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Homework Help: First Law of Thermodynamics and temperature

  1. Oct 19, 2013 #1
    1. The problem statement, all variables and given/known data

    A toy balloon is initially at a temperature of T = 20°C. Its initial volume is 0.0042 m3 (It is a 10 cm radius sphere). Assume that the pressure in the balloon always equals atmospheric pressure, 101.3 kPa.
    The new temp was found to be 52.23°C
    By how many Joules did the internal energy of the gas in the balloon increase? Air is almost entirely diatomic gas.
    ΔU =?

    2. Relevant equations

    3. The attempt at a solution


    I keep getting 1.17e-4, and it says i am off by a power of 10. I am not sure where I am messing up
  2. jcsd
  3. Oct 19, 2013 #2
    What is your equation for change in internal energy. I am trying to work this problem but I get a much different answer so I am probably using the wrong equation.

    What did you get for n, the number of moles?
    Last edited: Oct 19, 2013
  4. Oct 19, 2013 #3


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    What did you get for N?

  5. Oct 19, 2013 #4
    it is the degrees of freedom (5/2 for diatomic)*(Temperature)*(K which is a constant 1.381e-23)*(# of molecules)
  6. Oct 19, 2013 #5
    So, what did you get for N or n?
  7. Oct 19, 2013 #6
    n= 1.746e-7
  8. Oct 19, 2013 #7
    I think your value for n is incorrect. Did you get it from the equation

    PV = nRT

  9. Oct 19, 2013 #8
    Yes. I did 1*.0042e-3/(.08206*293)..converting the kPa's to atms and Celsius to k
  10. Oct 19, 2013 #9
    Using PV = nRT I have always used P in Pascals, V in meters^3 n in moles R = 8.32 and T = 293.

    If you use P in atm, then is the constant 0.082 correct?
  11. Oct 19, 2013 #10
    yes i think so
  12. Oct 19, 2013 #11
    My previous post is incorrect. I should have said the volume is in liters, not meter^3
  13. Oct 19, 2013 #12
    right, and I think that m^3=mL=.001L
  14. Oct 19, 2013 #13
    So when I change my calculation from m^3 to liters I get an answer closer to yours but not exact.

    so, to find n, we use pressure in Pascals, volume in liters, R = 8.3, and K = 293.
    so what do you get for n?
  15. Oct 19, 2013 #14
  16. Oct 19, 2013 #15
    I agree. Now, change in internal energy is (5/2)Rn(delta T)?
  17. Oct 19, 2013 #16
    I believe so
  18. Oct 19, 2013 #17
    I get (5/2)(8.31)(1.74E-4)(32.2) = I can't give the answer
  19. Oct 19, 2013 #18
    I get the same message with that answer that I do with my previous one. "It appears you have made a power of ten error"
  20. Oct 19, 2013 #19


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    In PV = nRT, [P] = N/m2, [n] = mol, [T] = K and with R = 8.31, [R] = J K-1mol-1 => [V] = J m2/N = m3, no?
  21. Oct 19, 2013 #20
    Are you using Quest? sometimes their answers are wrong. I get .116 if I use the equation above or if I use your original equation. I think you made a math error when you got 1.17 E-4.
  22. Oct 19, 2013 #21
    I am not 100% sure about your units for P and V (it may just be notation i am not familiar with)
    I usually see [P]=atm or pascals, [V]=L (m^3=mL)
  23. Oct 19, 2013 #22
    It's smart physics
  24. Oct 19, 2013 #23
    I think we agree on n = -1.74E-4, the units of p and v are done with.

    This means that N = (1.74E-4)(6.022E23) =1.04E20

    Just recalculate your first answer. I think you made a math error.

    1.17E-4 is not correct didn't you get .117?
  25. Oct 19, 2013 #24
    Here is everything I am doing

    =(5/2)*(1.381e-23)*(32.23)*( (1*.0042)/(.08206*293) )* (6.022e23)
  26. Oct 19, 2013 #25
    I don't recognize some of your constants and conversions but.....

    Take your equation from post #1, 5/2*(52.23-20)*1.381e-23*N and plug in the value for N

    N = (1.74E-4)(6.02E23) = 1.04E20 and what do you get? .116 or .117 I think this is the correct answer.

    Check your math and we will agree even though smart Physics says we are off. I don't believe it.
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