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Entropy of an adiabatic process

  1. Sep 28, 2008 #1
    1. The problem statement, all variables and given/known data

    A sample of 1.00 mol of an ideal gas at 300 K and 101325 Pa is expanded adiabatically in two ways: a) reversibly to .5 atm and b) against a constant external pressure of .5 atm. Determine the values of q, w, [tex]\Delta[/tex]U, [tex]\Delta[/tex]H, [tex]\Delta[/tex]S, [tex]\Delta[/tex]Ssurr, and [tex]\Delta[/tex]Stot for each path where the data permit. Take C_v,m = 3/2 R (HINT: For part b: use the fact that [tex]\Delta[/tex]U = w = nC_v[tex]\Delta[/tex]T = -Pext[tex]\Delta[/tex]V to find the final temperature, and then construct a reversible path of 2 steps, where P and T change only one at a time.



    2. Relevant equations

    q = 0 in all adiabatic processes
    C_p,m = 5/2*R
    C_v,m = 3/2*R
    [tex]\Delta[/tex]U = w = int C_p dT
    [tex]\Delta[/tex]H = int C_v dT
    [tex]\Delta[/tex]S = [tex]\Delta[/tex]H/T = intC_p/T dT

    for part a) reversible adiabatic: T_2/T_1 = (P_2/P_1)^[(gamma-1)/gamma] where gamma = C_p/C_v = 5/3


    3. The attempt at a solution

    for part a)

    I used the equation for an ideal gas in reversible adiabatic expansion to find T_2

    T_2/300 K = (50662.5 Pa/ 101325 Pa)^(2/5)
    T_2 = 227.357 K

    from there I got [tex]\Delta[/tex]U = w = -905.931 J/mol and [tex]\Delta[/tex]H = -1509.884 J/mol

    For part a I am unsure of how to find entropy of an adiabatic system expanding reversibly...my ideas:
    [tex]\Delta[/tex]S = 0 [tex]\Delta[/tex]S_sys = C_p*ln(T_2/T_1) = 5.76 J/k mol, and therefore [tex]\Delta[/tex]S_tot = 5.76 J/k mol but this seems like an unreasonable answer.

    And for part b I am sort of lost for where to start.

    1) I know all of the above stated equations and unknowns. And that hint to find T_2 but I don't know how to find it! I have plugged in Starting P, n, and T's into the ideal gas equation but this only yields V_1 to plug into my work equation to find T_2.

    Thank you for your time reading this and possibly responding :D
     
  2. jcsd
  3. Sep 28, 2008 #2

    Andrew Mason

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    Start with the definition. For reversible processes: dS = dQ/T

    [tex]\int_A^B dS = S_B - S_A = \Delta S = \int_A^B dQ/T[/tex]

    If dQ = 0 for the gas, then what is dQ for the surroundings? What, then is dS and, therefore, [itex]\Delta S_{surr} \text{ and } \Delta S_{gas}[/itex]?

    This is a little trickier. dS = dQ/T only for reversible processes.

    For expansion at constant pressure, it is a simple matter to determine the work done: [itex]W = P\Delta V[/itex]. Use the first law, dQ = dU + dW to determine what [itex]\Delta U[/itex] will be when expansion is complete and this will give you the temperature of the gas. Since entropy is a state function, the change in entropy in the non-reversible process will be equal to the entropy change resulting from a reversible path between those two states.

    AM
     
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