A meteorological balloon rises through the atmosphere until it expands to a volume of 1.0 × 10 6 m 3, wherethe pressure is 1.0 × 10 3 Pa. The temperature also falls from 17 °C to − 43 °C.The pressure of the atmosphere at the Earth’s surface = 1.0 × 10 5 Pa.Show that the volume of the balloon at take off is about 1.3 × 10 4 m 3
Question
A meteorological balloon rises through the atmosphere until it expands to a volume of 1.0 × 10 6 m 3, wherethe pressure is 1.0 × 10 3 Pa. The temperature also falls from 17 °C to − 43 °C.The pressure of the atmosphere at the Earth’s surface = 1.0 × 10 5 Pa.Show that the volume of the balloon at take off is about 1.3 × 10 4 m 3
Solution
To solve this problem, we can use the ideal gas law, which states that the product of the pressure and volume of a gas is directly proportional to the temperature of the gas. This can be written as:
P1V1/T1 = P2V2/T2
Where: P1 = initial pressure V1 = initial volume T1 = initial temperature P2 = final pressure V2 = final volume T2 = final temperature
Given in the problem: P1 = 1.0 × 10^5 Pa (pressure at Earth's surface) V2 = 1.0 × 10^6 m^3 (final volume of the balloon) P2 = 1.0 × 10^3 Pa (final pressure) T1 = 17 °C = 17 + 273.15 = 290.15 K (initial temperature in Kelvin) T2 = -43 °C = -43 + 273.15 = 230.15 K (final temperature in Kelvin)
We need to find V1 (the volume of the balloon at take off).
Substituting the given values into the ideal gas law equation, we get:
(1.0 × 10^5 Pa * V1) / 290.15 K = (1.0 × 10^3 Pa * 1.0 × 10^6 m^3) / 230.15 K
Solving for V1, we get:
V1 = (1.0 × 10^3 Pa * 1.0 × 10^6 m^3 * 290.15 K) / (1.0 × 10^5 Pa * 230.15 K)
V1 = 1.26 × 10^4 m^3
So, the volume of the balloon at take off is approximately 1.3 × 10^4 m^3.
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