A man heats a balloon in the oven. If the balloon initially has a volume of 0.8 liters and a temperature of 12°c, what will the volume of the balloon be after he heats it to a temperature of 300°c?

1.044 L

Explanation:

Data Given

initial volume V1 of gas in balloon = 0.8 L

initial Temperature T1 of gas in balloon = 12 °C

convert Temperature to Kelvin

T1 = °C + 273

T1 = 12°C + 273 = 285 K

final Temperature T2 of gas in balloon = 100°C

convert Temperature to Kelvin

T1 = °C + 273

T1 = 100°C + 273 = 373 K

final volume V2 of gas in balloon = ?

Solution:

This problem will be solved by using Charles' law equation at constant pressure.

The formula used

V1 / T1 = V2 / T2

As we have to find out volume, so rearrange the above equation

V2 = (V1 / T1) x T2

Put value from the data given

V2 = (0.8 L / 285 K) x 373 K

V2 = (0.0028 L/K) x 373 K

V2 = 1.044 L

So the final volume of gas in balloon = 1.044 L