Question

The gas inside of a container exerts `18 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

The new temperature will be `~~` `"630 K"`.

Explanation:

This is an example of Guy-Lussac's law which states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. So if the pressure increases, so will the temperature and vice versa.

The equation is:

`P_1/T_1=P_2/T_2`,

where:

`P` is pressure and `T` is temperature.

Organize the data.

Known

`P_1="18 Pa"`

`T_1 ="450 K"`

`P_2="25 Pa"`

Unknown

`T_2`

Solution

Rearrange the equation to isolate `T_2`. Plug in the known values and solve.

`T_2=(P_2T_1)/(P_1)`

`T_2=(25color(red)cancel(color(black)("Pa"))xx450"K")/(18color(red)cancel(color(black)("Pa")))="630 K"` rounded to two significant figures due to 18 Pa and 25 Pa