You have a weather balloon that is filled with Helium (He) and has a diameter of 3 ft. Find the mass in grams of the He in the balloon at 21 degrees Celsius and normal pressure. And the density of He with these conditions is .166 g/L.
My Work:
V=4/3 * pi * r^3 V = 14.1 ft^3 * (28.3 L/1 ft^3) = 399 L
T = (21 + 273)K = 294 K
R = .0821 L*atm/(K*mol) ---> Not sure if this is correct value to use.
P = 1 atm ---> Not sure if this is correct value to use.
M = 4.00 g/mol
Using ideal gas law
m = (PMV)/(RT)
m=(1 atm * 4.00 g/mol * 399 L)/(.0821 L*atm/(K*mol) * 294 K)
m = 66.1 g Helium
How does this look?
It's close to what I have but with a lot more work. I think you have converted to STP but I think the problem is asking for mass He at the temperature of 21 C and not 0 C.
Here is what I did. Check me out on this.
First, I didn't know the conversion for cu ft so
3 ft x (12 inches/ft)x (2.54 cm/in)/2 = radius = 45.72 cm
V = (4/3)*pi*(45.72)3 = 400,319 cc = 400.32 L
The problem states that the density under these conditions; i.e., normal pressure and 21 C) is 0.166 g/L so
400.32 x 0.166 = 66.453 which I rounded to 66.4 L.
oops. That's 66.4 grams, not L.
Oh so technically the ideal gas law wasn't needed. That makes more sense as to why the density was put in the problem.
Your work looks almost correct, but there are a few adjustments you can make to obtain the correct answer.
First, let's correct the volume conversion. You correctly calculated the volume of the balloon as 14.1 ft³. However, to convert this to liters, you should multiply by the conversion factor of 28.3 L/1 ft³. This gives us:
V = 14.1 ft³ * (28.3 L/1 ft³) = 399.03 L (rounded to 3 decimal places)
Next, you correctly converted the temperature from Celsius to Kelvin by adding 273. However, it seems that you made a mistake with the gas constant, R. The value of the ideal gas constant R is actually 0.0821 L*atm/(K*mol). So, you can use this value in your calculations.
Furthermore, you used the correct standard pressure of 1 atm for the ideal gas law equation, so no adjustment is needed there.
Finally, it seems you have the correct molar mass for helium, which is 4.00 g/mol.
Applying the ideal gas law:
m = (P * M * V) / (R * T)
m = (1 atm * 4.00 g/mol * 399.03 L) / (0.0821 L*atm/(K*mol) * 294 K)
Calculating this expression will give you the correct mass of helium in grams.