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Estimating the final pressure of a transformation

  1. Aug 7, 2016 #1
    1. The problem statement, all variables and given/known data
    If the Helmholtz Free Energy remains constant, estimate the final pressure of 1.0mol of an ideal gas in the following transformation: (1.0atm, 300k) → (pfinal, 600k). Given Sgas = R.

    2. Relevant equations
    A = U - TS
    dA = -SdT - pdV

    3. The attempt at a solution
    If the Helmholtz Free Energy is constant, then ΔA = 0.

    So from dA = -SdT - pdV I'd get that S⋅ΔT = -pΔV.

    But I don't see how that helps me figuring out the final pressure.
     
  2. jcsd
  3. Aug 7, 2016 #2
    I think they meant for you to assume that the initial value of S for the gas is R (S0=R). If this is the case then, in terms of pfinal and ##C_p## what is Sfinal? What is ##\Delta (TS)##? In terms of ##C_p##, what is ##\Delta U##? What is ##\Delta A##?

    Is the gas mono-atomic?
     
    Last edited: Aug 7, 2016
  4. Aug 8, 2016 #3
    It did not specify if the gas is mono-atomic nor did it give me the Cp or Cv values.

    It only asks me to find the final pressure, which is 4 atm.
     
  5. Aug 8, 2016 #4
    OK. Then please set up the equations as I suggested in post #2, and let's see where that takes us. I'm confident that this is the correct approach to use.
     
  6. Aug 8, 2016 #5
    My solution to this problem gives 4 atm. if Cp/R=2. This is the same as gamma = 2. It don't think that there are any gases with this ratio, however.

    Chet
     
  7. Aug 9, 2016 #6
    You're right.

    My solution was making ΔA = 0 and then setting SdT = - pdV.

    Since S = R:
    - SdT = (1/V)*nRT⋅dV
    - dT = (1/V)*nT⋅dV (and n = 1)
    - dT/T = (1/V)dV
    - ln(T2/T1) = ln(V2/V1)
    - ln(2) = ln(V2/V1)
    1/2 = V2/V1
    V1 = 2V2

    (P1⋅V1)/T1 = (P2⋅V2)/T2
    (1 atm ⋅2V2)/300 = (P2 ⋅ V2)/600K
    P2 = 4 atm
     
  8. Aug 9, 2016 #7
    I didn't assume that R was constant throughout the change. I assumed that only the initial value of S was equal to R.

    For the final entropy I got:
    $$S_2=R+C_p\ln(T_2/T_1)-R\ln(P_2/P_1)$$
    So, ##\Delta (TS)## was:
    $$\Delta (TS)=R(T_2-T_1)+C_pT_2\ln(T_2/T_1)-RT_2\ln(P_2/P_1)$$
    For the change in internal energy, I got:$$\Delta U=C_v(T2-T_1)=(C_p-R)(T2-T_1)=C_p(T2-T_1)-R(T2-T_1)$$
    So, $$\Delta A=\Delta U-\Delta (TS)=(C_p-2R)(T2-T_1)+C_pT_2\ln(T_2/T_1)-RT_2\ln(P_2/P_1)$$
    Setting ##\Delta A## equal to zero and solving for ##\ln(P_2/P_1)##, we get:
    $$\ln(P_2/P_1)=\frac{C_p}{R}\ln(T_2/T_1)+\left(2-\frac{C_p}{R}\right)\left[1-\frac{T_1}{T_2}\right]$$
    Only if ##\frac{C_p}{R}=2## does the second term on the right vanish, and do we obtain $$\frac{P_2}{P_1}=\left(\frac{T_2}{T_1}\right)^2=4$$
    But, even for a mono-atomic gas, ##\frac{C_p}{R}>2##.
    Our two analyses are consistent with one another only of ##\frac{C_p}{R}=2##. That's the only way that S can be constant throughout the change.

    In my opinion, this is a pretty hinky problem.
     
    Last edited: Aug 9, 2016
  9. Aug 10, 2016 #8
    Here is a simpler analysis of the situation. If the entropy is constant at a value of R, then the change in free energy is given by:
    $$\Delta A=C_v(T_2-T_1)-R(T_2-T_1)=(C_v-R)(T_2-T_1)$$
    The only way this can be zero is if ##C_v=R##.

    The change in entropy for an ideal gas is given by:$$\Delta S=C_p\ln(T_2/T_1)-R\ln(P_2/P_1)$$
    But, if ##\Delta S = 0##, we have:
    $$C_p\ln(T_2/T_1)=R\ln(P_2/P_1)$$
    But, if ##C_v=R##, then ##C_p=2R##
    So, $$\ln(P_2/P_1)=2\ln(T_2/T_1)$$
    But, for all this to hold together, we must have that ##C_v=R##, which is physically unrealistic.
     
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