Question

The gas inside of a container exerts `30 Pa` of pressure and is at a temperature of `450 ^o K`. If the pressure in the container changes to `25 Pa` with no change in the container's volume, what is the new temperature of the gas?

Answer 1

`T_2 = 375` `"K"`

Explanation:

To solve this problem, we can use the pressure-temperature relationship of gases, illustrated by Gay-Lussac's law:

`(P_1)/(T_1) = (P_2)/(T_2)`

Our known values are

• `P_1 = 30` `"Pa"`

• `T_1 = 450` `"K"`

• `P_2 = 25` `"Pa"`

We're asked to solve for the new temperature, so let's rearrange the equation to solve for `T_2`:

`T_2 = (P_2T_1)/(P_1)`

Plugging in the known variables, we have

`T_2 = ((25cancel("Pa"))(450color(white)(l)"K"))/(30cancel("Pa")) = color(red)(375` `color(red)("K"`

The new temperature is thus `color(red)(375` `color(red)("K"`.