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How is density linked to mass and mols?

  1. Apr 13, 2009 #1
    Apologies for numerous posts today, I'm trying to catch up from work I missed last term, and my uni is on holiday at the moment so cant get help from lecturers.

    I've got these two questions. I think I've correctly worked them out however I'm currently stuck on the last part. How is density linked to mass and mols?

    1. The problem statement, all variables and given/known data

    1 A gas cylinder contains 0.12 m3 of a gas at a pressure of 4545 kPa. What volume would it occupy if it was all released into a pressure of 101 kPa? Assume the temperature remains constant.

    2 A car tyre of volume 1.0 * 10-2 m3 contains air at a pressure of 300 kPa and at a temperature of 17oC. The mass of one mole of air is 2.9 *10-2 kg. Assuming that the air behaves as an ideal gas, calculate;
    i. the amount of air in moles,
    ii. the mass of the air,
    iii. the density of the air.


    2. Relevant equations

    pV=k

    pV=nRT

    3. The attempt at a solution

    1) pV=k

    4545kPa x 0.12m3=545.4

    Therefore:

    101Pa x V = 545.3

    V = 5.4m3

    2)i) Assuming ideal gas.

    pV=nRT

    n=[(4545x103Pa)(1x10-2m3)] / [(8.31Jmol-1K-1)(290.15K)]

    n=18.84995375mol

    ii) mass of gas = 18.84995375mol x 2.9x10-2

    =0.546648658

    =0.547 (3dp) kg

    iii) Please see above comments.

    Thanks,
     
  2. jcsd
  3. Apr 13, 2009 #2

    Kurdt

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    How do you normally work out density? You're given the mass of one mole and you've worked out how many moles there are.
     
  4. Apr 13, 2009 #3
    Density = Mass / Volume

    I've not really read through all of your work but seems right

    edit: sorry >.< i'll try to be more constructive next time
     
  5. Apr 13, 2009 #4
    Is it just the mass of the total gas / volume. So;

    Density = 0.546648658 / 1.0x10-2

    =54.66486586kg/m3

    Have I got the unit correct?

    Thanks
     
  6. Apr 13, 2009 #5
    correcto
     
  7. Apr 13, 2009 #6

    Kurdt

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    Yes units are fine.
     
  8. Apr 13, 2009 #7
    Brilliant. Thanks for the help.
     
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