Kinetic Energy of gas Molecules

  1. Hi, the question, I'm having problems with is this:

    A 1.0 mol sample of hydrogen gas has a temperature of 30 C. What is the total kinetic energy of all the gas molecules in the sample? How fast would a 75 kg person have to run to have the same kinetic energy?



    Relevant equations:
    K = (3/2)kT
    or K = (3/2)nRT




    Ok, so I tried using both formulas:

    (3/2)(1.38*10^-23)(30+273) = 6.2721 * 10^-21

    Then, since its one mole and the question asks for the total KE, I multiplied this answer by Avogadro's number (6.02*10^23) and got

    = 3775.8042 Joules.

    For the second part, I would just set this number equal to 0.5mv^2 and solve for v but since I got the first one wrong, it doesn't quite work :(


    I really need help; I thought I had the right idea but it's not working out. Thanks
     
    Last edited: Dec 7, 2008
  2. jcsd
  3. Andrew Mason

    Andrew Mason 6,856
    Science Advisor
    Homework Helper

    Your method is right. Using [itex]U = \frac{3}{2}nRT[/itex], U = 1.5*8.314*303 = 3778.7 Joules.

    Think of one mole or 1 gram of hydrogen possessing 3778.7 J of kinetic energy. The effective average (rms) velocity is:

    [tex]v = \sqrt{2E/m}[/tex]

    AM
     

  4. I thought my method was right but my homework web-assign isn't accepting that answer :(
    It asked for it in 2 sig figs so i put it in as 3.8 *10^3 but it still said no...
     
  5. Oh I figured it out lol, it needed (5/2)kT to work and then multiply by Avogadro's number.
     
  6. Andrew Mason

    Andrew Mason 6,856
    Science Advisor
    Homework Helper

    Of course. H2 is diatomic.

    AM
     
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