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A three step cycle problem check

  1. Aug 5, 2009 #1
    THE PROBLEM: A three step-cycle of an ideal monoatomic gas (3.40 mol) undergoes an (1) increase in temperature from 200 to 500K at a constant volume, is then (2) adiabatically expanded back to it's original pressure, then is (3) contracted at a constant pressure back to it's original volume. What is the efficiency of this cycle?


    I'm very shaky on my thermodynamics - If someone could check over this answer to see if my calculations seem logical that would be very helpful, because there's no answer for this question in the back of my textbook. My original question about this problem was "I'm getting a negative number for total work, is this possible?" but then when I solved for heat gained by using the first law of thermodynamics it was also a negative number which gave me a positive percentage. Am I doing anything wrong, or is this correct?



    my work goes as follows:
    T(1) = 200
    T(2) = 500
    n = 3.40 mol
    R = 8.31

    Q = W + (delta)U
    ISOMETRIC W = 0
    ADIABATIC W = - (delta)U
    ISOBARIC W = P x (delta)V

    W(12) + W(23) + W(31) = W(total)
    *because there is no work in an isometric process, W(12) = 0

    W(23) is based on the negative change in internal energy, and with a monoatomic gas looks like -(3/2)(3.4 mol)(8.31)(500-200) = -12714.3

    W(31) uses the constant pressure times the change in volume, or P(V(2) - V(1)), but because we don't have a number for pressure or volume, we take the ideal gas law and solve for the expanded equation PV(2) - PV(1), which ends up looking like: nRT(2) - nRT(1) = nR(T(2) - T(1)) = 8476.2

    however the total work would then be negative, -4238.1


    Then efficiency, which is the work divided by the total heat added/gained: W/Q(h) can be calculated. Q(h) is easily calculated because only one of the three processes adds heat - the isometric one. Taking the first law of thermodynamics, W = Q + (delta)U, so if work is equal to zero, Q = -(delta)U , which again is - (3/2)(3.4)(8.31)(500-200) = -12714.3

    so W/Q(h) = -4238.1/-12714.3 x 100 = 33%
     
  2. jcsd
  3. Aug 6, 2009 #2
    A negative answer for total work implies that the system is doing work, since the W in the first law of thermodynamics refers to the work done on the system. This should be the case since the cycle is actually a heat engine. Similarly, a negative Q implies that the system is giving off heat.
     
  4. Aug 6, 2009 #3

    kuruman

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    Homework Helper
    Gold Member

    Not true. The amount of heat added during the isobaric process is

    Q = n CP ΔT.
     
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