3. Charles Law: What is the volume of the air in a Helium balloon that has 135 mL at 67.6 degrees Fahrenheit if the temperature is raised to 100. degrees Fahrenheit?

199.7mL

Explanation:

Using Charles law equation as follows:

V1/T1 = V2/T2

Where;

V1 = initial volume (mL)

T1 = initial temperature (°F)

V2 = final volume (mL)

T2 = final temperature (°F)

According to the information provided in this question, V1 = 135mL, V2 = ?, T1 = 67.6°F, T2 = 100°F

Using V1/T1 = V2/T2

135/67.6 = V2/100

Cross multiply

135 × 100 = 67.6 × V2

13500 = 67.6V2

V2 = 13500/67.6

V2 = 199.704

The new volume in the air in the helium ballon is 199.7mL.