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Calculate the heat absorbed

  1. May 10, 2014 #1
    1. The problem statement, all variables and given/known data

    An ideal gas(γ=1.4) was expanded under constant pressure. The work done was 80 Joule.Calculate the heat absorbed and increase in internal energy of the system.
  2. jcsd
  3. May 10, 2014 #2
    Well, what's the equation for a gas undergoing expansion under constant pressure?
    [itex]w= P \int dV[/itex]
  4. May 10, 2014 #3
    what is the solution?
  5. May 11, 2014 #4
    I'm not just going to give you the solution. You won't learn anything. You need to show some work and initiative and I can guide you through it.

    Start by thinking about this: is the change in energy path dependent or path independent?
  6. May 11, 2014 #5
    W = Q-ΔU is known from from first law of thermodynamics and w is given here . Now how shall i proceed using γ=1.4 ? how to apply your equation ?
    please help.
    Last edited: May 11, 2014
  7. May 11, 2014 #6
    Is this calculus based physics or algebra based?

    See this is why you need to show some work, I have no idea at what level you're doing this. If the integral is what's confusing you then simply change it to: [itex]w = P \Delta V[/itex].

    You want to find [itex]Q_{in}[/itex] for a constant pressure problem.

    [itex]\Delta E = W + Q[/itex]

    We know that w= -80J. Because the system did work on the piston it has to be negative - energy left the system.

    [itex]\Delta E + 80 = Q[/itex]

    [itex]\gamma = 1.4[/itex] Therefore you know it's a diatomic gas.

    And I actually get stuck there too.. Hmm. You aren't given either moles of gas or change in temperature? With your gamma you can find Cp and Cv but you don't seem to have enough information.

    [itex]Q = nC_{p} \Delta T[/itex]

    and [itex]\Delta E = nC_{v} \Delta T[/itex]

    Edit: It's ambiguous if they mean that the work done was by the system or on the system. Is this the whole original question?
    Last edited: May 11, 2014
  8. May 11, 2014 #7


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    Divide these two equations and you will get Q/ΔE = cp/cv = γ = 1.4. Along with the equation from conservation of energy, you now have two equations with two unknowns.
  9. May 11, 2014 #8
    Ah! Genius. I would never have thought of that :).
  10. May 13, 2014 #9
    well done. Thanks to both of you.
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