# A container with a volume of 7 L contains a gas with a temperature of 420^o K. If the temperature of the gas changes to 300 ^o K without any change in pressure, what must the container's new volume be?

Jun 10, 2016

The new volume is $5 L$.

#### Explanation:

Let's start off with identifying our known and unknown variables.
The first volume we have is $\text{7.0 L}$, the first temperature is $420 K$, and the second temperature is $300 K$. Our only unknown is the second volume.

We can obtain the answer using Charles' Law which shows that there is a direct relationship between volume and temperature as long as the pressure and number of moles remain unchanged.

The equation we use is

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

where the numbers $1$ and $2$ represent the first and second conditions. I must also add that the volume must have units of liters and the temperature must have units of Kelvins. In our case, both have good units!

Now we just rearrange the equation and plug and chug.

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

${V}_{2} = \left(300 \cancel{\text{K") * "7 L") / (420 cancel("K}}\right)$

${V}_{2} = \text{5 L}$

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