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Homework Help: Thermodynamics - Carnot Engine

  1. Feb 13, 2010 #1
    1. The problem statement, all variables and given/known data
    A Carnot engine has 40 g of helium gas as its working substance. It operates between two reservoirs of temperatures T1 = 400°C and T2 = 60°C. The initial volume of the gas is V1 = 2.0 X 10-2 m3, the volume after isothermal expansion is V2 = 4.0 X 10-2 m3, and the volume after the subsequent adiabatic expansion is V3 = 5.0 X 10-2 m3.

    (a) Calculate the work generated by this engine during one cycle.

    (b) Calculate the heat absorbed and the waste heat ejected during one cycle.

    2. Relevant equations

    3. The attempt at a solution

    I'm not sure how tor eally start this problem...

    I calculated the effiency of the carnot engine via this:

    1-333K/673K-->0.505201

    At this point, I wasn't sure how to treat an isothermal expansion. I know that delta U=0, and delta Q= delta W.

    I'm not looking for someone to do this for me. I would really appreciate it though, if someone could give me a clue where to start. Thanks.
     
  2. jcsd
  3. Feb 13, 2010 #2

    rock.freak667

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    You should start by finding the P,V at all the four states. You can easily find P1 using the gas equation.

    They tell you that it expands isothermally, so that means

    PV=constants or P1V1=P2V2

    You can now find P2.

    For adiabatic expansion PVγ=constant. You will need to get γ for helium.

    Then it just becomes a case of using Q-W=ΔU
     
  4. Feb 14, 2010 #3

    Andrew Mason

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    You have worked out the efficiency. All you need to know is Qh. - the heatflow INTO the gas. This occurs only during isothermal expansion. Work out the Qh and you will solve this problem. (Hint: apply the first law to determine Qh).

    AM
     
  5. Feb 14, 2010 #4
    Thanks for the replies...

    Andrew Mason, your response really helped. I tried rock.freak667's solution first, but I my pressure values were off.

    Then I tried an equation that was buried in the chapter : Qh=nRT1 * ln (V2/V1), which was exactly what I needed. The rest was a piece of cake.

    Thanks again for both of your replies.
     
  6. Feb 15, 2010 #5

    Andrew Mason

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    For an isothermal process, there is no change in internal energy. So [itex]\Delta Q = W[/itex]

    So to find the heat flow into the gas, determine the work done at constant temperature (substituting P = nRT/V):

    [tex]W = \int_{V_i}^{V_f} Pdv = nRT\int_{V_i}^{V_f} dV/V = nRT\ln\left(\frac{V_f}{V_i}\right)[/tex]

    That is where your equation comes from.

    AM
     
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