Question

The gas inside of a container exerts `48 Pa` of pressure and is at a temperature of `270 ^o K`. If the pressure in the container changes to `60 Pa` with no change in the container's volume, what is the new temperature of the gas?

Answer 1

`337.5K`

Explanation:

Recall that Gay-Lussac's Law states that pressure is directly proportional to temperature. The formula that represents this is:

`color(brown)(P_1)/color(blue)(T_1)=color(purple)(P_2)/color(orange)(T_2)`

where:
`color(brown)(P_1)=`initial pressure
`color(blue)(T_1)=`inital temperature (Kelvin)
`color(purple)(P_2)=`final pressure
`color(orange)(T_2)=`final temperature (Kelvin)

Finding the Final Temperature
`1`. Since all other values have been given, the only variable you need to solve for is `T_2`, the final temperature. Thus, start by rearranging the formula to solve for `T_2`.

`P_1/T_1=P_2/T_2`

`T_2=(P_2T_1)/P_1`

`2`. Substitute your known values into the rearranged formula.

`color(purple)(P_2=60Pa)color(white)(XXX)color(blue)(T_1=270K)color(white)(XXX)color(brown)(P_1=48Pa)`

`T_2=((60Pa)(270K))/((48Pa))`

`3`. Solve for `T_2`.

`color(green)(T_2=337.5K)`

`:.`, the new temperature of the gas is `337.5K`.