# The gas inside of a container exerts 42 Pa of pressure and is at a temperature of 250 ^o K. If the pressure in the container changes to 30 Pa with no change in the container's volume, what is the new temperature of the gas?

Jun 12, 2016

The new temperature will be 180 K.

#### Explanation:

This is an example of Gay-Lussac's law, which states that the pressure of a given amount of a gas held at constant volume varies directly with the temperature in Kelvins. This means that when the pressure increases so does the temperature, and vice versa. The equation to use when solving these problems is ${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$.

In order to solve this problem, you need to identify what is known and what is unknown.

Known
${P}_{1} = \text{42 Pa}$
${T}_{1} = \text{250 K}$
${P}_{2} = \text{30 Pa}$

Unknown
${T}_{2} = \text{?}$

Solution
Rearrange the equation to isolate ${T}_{2}$. Then substitute the known variables into the equation and solve.

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

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

${T}_{2} = \left(250 \text{K"*30cancel"Pa")/(42cancel"Pa}\right)$

${T}_{2} = \text{180 K}$ (rounded to two significant figures)