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

Answer 1

`T_2 = 170` `"K"`

Explanation:

We'r asked to find the new temperature of a gas,after it is subjected to a pressure change with no change in volume.

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

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

Here,

• `P_1 = 27` `"Pa"`

• `T_1 = 150` `"K"`

• `P_2 = 30` `"Pa"`

• `T_2 = ?`

Plugging in known values and solving for `T_2`, we have

`T_2 = (P_2T_1)/(P_1) = ((30cancel("Pa"))(150color(white)(l)"K"))/(27cancel("Pa")) = color(red)(170` `color(red)("K"`

rounded to `2` significant figures.