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Homework Help: First law of thermodynamics: why some equations can't be used

  1. Jan 12, 2017 #1
    1. The problem statement, all variables and given/known data
    A cylinder fitted with a frictionless piston contains 5.0×10-4m3 of an ideal gas at a pressure of 1.0×105 Pa and temperature of 300K.
    The gas is then
    (i) heated at constant pressure to 450K, and then
    (ii) cooled at constant volume to the original temperature of 300K. The heat extracted in this stage is 63J.

    Calculate the total heat input in stage (i)

    2. The attempt at a solution
    Ok, so firstly, I determined the work done by the gas in stage (i), which is 25J. (this is the correct answer)
    After, as we know that internal energy of an ideal gas U = (3/2)PV, I thought I could find the change in internal energy ΔU of stage (i) by using ΔU=(3/2)pΔV, then use the first law of thermodynamics ΔU=Q+W to determine the heat input. However, my answer was wrong. :nb)

    Apparently you can't do that. And instead, the answer key made use of the total change in internal energy from stage (i) to (ii) instead. However, I am really confused, as now I am unsure of when exactly I should or should not make use of the equation U=(3/2)pV

    Would really appreciate some help! Thank you!:smile:
  2. jcsd
  3. Jan 12, 2017 #2


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    Science Advisor

    That's only for monatomic ideal gases. Do you know how U varies with T for molecules with degrees of freedom other than translation? You don't actually need to know this; you can get Cv from part (ii), work out Cp and use it in part (i).
  4. Jan 12, 2017 #3
    From the problem statement, how many moles of gas do you have?
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