posted by Anonymous .
You are completing tests using a weather balloon and want to check the temperature at 3000 meters into the air. You check if the balloon will not burst before it gets to that altitude. You will need to perform a few calculations. The manufacturer of the balloon guarantees it up to 47.0 Liters in size. Will the balloon make it to the mentioned altitude without bursting? The temperature at ground level is 20 C. The pressure is 765 mmHg. The volume of the balloon prior to release is 30.0 Liters. It is filled with helium. Use PV = nRT for both the calculations below.
1. Calculate the number of grams of helium in the balloon.
2. Find if the balloon can reach 3000 meters without bursting. The temperature and pressure at 3000 meters was measured at Temp = 6.0 C, Pressure = .565 atm
I found the number of grams to be 5.0251. However, I don't see how the grams can be used to answer the second question. Is there something I'm missing?
n = 1.25 and 5.025 g He is correct for #1.
PV = nRT
P is 0.565 atm
T is 6.0. Convert to kelvin first.
n from above is 1.25
R you know
Solve for V and compare it with the manufaturer's value of not bursting up to 47.0 L
So the grams are not needed to answer the second question?
Yes grams are and no grams are not. You calculated grams in part 1 by using PV = nRT, then since n = grams/molar mass, you substituted molar mass of 4 for He and calculated grams.
For part 2, you used PV = nRT again and if you wanted to use grams, then mols = n = grams/molar mass (n = 5.02/4) and you substitute the n value for mols BUT since you already had mols of 1.25 from part a, why go through the extra step.