If the temperature of 141.9 L of a gas at 58.9 ºC falls by 6.6 Cº, what is the new volume of the gas if the pressure is constant?

Answer 1

This problem involves Charles Law where the volume is directly proportional to the temperature change in degrees Kelvin.

The law is written: V1/T1 = V2/T2

You are given Vi = 141.9 L and T1 = 58.9 0C and told the temperature falls by 6.6 0C. First find T1 by adding the original temperature, 58.9 0C to 273 to find T1 in K = 331.9. To find T2 subtracting 58.9 0C - 6.6 0C the, new temperature is 52.3 0C. Converting this new temperature, T2 to K by adding 273 to the 52.3 0C = 325.3K

Putting this information into the equation: 141.9 L/331.9K. = V2/325.3K

Solving for V2 you obtain 139..1 L. This makes sense as the temperature drops the the kinetic energy drops and therefore less collisions of the gas molecules against the container wall, less expansion of the container and the Volume will therefore go down.