I am having a really hard time figuring this question out, would somebody be able to help me?
Imagine a giant dry-cleaner's bag full of air at a temperature of -35∘C floating like a balloon with a string hanging from it 12km above the ground.
Estimate what its temperature would be if you were able to yank it suddenly back to Earth's surface.
since V stays constant,
figure the pressure change.
T(°K) is directly proportional to pressure.
Of course, I'd be happy to help you with this question!
To estimate the temperature of the giant dry-cleaner's bag if it were suddenly yanked back to Earth's surface, we can make some simplifying assumptions.
First, let's assume that the giant bag behaves like an ideal gas. This means that we can use the ideal gas law to estimate the temperature change. The ideal gas law equation is as follows:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant
T is the temperature of the gas
Now, let's consider the scenario. The giant bag contains air at a temperature of -35∘C. We can convert this temperature to Kelvin by adding 273.15 to it, which gives us approximately 238.15K.
At an altitude of 12km above the ground, the air pressure is significantly lower compared to the Earth's surface. To simplify the calculation, let's assume the pressure outside the bag is negligible (close to a vacuum) compared to the pressure inside the bag. Therefore, we can consider the bag as an isolated system.
When the bag is suddenly yanked back to Earth's surface, the volume of the bag remains the same, so we can consider it a constant in the equation. As a result, the equation simplifies to:
nRT = constant
Since the number of moles (n) and the volume (V) are constant, we can rewrite the equation as:
T1/T2 = P1/P2
Where:
T1 is the initial temperature of the gas (-35∘C = 238.15K)
T2 is the final temperature of the gas (what we want to find)
P1 is the initial pressure of the gas (inside the bag at 12km)
P2 is the final pressure of the gas (inside the bag at Earth's surface)
Since we assumed the pressure outside the bag is negligible, we can assume the pressure inside the bag would increase to the atmospheric pressure at the Earth's surface. The atmospheric pressure near the Earth's surface is approximately 1 atm or 101,325 Pa.
By plugging in these values into the equation:
T1/T2 = P1/P2
238.15K/T2 = P1/101325
Now, we need to find the initial pressure (P1) inside the bag at 12km above the ground. For this, we can use a simplified atmospheric pressure model, such as the barometric formula, which states that the atmospheric pressure decreases exponentially with altitude.
Using the barometric formula, we can approximate the pressure at 12km above the ground to be around 22,000 Pa.
Plugging this value into the equation:
238.15K/T2 = 22000/101325
Simplifying further, we can find:
T2 = 238.15K * (101325/22000)
Calculating this, we get:
T2 ≈ 1095K
So, if you suddenly yank the giant dry-cleaner's bag back to Earth's surface, its temperature would be approximately 1095 Kelvin.
Please note that this calculation is a simplified estimation and may have some uncertainties, but it should give you a reasonable estimate.