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Ideal gas using carnot cycle

  1. Jun 1, 2008 #1
    1. The problem statement, all variables and given/known data
    an ideal gas operates in a carnot cycle so that it produces a net positive work of 400 joules per cycle. the maximum temperature during the cycle is 300 deg celsius and the heat lost to a low temperature bath is 600 joules per cycle.

    a) what must be the temperature of the low temp bath?

    b) what is the change in entropy of the gas during the isothermal expansion

    2. Relevant equations

    Q = W = nRT ln (Vf/Vi) i don't think is use this equation as moles is not stated, volumes not given either
    (entropy) deltaS = deltaQ/T

    3. The attempt at a solution

    i am not sure what equations to use as i think i need to consider the entire cycle not just the individual isothermic/adiabatic expansions/contractions

    will the same W equation from (a) be used in part (b) because in an isothermic expansion Q = W, but again moles and volume is not given?

    any help with which equation to use / approach is appreciated...
  2. jcsd
  3. Jun 1, 2008 #2
    What is the carnot efficiency?
    [tex]\eta = \frac{W}{Q_{h}}=1-\frac{T_{c}}{T_{h}}[/tex], W=work, Q=heat

    where [tex]T_{c}[/tex] and [tex]T_{h}[/tex] are in absolute temperature, Kelvins. So you must convert from celsius to Kelvins.

    For the second part, you are given that the heat is lost in the cold temperature so use the formula for entropy.
    [tex] s = \frac{Q_{c}}{T_{c}}[/tex]
    Last edited: Jun 1, 2008
  4. Jun 2, 2008 #3
    using the equation you specified konthelion

    do i sub in 400 for W, 600 for Qh, and 573K = 300 deg cels for Th, then do i solve for Tc?

    400/600 = 1 -(Tc/573)
    573(400/600) = 573 - Tc
    -191 = -Tc
    191K = Tc

    so Tc = -82 deg celsius --> does it sound reasonable?

    and for part b, finding the change in entropy of the isothermal expansion using the eq you provided
    i got
    s = Qc/Tc where Qc = 600joules, and Tc from part a was 191K so s = 6.59 joules/kelvin


    Last edited: Jun 2, 2008
  5. Jun 2, 2008 #4


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    Homework Helper

    Hi scholio,

    The number for Qh does not look right to me. The problem said that was the heat transferred to the low temperatur bath, so 600 is Qc. How can you find Qh from what is given in the problem? Once you have that, I think the rest of it looks right.
  6. Jun 2, 2008 #5
    i looked through my textbook and found this equation

    W/Q_h = (Q_h - Q_c)/Q_h

    where W = 400, and Q_c = 600, i solved for Q_h and got 1000

    so subsituting 1000 into where i incorrectly entered 600, i solved for T_c and got 343.8K = 70.8 deg celsius

    does that sound better now?

    also i solved for the change in entropy and got deltaS = Q_c/T_c = 600/343.8 - 1.75 joules/kelvin
  7. Jun 2, 2008 #6
    Looks correct to me.
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