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Thermodynamics - Change in Density due to Change in Height

  1. Feb 20, 2010 #1
    1. The problem statement, all variables and given/known data

    A column of water contains fine metal particles of radius 20nm, which are in thermal
    equilibrium at 25°C.

    The density of the metal is [tex]2\times10^{4} kg m^{-3}[/tex].

    If there are 1000 particles per unit volume at given height, what will the particle density per unit volume be at a position of 1 mm higher?

    2. Relevant equations

    Stated within the question / solution attempt.

    3. The attempt at a solution

    For each particle:

    Radius = [tex]20\times10^{-9}m[/tex]

    Hence Volume = [tex]\frac{4}{3} \pi (20\times10^{-9}m^{3} = 3.351\times^{-23}m^{3}[/tex]

    Therefore since density is mass divided by volume, [tex]\rho = \frac{m}{V}[/tex]:

    [tex]m_{total} = \rho m = (2\times10^{4} kg m^{-3})(3.351\times^{-23}m^{3}) = 6.702\times^{-19}kg [/tex] which is the total contained mass.

    It is given that at any height there are 1000 particles, therefore:

    [tex]m_{particle} = \frac{6.702\times^{-19}kg }{1000} = 6.702\times^{-22} kg[/tex]

    I thought this would probably help somehow, not sure how anymore though.
     
    Last edited: Feb 20, 2010
  2. jcsd
  3. Feb 21, 2010 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    Ask yourself, 'What's keeping the metal particles from falling to the bottom?'.
     
  4. Feb 21, 2010 #3
    .. little more help? :blushing:
     
  5. Feb 22, 2010 #4
    1. Using known radius of a particle have found the volume of a particle.
    2. Using this value of volume, and the known pressure, have found the mass.
    3. Using this value of mass, and known values of g and h, have found the energy:

      [tex]E = mgh = (m)(9.81)(1\times10^{-3}) = 6.6\times10^{-21}[/tex]

    4. Using Boltzmann constant [tex]k_{B}[/tex] and T (converted into Kelvin), can now put all these values into the Stefan-Boltzmann equation:

      [tex]n = n_{0}exp\left(\frac{-E}{k_{B}T}\right) = (1000)(0.202) = 202[/tex]


    5. Therefore at position of 1mm higher, the particle density is: 202 particles per unit volume.

    .. hopefully correct :smile:
     
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