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?

1 Answer
May 13, 2017

50.L50.L

Explanation:

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

(V_1)/(T_1) = (V_2)/(T_2)V1T1=V2T2

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

25 ^o C + 273 = 298K25oC+273=298K
55 ^o C + 273 = 328 K55oC+273=328K

Now, rearrange the equation to solve for V_2V2 (the new volume):

V_2 = (V_1)(T_2)/(T_1)V2=(V1)T2T1

and plug in the known values:

V_2 = 45 L(328 K)/(298 K) = 50.LV2=45L328K298K=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.)