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Understanding Moles From Ideal Gas Law

  1. Feb 25, 2013 #1
    Hi All,

    I would like to understand a little more about the following equation

    [tex] moles = \frac{Mass(g)}{Molar Mass}[/tex]

    I know that this is to calculate the Moles for use in the ideal gas law equation

    [tex] PV=n \times R \times T [/tex]

    Lets say I had to find two variable in the ideal gas law one being Pressure the other moles. If I know that the volume of a tank is 120L of Air and the molar mass is 16 is this enough to work out this equation??

    [tex] moles = \frac{Mass(g)}{Molar Mass}[/tex]

    [tex] moles = \frac{120000}{16}= 7500[/tex]

    Now can this be substituted back into the ideal gas equation with the following

    V = 120L
    n = 7500
    R = 0.0821 L.Atm/mol.K
    T = 303 K

    [tex] P=\frac{7500 \times 8.314 \times 303}{120} = 1554 kPa [/tex]

    Is this correct?
     
  2. jcsd
  3. Feb 25, 2013 #2

    jtbell

    User Avatar

    Staff: Mentor

    How do you figure that 120 L of air has a mass of 120000 g (120 kg)? That seems a bit heavy to me. :uhh:
     
  4. Feb 25, 2013 #3
    No, isn't.

    When you say 120L you're talking about volume, not about mass.
     
  5. Feb 25, 2013 #4
    Ok yes you are correct that is volume. Does this mean that the equation is correct??

    Put another way if it was 20 Liters of water

    20Litre = 20 Kg = 20 000grams then

    [tex] Moles = \frac{20 000}{16 }= 1250 [/tex]

    I guess I am trying to get my head around what the weight of air is when compressed into a 120L cylinder and if this is proportional to the volume of the cylinder.
     
    Last edited: Feb 25, 2013
  6. Feb 25, 2013 #5
    The ideal gas law applies to gas, not to liquid. Liquid water may weigh 1000 kg per liter, but gaseous water weighs much less. If you know the volume of the tank, you also need to know the pressure and the temperature in order to calculate the number of moles of air in the tank (using the ideal gas law). Once you know the number of moles, you can calculate the mass of air in the tank by multiplying by the molar mass. Incidentally, the molar mass of air is 29 (a weighted average of the molecular weight of oxygen and nitrogen), not 16.
     
  7. Feb 25, 2013 #6
    I have been doing some more digging and would like to test my theory

    The molar mass of air is a make up of 21% Oxygen & 78% Nitrogen

    [tex] O_2 = 2\times 16 = 32 g/mol[/tex]
    [tex] N_2 = 2\times 14 = 28 g/mol[/tex]

    Therefore
    [tex] O_2 = 32 g/mol \times 0.21 = 6.7 g/mol[/tex]
    [tex] N_2 = 28 g/mol \times 0.78 = 21.84 g/mol [/tex]
    [tex] Air = 21.84 + 6.8 g/mol = 28.59 g/mol[/tex]

    If I have a volume in a tank of 120L this works out to be 120000grams and I have the Air Molar mass can I now calculate the moles?

    [tex] Moles =\frac{120000 grams}{28.59 grams/mole} = 4197 moles [/tex]

    If this is correct then I should be able to substitute any mass in there to work out the moles.

    I then should be able to work out the ideal gas law if I want to find the pressure
    lets say V = 120L, R= 0.0821atm.L/K.Mol T =303K n =4197 mol
    [tex] P = \frac{n \times R \times T}{V}[/tex]
    [tex] P = \frac{4197 \times 0.0821 \times 303}{120} [/tex]
    = 870atm or 88149 kPA or 12785 PSI

    Any guidance is appreciated
     
  8. Feb 25, 2013 #7
    Several people (including myself) have already said that the mass of gas in 120 L is not generally 120 kg. Why do you keep assuming that it is? In order for air to have that density, you have shown in your calculations that its pressure would have to be 870 atm. So what? In addition, this is outside the ideal gas region for air, so the estimate of 870 atm. is inaccurate.
     
  9. Feb 25, 2013 #8
    Chestermiller,

    I did not see this reply when I posted my last post so this was not intentional.
     
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