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Homework Help: Stuck on Thermodynamics

  1. Jul 30, 2005 #1
    These problems are from Giancoli 5th edition (principles with applications)

    A cubic box of volume 0.039 m^3 is filled with air at atmospheric pressure at 20 celsius. The box is closed and heated to 180 celsius. What is the net force on each side of the box?

    I first used the ideal gas law to figure out the pressure after the box is heated. So P1/T1 = P2/T2 because the box is closed and the volume doesn't change, and nR is constant. I then got P2 = 1.56 * 10^5 N/m^2 - but I'm not sure how to use this to figure out the forces on the sides of the box

    Calculate approximately the total translational kinetic energy of all the molecules in an E.Coli bacterium of mass 2.0 * 10^-15 kg at 37 celsius. Assume 70% of the cell, by weight, is water, and the other molecules have an average molecular weight on the order of 10^5.

    I don't know how to start this problem - Giancoli shows us how to calculate the average KE and how it's proportional to temperature, but not total translational KE. I'm not sure how to use the composition of the E.Coli cells either.

    Any help is greatly appreciated! Thank you
  2. jcsd
  3. Jul 30, 2005 #2
    force = pressure x area.
    force on each side = pressure x area on each side of cube.
    in kinetic model, T is proportional to KE.
    precisely KE = (3/2)NkT for total N particles.
    an e.coli consists of water and other molecules, as you said.
    your job is to find N.
    mass of water in one e.coli = 0.7 * mass of e.coli.
    we can get number of mole of water from this.
    divide the mass by molecular mass of water (i think 18), you get mole.
    thus you can get N for water in one e.coli. (avogadro)
    the rest of the molecules you said have mass of the order 10^5.
    what does it mean ? look at the book again. write the complete sentence.
    i hope you make mistake in this last figure, because it does not look sensible to me.
  4. Jul 30, 2005 #3
    Thanks sniffer

    The answer to the first question was 6400 N, but I'm getting something different. Since Volume is 0.039, I should take the cube root of that number to figure out the length of one side of the cube, right? Then to find the area of one of the faces of the cube, I square the resulting number (Area of one face is about 0.115 m^2). So now, Force of one side = Pressure of one side * Area of one side. The pressure inside this box is 1.56 * 10^5, and there are 6 sides inside the box. So the pressure on one side is (1.56 * 10^5) /6. Multiplying that by 0.115 m^2 gets me 2990 Newtons

    For the second one, I found there were 7.78 * 10^-14 moles of water, meaning there are 4.69 * 10^10 molecules of H2O. The part about order of 10^5 is exactly quoted from the textbook, so I assumed this meant that the molecular weight of the other part of the cell is 1 * 10^5 kg. So, there are 6 *10^-21 moles of the other parts, or 3612 molecules. So together there are 4.69 * 10^10 + 3612 molecules. Multiplying by Boltzman's constant 1.38 * 10^-23 and then multiplying by temperature, 310 K gets me 2 * 10^-10 Joules, when the answer is 3 * 10^-10 Joules

    Thanks for the help
  5. Jul 31, 2005 #4


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    Homework Helper

    for first
    What about Pascal' law. Pressure will be same for all sides and equal to 1.56*10^5,Nm^-2

    For net force dont forget the force due to atmosphere.
  6. Jul 31, 2005 #5
    ooo... ya correct. atmospheric pressure!
    thanks for mukundpa

    so resultant pressure is 1.56*10^5 - 1*10^5 = 0.56 * 10^5 Pa.
    the force on each surface is 0.56 * 10^5 * 0.115=6400 N (2 significant figures).

    the last bit is ok, but what about multiplying with 3/2, as i said in the formula? EK=(3/2)NkT.

    it gives you exactly the asnwer.
  7. Jul 31, 2005 #6


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    Homework Helper

    Air is a mixture, but the experimental value of Cp/Cv is 1.4 gives the idea that it may be considered as diatomic wigh degrees of freedom 5 and hence it should be (5/2)NkT but still in some of the books I have seen (3/2)NkT for diatomic gases which does not appears correct to me. Any expert view will be welcomed.
    Last edited: Jul 31, 2005
  8. Jul 31, 2005 #7
    Oooh, I got the answers now. Thanks to you both for the help!
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