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Enthalpy change in ideal gas

  1. Feb 12, 2014 #1
    1. The problem statement, all variables and given/known data

    Argon is heated from T1 to T2. Assume ideal gas behaviour. Calculate the energy input. Assume a constant volume, mass and constant specific heat capacity. Pressure variable

    T1=523.15 K
    T2=823.15 K
    M= 0.03995 kg
    m=50 kg
    R=8.31
    Cv=0.0125 kJ/(mol.K)

    2. Relevant equations

    Enthalpy change in gas
    ΔH = ΔU + nRΔT

    ΔU=CvΔT
    n= m/M
    ΔH=CvΔT+ (m/M)RΔT

    3. The attempt at a solution

    ΔH=((0.0125)*(300))+(50/0.03995)*( 8.31)*(300)…kj

    Is this equal to the heat energy input?

    Would this equal the amount of energy it would take to cool the gas, i.e. the reverse process?
     
  2. jcsd
  3. Feb 12, 2014 #2

    ehild

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    Why do you use enthalpy when the change of energy is the question? And the volume is constant, the pressure changes. Your equation for ΔH is not correct.


    ehild
     
  4. Feb 12, 2014 #3

    BvU

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    General advice, well meant: Always hang on the dimensions until done. Anything physical has a dimension (unless expressed as a ratio, and often even then!).
    1. You catch mistakes in expressions a lot earlier
    2. You catch unit conversion problems a lot earlier
    3. You catch things that cancel each other a lot earlier
    4. Accuracy improves almost automatically
    5. You need less time to re-calculate after discovering an error on the way
    6. You score more points if you nevertheless make a calculation error

    And a few more things. In this case, you would have prevented ehilds pointing out that ΔH expression is wrong, because you would have seen it yourself already. On two counts:

    You would have seen that Cv given is the molar heat capacity, not the specific heat of your (huge) sample of Argon, which is unfortunately the Cv in ΔU = CvΔT.

    You would also have questioned the second term, or rather the combination of that and the ...kj (please use kJ, not kj) -- and then corrected with a factor of 1000.


    And this good practice of keeping track of the units begins with summing up the given data:

    T1=523.15 K
    T2=823.15 K
    M= 0.03995 kg/mol
    m=50 kg
    R=8.31 J/(mol.K)
    Cv=0.0125 kJ/(mol.K)

    Your rendering of the original problem gives me the impression that this is only part of the whole exercise, but I can't be sure. It does break off somewhat strangely.

    Reason: even in the 1.5 line of the problem statement there is too much already, so you have to select the useful stuff.

    You either have learned already, or will learn soon (for instance in this exercise, if it is as long as I suspect), that H is useful in dealing with constant pressure cases, because its d/dT includes pV work. U, on the other hand, is very useful when no work is done, which is often the case in constant volume cases. There dU = dq + dw becomes dU = dq and with your ΔU=CvΔT equation you are fully in business -- provided you keep a keen eye on the units!
     
  5. Feb 16, 2014 #4
    Thanks for the help.

    Ok, so I was using the incorrect equations for constant pressure. ΔU=CvΔT should give me an answer for the work done.

    The Cv here being the molar heat capacity, so this will give me a solution for the work input for each mole? To find the total work input can I just multiply for the total amount of moles in the system?
     
  6. Feb 16, 2014 #5

    BvU

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    Sounds reasonable, doesn't it ?
     
  7. Feb 17, 2014 #6
    For
    ΔT=300 K
    Cv=0.0125 kJ/mol.K
    Atomic Mass= 0.03995 kg
    Mass=50 Kg

    ΔU = CvΔT

    ΔU=(0.0125*300)*(50/0.03995)

    ΔU=4639.367 kJ

    Is this a resonable estimate for the energy required to heat up 50 kg of this gas? Also will the energy needed to cool the gas be the same?
     
  8. Feb 18, 2014 #7

    BvU

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    Estimate ? It is a calculation ! By the way, with the mass given in 2 decimals, I wouldn't express the energy in 7 decimals. An answer like 4.64 MJ would be more than enough accuracy.

    But I am not giving you enough credit for your careful attitude: it is indeed very wise to ask yourself this kind of questions. It helps you find small things like mixing up orders of magnitude in units, mistyping on the calculator, using a calculator on radians setting instead of degrees, etc.

    So take 50 kg of water, apply 4.6 MJ of heat. Heats it up how much ? But that specfic heat capacity (in kJ/kg) is about 13 times higher. I find that in checking this I make about all hurried mistakes that are possible (and then correct them). By now I have a solid faith in your answer!
     
  9. Feb 18, 2014 #8
    In your problem, there is no work done because the volume of the system is constant. From the first law of thermo, that means that ΔU=CvΔT=Q, where Q is the (heat) energy input.

    Chet
     
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