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Homework Help: Problem on thermodynamics first law again

  1. May 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider a system consisting of 2.0 mol CO2 (assumed to be a perfect gas) at 298K confined to a cylinder of cross-section 10cm^2 at 10 atm. The gas is allowed to expand adiabatically and reversibly against a constant pressure of 1 atm. Calculate W, Q, ΔU, ΔH, and ΔT when the piston has moved 200cm.

    (Ans: W=-1.6KJ, Q=0, ΔU=-1.6KJ, ΔT=-28, ΔH=-2.1KJ)

    2. Relevant equations
    The thermodynamics equations

    3. The attempt at a solution
    as it is adiabatic, Q=0.
    ΔV=0.001m(0.2m)=2x10^-4 m3
    W=-PexΔV=-101325(2x10^-4)=-20.27 J
    ΔU=W+Q=-20.27 J

    and i found that my W and ΔU are wrong
    what's wrong with my answer?

    should Pgas=Pex if the system is in equilibrium?
    i just found that it isn't
    but in my book :
    "To achieve reversible expansion we must match Pex to P at each state: dw=-Pex dV=-P dV"

    i am so confused.
    please help!
  2. jcsd
  3. May 16, 2009 #2

    D H

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    Calculate the work done by the gas on the piston:

    [tex]W_{\text{on piston}} = \int_0^{l_f} P*A \,dl[/tex]

    The work done on the gas is just the additive inverse of the work done by the gas. The piston moves 200 cm, given. The area is a constant, so you essentially need to integrate P*dl. The process is adiabatic. What is the relation between pressure and volume for an adiabatic process? (Hint: You need to use the fact that the gas is CO2.)
  4. May 16, 2009 #3

    Andrew Mason

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    Homework Helper

    You make a good point. The problem is with the question. The gas can expand adiabatically but not reversibly. In order for it to be a reversible expansion, there has to be an infinitessimal pressure difference.

    In this case there is a significant pressure difference. So what happens here is that some of the energy in the gas does work on the atmosphere and some does work on the piston. The work done on the piston means the piston acquires kinetic energy (or a combination of kinetic and gravitional potential energy if it is not horizontal). So it is a dynamic rather than a quasistatic or reversible process.

    In order to calculate that the work that the gas does on the atmosphere and piston, use the internal pressure of the gas. As DH says, you have to use the relationship between P and V in an adiabatic process (adiabatic condition). Express P as a function of V and then as a function of A and length.

  5. May 16, 2009 #4
    for an adiabatic process, Q=0,

    that's all i can think of

    i have no idea lol
  6. May 16, 2009 #5
    thank you!
    let me think about it...
  7. May 16, 2009 #6

    Andrew Mason

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    But you also know that

    [tex]PV^{\gamma} = P_0V_0^{\gamma} = K[/tex]

    Since PV = nRT, P = nRT/V, so

    [tex]TV^{\gamma-1} = K/nR[/tex]

    which is also constant, ie [itex]TV^{\gamma-1} = T_0V_0^{\gamma-1} [/itex]

    So work out the change in T after an adiabatic expansion of 200 cm (you have to first determine the initial volume and the expanded volume) using this expression. (What is the [itex]\gamma[/itex] for CO2?)

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