A container with a volume of 7 L contains a gas with a temperature of 420^o C. If the temperature of the gas changes to 570 ^o K without any change in pressure, what must the container's new volume be?

Jan 3, 2017

The new volume is $5.76 L$

Explanation:

Let's start off with identifying our known and unknown variables:

$\textcolor{g o l d}{\text{Knowns:}}$
- Initial Volume
- Initial Temperature
- Final Temperature

$\textcolor{m a \ge n t a}{\text{Unknowns:}}$
- Final 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 where the numbers $1$ and $2$ represent the first and second conditions, respectively.

I must also add that the volume must have units of liters and the temperature must have units of Kelvin. In our case the volume is fine, we just have to change the first temperature from centigrade to Kelvins.

We do this by adding $\text{273.15K}$ to the given temperatures. Thus the first temperature is

$\text{273.15K" + 420^@"C" = "693.15K}$

Now we just rearrange the equation and plug and chug.

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

${V}_{2} = \left(570 \cancel{\text{K") * "7L") / (693.15 cancel("K}}\right)$

${V}_{2} = \text{5.76 L}$