pa") = T_2/(40"Pa")`

`T_2 = 300"K"`