Chemistry - Ideal Gas Law
posted by Black Reaper on .
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.
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.