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.
#2.
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.
To answer the second question regarding whether the balloon can reach 3000 meters without bursting, you need to consider the ideal gas law equation (PV = nRT) and compare the initial volume of the balloon to the volume at 3000 meters.
Firstly, let's calculate the number of moles of helium gas in the balloon using the given information. We can rearrange the ideal gas law equation to solve for n (number of moles):
n = PV / RT
where:
P = pressure in atmospheres
V = volume in liters
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (20 °C = 293.15 K)
Using the initial volume of the balloon (30.0 L) and the given pressure (765 mmHg = 1.011 atm), we can calculate the initial number of moles:
n1 = (1.011 atm * 30.0 L) / (0.0821 L·atm/(mol·K) * 293.15 K)
Calculating this yields n1 = 1.351 mol.
Next, we need to calculate the volume at 3000 meters altitude. The temperature at 3000 meters was measured at 6.0 °C = 279.15 K, and the pressure was measured at 0.565 atm. Since the balloon is assumed to be at the same temperature and pressure as the surrounding air at that altitude, we can use those values.
To find the final volume (V2) at 3000 meters, we rearrange the ideal gas law equation again:
V2 = n2RT / P2
where:
P2 = pressure at 3000 meters altitude in atmospheres
n2 = number of moles (which remains constant)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (279.15 K)
Using the given pressure at 3000 meters (0.565 atm), we can calculate the final volume of the balloon:
V2 = (1.351 mol * 0.0821 L·atm/(mol·K) * 279.15 K) / 0.565 atm
Calculating this yields V2 = 37.9 L.
Now compare the final volume (V2 = 37.9 L) to the manufacturer's guaranteed volume (47.0 L). If the final volume is less than or equal to the guaranteed volume, then the balloon will not burst. In this case, the balloon will make it to the mentioned altitude without bursting.
Therefore, the answer to the second question is yes, the balloon can reach 3000 meters without bursting.