# The gas inside of a container exerts 24 Pa of pressure and is at a temperature of 210 ^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?

Aug 21, 2016

The new temperature is $525 K$

#### Explanation:

The answer can be determined by using Gay-Lussac's law:

${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$

The number 1 represents the initial conditions and the number 2 represents the final conditions.

Identify your known and unknown variables:

color(indigo)("Knowns:"
${P}_{1}$= 24 Pa
${P}_{2}$= 60 Pa
${T}_{1}$= 210K

color(gold) ("Unknowns:"
${T}_{2}$

Rearrange the equation to solve ${T}_{2}$

${T}_{2} = \frac{{P}_{2} \times {T}_{1}}{P} _ 1$

Plug in your given values to obtain the final temperature:

${T}_{2} = \left(60 \cancel{\text{Pa"xx210K)/(24\cancel"Pa}}\right)$ = $525 K$

I should also mention that when you are dealing with the Kelvin scale, don't write the degrees symbol. Just write the letter K.