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Buoyant Force and air molecules

  1. Apr 30, 2009 #1
    1. The problem statement, all variables and given/known data
    It is hard to imagine that there can be enough air between a book and a table so that there is a net upward (buoyant) force on the book despite the large downward force on the top of the book. About how many air molecules are there between a textbook and a table, if there is an average distance of about 0.006 mm between the uneven surfaces of the book and table?

    Additionally, assume the book's area (width x height) to be 138 in^2 (converted suitably to SI units).

    Given from previous problems, density of air is 1.3 kg per m^3

    2. Relevant equations

    Fbuoy = pVg
    where Fbuoy is the buoyant force
    p is the density
    V is the volume
    and g is gravity

    3. The attempt at a solution
    Converted 138 in^2 to appropriate SI units. Came out to 0.08903208 m^2.
    V = (.006*10^-3 m) * (0.08903208 m^2) = 5.3419248*10-7 m^3

    Then....
    Fbuoy = (1.3 kg/m^3)*(5.3419248*10-7 m^3)*(9.8 m/s^2) = 6.805612195*10-6 N

    I'm not sure if I did that right and if I did, I don't have an equation to convert N to number of air molecules.
    I'm pretty sure I need Avogadro's number but I don't see how it ties in to converting.
     
  2. jcsd
  3. May 3, 2009 #2

    nvn

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    Science Advisor
    Homework Helper

    jtestuer: There is no net upward force on the book; otherwise, the book would be rising. The first sentence in post 1 appears to be a diversion. The problem doesn't ask for buoyancy force; it asks you to compute the number of air molecules. Assume air is an ideal gas. Look for the equation of state for an ideal gas, sometimes called the ideal gas law. Hint: It is a very popular formula, which some people know from memory. Solve it for number of kilogram-moles (kg*mol), n. Hint: Assume standard atmospheric pressure, p = 101 325 Pa. Assume a typical room temperature, in kelvin (K). R = universal gas constant = 8314.472 J/(kg*mol*K). Hint: Multiply n by Na = Avogadro constant = 6.022 141 79e26 molecules/(kg*mol).

    By the way, the air density given in post 1 is essentially wrong; perhaps another diversion. For typical indoor temperatures, at standard atmospheric pressure, air density is close to 1.18 or 1.19 kg/m^3, not 1.3.
     
  4. May 3, 2009 #3
    Thanks. That helped (even though the assignment was due a couple days ago :P ).

    I'll keep this in mind while studying for midterms. :D
     
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