# The gas inside of a container exerts 15 Pa of pressure and is at a temperature of 450 ^o K. If the temperature of the gas changes to 90 ^oK with no change in the container's volume, what is the new pressure of the gas?

Jun 13, 2016

The new gas pressure is $3 P a$.

#### Explanation:

For this type of question, you would use Gay-Lussac's Law

Gay-Lussac's Law shows that there is a direct relationship between pressure and temperature as long as the volume and number of moles of gas remain unchanged.

Let's start off with identifying our known and unknown variables.
The first pressure we have is $\text{15 Pa}$, the first temperature is $450 K$, and the second temperature is $90 K$. Our only unknown is the second pressure.

Now we just rearrange the equation in order to solve for ${P}_{2}$ and plug in the appropriate values like so:

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

${P}_{2} = \left(90 \cancel{\text{K"xx15Pa)/(450cancel"K}}\right)$

${P}_{2} = 3 P a$

P.S. When using the Kelvin scale, you do not put the degree symbol. You just write K.