Question

The gas inside of a container exerts `18 Pa` of pressure and is at a temperature of `360 ^o K`. If the pressure in the container changes to `27 Pa` with no change in the container's volume, what is the new temperature of the gas?

Answer 1

The final temperature, `T_2`, will be `"540 K"`.

Explanation:

This question is an example of Gay-Lussac's gas law, which states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. This means that if the pressure increases, so does the temperature, and vice-versa. The equation to use in order to answer your question is:

`P_1/T_1=P_2/T_2`

Organize your data.

Given

`P_1="18 Pa"`
`T_1="360 K"` `lArr` The degree symbol `(""^@")` is not used with Kelvins.
`P_2="27 Pa"`

Unknown: `T_2`

Solution
Rearrange the equation to isolate `T_2`. Insert your data into the equation and solve.

`T_2=(P_2T_1)/(P_1)`

`T_2=(27color(red)cancel(color(black)("Pa"))xx360"K")/(18color(red)cancel(color(black)("Pa")))="540 K"`

As you can see, as the pressure increased, so did the temperature.

Answer 2

`540"K"`

Explanation:

We can use the temperature-pressure relationship of gases illustrated by Gay-Lussac's law:

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

At constant volume, the temperature and pressure of a fixed quantity of gas are directly proportional to each other, which is explained by the kinetic-molecular theory.

We can rearrange this equation to solve for the final temperature `T_2`:

`T_2 = (P_2T_1)/(P_1)`

and finally, plug in the known variables to find the temperature.

`T_2 = ((27"Pa")(360K))/(18"Pa") = color(red)(540"K"`