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Thermodynamics (density)

  1. Dec 16, 2003 #1
    I'm stumped on some of the thermodynamic problems involving density.

    Heat a 1.0kg bar of lead at atmospheric pressure from 25C to 60C and find the work done by the lead. (density of lead = 1.13*10^4 kg/m^3)
    The only equations I was given was the Ideal Gas Law... which lead is not a gas.. Someone also told me that apparently work=pressure*volume, which also doesn't help. Density is molar mass/volume (I think), which does nothing in this case, since I basically already can determine both those components. Where do I start?

    [Answer=.02729J]

    A droplet of silver has a radius of .6mm. How many silver atoms are in the droplet. The density of silver is 1.05*10^4 kg/m^3
    This sounds simple, but I must be missing a conversion. I'm sure I'm suppose to multiply something by Avogradro's number, but not sure how to get the info from the density and radius.

    (Answer=5.3*10^19 atoms)
     
  2. jcsd
  3. Dec 16, 2003 #2

    ShawnD

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    Science Advisor

    I'm thinking you would need the coefficient for linear expansion for the lead. Then you would have the change in volume. Since the pressure is N/m^2. If you multiply the pressure by the change in volume (m^3), you get Nm which is J.

    find volume:
    v = (4/3)[tex]\pi[/tex]r^3
    v = (4/3)[tex]\pi[/tex](0.0006)^3
    v = 9.0478*10^-10 m^3

    find mass:
    (9.0478*10^-10 m^3) * (1.05*10^4 kg/m^3) = 9.5*10^-6 kg = 0.0095g

    use molar mass to find mols:
    0.0095g * 1mol/107.87g = 8.807*10^-5 mols

    use avogadro's number to find atoms:
    8.807*10^-5 mols * 6.02*10^23 atoms/mol = 5.30185*10^19 atoms


    Yep, still got it :wink:
     
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