In an air-conditioned room at 19.0 ∘C
, a spherical balloon had the diameter of 50.0 cm
. When taken outside on a hot summer day, the balloon expanded to 51.0 cm
in diameter. What was the temperature outside in degrees Celsius? Assume that the balloon is a perfect sphere and that the pressure and number of moles of air molecules remains the same.
To solve this problem, we can use the formula for the volume of a sphere:
V = (4/3) * π * r^3
Where r is the radius of the sphere.
Given that the balloon inside had a diameter of 50.0 cm, the radius inside would be 25.0 cm. Therefore, the initial volume inside would be:
V_inside = (4/3) * π * (25.0)^3 = (4/3) * π * 15625 = 65208.33 cm^3
Similarly, the volume outside when the balloon expanded to 51.0 cm in diameter would be:
V_outside = (4/3) * π * (25.5)^3 = (4/3) * π * 16581.25 = 69321.67 cm^3
Since the number of moles and pressure are constant, we can assume that the temperature and volume are directly proportional, according to the ideal gas law. Therefore, we can set up the following equation:
V_outside / V_inside = T_outside / T_inside
69321.67 / 65208.33 = T_outside / 19.0
T_outside = (69321.67 / 65208.33) * 19.0 = 20.20 °C
Therefore, the temperature outside on the hot summer day was approximately 20.20 °C.