How Many Helium Atoms Fill a 24 cm Balloon at 76°C and 0.789 atm?

[Punjabi.Sher]

I tried this problem a lot of times, but just didnt get the answer.

The molecular mass of helium is 4 g/mol,
the Boltzmann’s constant is 1.38066 × 10−23 J/K, the universal gas constant is
8.31451 J/K · mol, and Avogadro’s number
is 6.02214 × 1023 1/mol. Given: 1 atm =
101300 Pa.

How many atoms of helium gas are required
to fill a balloon to diameter 24 cm at 76◦C and
0.789 atm?

What I did was find the volume and then use pv = nrt to find the moles. From the moles I found number of atoms. But it was wrong. Help Please.

 

[Ygggdrasil]

Can you show your work? Your approach sounds like it is correct, and it's hard to see where you went wrong if you don't show us what you did.

 

[Punjabi.Sher]

pv = nrt

0.789 atm (4/3*3.14*12^3) = n (0.082)(349.15 K)
n = 199.47 moles = 1.20e26 atoms

 

[Ygggdrasil]

Check the units on your calculations. You'll see where you went wrong.

Remember, always include units when you do these types of calculations. They can help avoid errors like this.

 

[Punjabi.Sher]

V = 7238.23 cm^3 x (0.001 L / 1 cm^3) = 7.24 L

0.789 atm (7.24 L) = n (0.082 L·atm·K−1·mol−1) (349.15 K)

n = 0.1995 mol x (6.022e23 / 1 mol) = 1.20e23 atoms

Still this is not the right answer I think.

 

[Ygggdrasil]

That's the answer that I get when I do the calculation.