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.
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?
3 answers
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.
1 answer