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Calculating Thermodynamic Heat and Work

  1. Aug 18, 2013 #1
    1. The problem statement, all variables and given/known data

    One mole of ideal gas expands isothermally at 20°C against a constant pressure of 1 atm from 20 to 30L. (1 atm = 1.013 x 105Pa; 1 m3 = 1000L

    Find q, w, ΔE, and ΔH in Joules

    2. Relevant equations

    ΔE = q + w
    w = -[itex]\int[/itex]PdV
    H = E + PV

    3. The attempt at a solution

    I solved for work using the second equation listed above. Since pressure is constant, I removed it from the integral and simply integrated the volume from 20L to 30L. This resulted in -10 atm x L.

    Converting that to Joules using the conversion factors given, I obtained -1013 J as the thermodynamic work.

    But then how do I obtain the thermodynamic heat?
     
  2. jcsd
  3. Aug 19, 2013 #2

    DrClaude

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    Staff: Mentor

    You need to figure out the change in the internal energy of the gas, ##\Delta E##, considering it is ideal.
     
  4. Aug 20, 2013 #3
    Okay, so what I'm getting is that since it's an isothermal process, the change in internal energy is 0.

    That would mean q is just the reverse of w. Does this sound right? That would mean this is reversible too, yes?
     
  5. Aug 20, 2013 #4

    DrClaude

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    Staff: Mentor

    Yes. Just to be clear, the energy lost by the gas as it does work is compensated by heat coming in from the environment.

    I guess that it is reasonable to assume that this is the result of a slow (quasistatic) process, for the temperature to be maintained constant during the entire process, and therefore it is reversible.
     
  6. Aug 20, 2013 #5
    Ah yes, thank you.

    I have one more item I'd like to clarify.

    I have two equation for thermodynamic work listed in my notes.

    1. w = -∫PdV
    2. w = -nRT ln[itex]\frac{V_{2}}{V_{2}}[/itex]

    I used the first equation to obtain -1013J, but the second equation provides a different value of -987.7J. This is a pretty significant difference, so I'm curious what the differences between the equations are, if any.
     
  7. Aug 20, 2013 #6

    DrClaude

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    Staff: Mentor

    I didn't pay enough attention to the original statement. The situation described is simply impossible: an ideal gas cannot expand when both temperature and pressure are held constant. Obviously, since ##PV = nR T##, ##V## can't change if ##P##, ##T##, and ##n## are all kept constant.

    Otherwise, you get equation 2 from equation 1 by replacing ##P## by ##n R T/V##, and then integrating over ##V## for ##T## constant. Equation 2 is therefore useful for isothermal processes (for an ideal gas only).
     
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