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Gases and avogadros number

  1. Sep 6, 2008 #1
    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.
     
  2. jcsd
  3. Sep 6, 2008 #2

    Ygggdrasil

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    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.
     
  4. Sep 6, 2008 #3
    pv = nrt

    0.789 atm (4/3*3.14*12^3) = n (0.082)(349.15 K)
    n = 199.47 moles = 1.20e26 atoms
     
  5. Sep 6, 2008 #4

    Ygggdrasil

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    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.
     
  6. Sep 6, 2008 #5
    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.
     
  7. Sep 6, 2008 #6

    Ygggdrasil

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