# If I have 45 liters of helium in a balloon at 25 degress celcius and increase the tempurate to 55 degrees celcius what will the new volume be?

May 13, 2017

$50. L$

#### Explanation:

This is a problem involving the temperature-volume relationship of gases, which is represented by Charles's law:

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

Remember that $T$ is the absolute temperature, so you must convert the temperature from Celcius to Kelvin:

${25}^{o} C + 273 = 298 K$
${55}^{o} C + 273 = 328 K$

Now, rearrange the equation to solve for ${V}_{2}$ (the new volume):

${V}_{2} = \left({V}_{1}\right) \frac{{T}_{2}}{{T}_{1}}$

and plug in the known values:

${V}_{2} = 45 L \frac{328 K}{298 K} = 50. L$

(The decimal place is present to emphasize there are two significant figures, otherwise it's arbitrary whether you're using one or two.)