Question

The gas inside of a container exerts `"24 Pa"` of pressure and is at a temperature of `"320 K"`. If the pressure in the container changes to `"32 Pa"` with no change in the container's volume, what is the new temperature of the gas?

Answer 1

The final temperature is 430 K.

Explanation:

This is an example of Guy-Lussac's law, which states that as long as volume is constant, the pressure and temperature of a gas are directly proportional. This means that if the pressure goes up, so does the temperature and vice-versa. The equation for Gay-Lussac's law is `P_1/T_1=P_2/T_2`, where `P_1` is the initial temperature, `T_1` is the initial temperature, `P_2` is the final temperature, and `T_2` is the final temperature.

Given
`P_1="24 Pa"`
`T_1="320 K"`
`P_2="32 Pa"`

Unknown
`T_2`

Solution
Rearrange the equation to isolate `T_2`. Substitute the given values into the equation and solve.

`P_1/T_1=P_2/T_2`

`T_2=(P_2T_1)/P_1`

`T_2=(32cancel"Pa"xx320"K")/(24cancel"Pa")="430 K"` rounded to two significant figures