A Weather balloon carries instruments that measure temperature, pressure, and humidity as it rises through the atmosphere. Suppose such a balloon has a volume of 1.2 m3 at sea level where the pressure is 1 atm and the temperature is 20oC. When the balloon is at an altitude of 11 km (36,000 ft) the pressure is down to 0.5 atm and the temperature is about negative 55oC (-55oC). What is the volume of the balloon then?
To solve this problem, we can use the ideal gas law, which relates pressure, volume, and temperature of a gas. The ideal gas law equation is:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles of gas
R = Ideal gas constant
T = Temperature in Kelvin
In this case, we don't know the number of moles of gas, but we can assume it remains constant as the balloon rises.
Let's find the initial number of moles of gas at sea level:
P1 = 1 atm
V1 = 1.2 m^3
T1 = 20°C = 20 + 273 = 293 K
We can rearrange the ideal gas law equation to solve for n:
n = PV / RT
Substituting the given values, we get:
n1 = (1 atm) (1.2 m^3) / (0.0821 atm·m^3/mol·K) (293 K)
n1 ≈ 0.058 moles
Now, let's find the final volume of the balloon at an altitude of 11 km:
P2 = 0.5 atm
T2 = -55°C = -55 + 273 = 218 K
Again, using the ideal gas law equation:
V2 = (n1 * R * T2) / P2
Substituting the values, we get:
V2 = (0.058 moles) (0.0821 atm·m^3/mol·K) (218 K) / (0.5 atm)
V2 ≈ 1.49 m^3
Therefore, at an altitude of 11 km, the volume of the balloon is approximately 1.49 m^3.