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Adiabatic, isobaric & isothermal expansions

  1. Mar 2, 2012 #1
    1. The problem statement, all variables and given/known data
    An ideal gas of N particles is reversibly expanded from V1 to V2 = 4V1. The starting temperature T1 is known and [itex]E_1 = \frac{3}{2} N k_B T_1[/itex]. As of yet [itex]E_2[/itex] is unknown.

    a) Express [itex]\Delta E = E_1 - E_2[/itex] in terms of the added heat Q and work done on the gas W.

    b) Calculate the work done W if the expansion is isobaric.

    c) Calculate Q, W and [itex]E_2[/itex] if the expansion is isothermal.

    d) Calculate Q if the expansion is adiabatic.

    e) If the expansion in question d) was irreversible instead of reversible, would the added heat be less, more or equal?

    3. The attempt at a solution

    a) The gas does positive work, so the work done on the gas is negative: [itex]\Delta E = Q - W[/itex]

    b) For an isobaric expansion: [itex]W = p(V_2 - V_1) = 3V_1p[/itex]

    c) [itex]dW = pdV[/itex], so [itex]W = \int_{V_1}^{V_2} \frac{N k_B T_1}{V} dV = N k_B T_1 \ln{\frac{4 V_1}{V_1}} = N k_B T_1 \ln{4}[/itex]

    [itex]Q = W = N k_B T_1 \ln{4}[/itex]

    [itex]dU = 0[/itex], so there is no change in energy: [itex]E_1 = E_2 = \frac{3}{2}N k_B T_1[/itex]

    d) In an adiabatic expansion Q = 0.

    e) If the expansion was irreversible, then heat was lost on friction or something else, so to make up for that, the added heat would have to be more than 0.

    Can someone check these answers? I am doubting most of them.
  2. jcsd
  3. Mar 2, 2012 #2


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    b) Give W with N and V1.

    e.) "Adiabatic" means no heat exchange. Does it depends if the process is reversible or irreversible?

  4. Mar 2, 2012 #3

    Andrew Mason

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    Careful! It says W = work done ON the gas. The first law is:

    Q = ΔU - W where W = work done ON the gas (ie. W = -∫PdV if process is reversible).

    Are you finding the work done ON the gas? or work done BY the gas? Can you express this in terms of T1?

    Correct if W = work done BY the gas. But in a) it says W = work done ON the gas.
    If the expansion is still adiabatic, why would there be any heat flow at all?

  5. Mar 2, 2012 #4
    Thanks for the help.

    a) So this should be [itex]\Delta E = Q + W[/itex], I understand why.

    b) Can I simply say [itex]W = -p(V_2 - V_1) = -3V_1p = -\frac{3V_1 N k_B T_1}{V_1} = -2 V_1 N k_B T_1?[/itex]

    c) [itex]Q = W = -N k_B T_1 \ln{4}[/itex], I understand.

    e) So it doesn't depend on reversibility and it's still 0.
  6. Mar 2, 2012 #5

    Andrew Mason

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    Check your division.

  7. Mar 2, 2012 #6
    Doh, that should be [itex]-3NK_B T_1[/itex] of course, how stupid. Is this correct then?
  8. Mar 2, 2012 #7


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    The change of the energy is ΔE=E2-E1. It was written in the opposite way in the OP. If it was that way in the original text, everything gets the opposite sign.

    When W is the work done on the gas, and Q is the added heat,

    In an isothermal process, ΔE=0, Q+W=0, Q=-W. W=−NkBT1ln(4), but Q=NkBT1ln(4), positive.

    The answer for b.is correct now.

  9. Mar 3, 2012 #8
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