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Heat capacity ratio yields inconsistent results

  1. Jan 6, 2015 #1
    1. The problem statement, all variables and given/known data
    A Gas is in a Volume V0 = 1 Liter at Pressure p0 = 3 bar.
    1. Isochoric Heating using the Heat Q1 = 182 J, the pressure raises to p1 = 6.34 bar.
    2. Gas is reset to inital state. Isobaric Heating using the Heat Q2 = 546 J, the Volume increases to V2 = 3 Liter.
    Calculate Cp/CV. Calculate the degrees of freedom f.

    2. Relevant equations

    pV = nRT
    CV = ΔQ/ΔT (for first step)
    Cp = ΔQ/ΔT (for second step)
    Cp/CV = 1+ 2/f
    and optionally
    U = ΔQ + ΔW = f/2 n R T

    3. The attempt at a solution
    The task as described above is pretty straightforward and yields the expected f = 3 solution.

    However, I thought about another way of solving this exercise, but it keeps on yielding different results:
    If I take the last equation (Joules Law: U = f/2 n R T) together with the ideal Gas Equation for the isochoric process, I get f = 1.09. (Again straightforward inserting and solving for f.)

    Those results don't match. Neither do they, if I take the isobaric process (taking into account, that work is done). Just by taking the quotient of those two, I get the expected result. Is this a flaw in the exercise or is there any mistake within my formulas?
     
  2. jcsd
  3. Jan 6, 2015 #2
    the last equation provided by you has an inherent sign convention which is where you are going wrong . hence check the feasability of the 2nd statement of the problem
     
    Last edited: Jan 6, 2015
  4. Jan 6, 2015 #3
    Well in the case of an isochoric heating the sign convention wouldn't affect the result, would it? I mean ΔW = 0 anyways.
     
  5. Jan 6, 2015 #4
    Show us the details of your work for the second method, which is the method I envisioned using. This method seems very straightforward.

    Chet
     
  6. Jan 7, 2015 #5
    In the case of isochoric heating we have ΔW = 0 which gives us from Joule's law and the first law of TD: ΔQ = f/2 n R ΔT.
    Furthermore we know from the ideal gas equation, that Δp V = n R ΔT, which we can solve for ΔT = Δp V / (R n).
    We insert the latter equation into the first one and get ΔQ = f/2 Δp V, which we can solve for f = 2 ΔQ / (Δp V).
    However, inserting the above values (ΔQ = 182 J, Δp = 3.34e5 Pa, V = 1e-3 Liter) results in f = 1.09.

    In the case of isobaric heating, Joule's law and the first law of TD give us: ΔQ = f/2 n R ΔT + p ΔV.
    Again, using the ideal gas equation with pΔV = n R ΔT gives inserted in the previous equation for ΔT:
    ΔQ = (f/2 + 1) pΔV.
    Solving for f and inserting the according values results in f = -0.04 (or something around that, calculated it several times yesterday).

    All this left me pretty confused…
     
  7. Jan 7, 2015 #6
    this is where the sign convention matters .

    and 1.09 looks like the correct answer
     
  8. Jan 7, 2015 #7
    Alright, but why does it differ from the value I get using Cp/CV ?
     
  9. Jan 7, 2015 #8
    I approached this a little differently:

    Let the subscript 1 refer to the case of constant volume and the subscript 2 refer to the case of constant pressure. So, for constant volume:

    ##nC_v(T_1-T_0)=Q_1##

    ##\frac{p_0v_0}{RT_0}C_v(T_1-T_0)=Q_1##

    From the ideal gas law,

    ##T_1=\frac{p_1}{p_0}T_0##

    Combining these equations, I get:

    ##\frac{C_v}{R}=\frac{Q_1}{v_0(p_1-p_0)}##

    By a similar procedure for the constant pressure case, I get:

    ##\frac{C_p}{R}=\frac{Q_2}{p_0(v_2-v_0)}##

    So, from these results:

    ##\frac{C_p}{C_v}=\frac{Q_2}{Q_1}\frac{(\frac{p_1}{p_0}-1)}{(\frac{v_2}{v_0}-1)}##

    Chet
     
  10. Jan 7, 2015 #9
    Hi Chet,

    thanks for your detailed explanation. That gives us exactly the expected result of 5/3 -> f = 3. But what's wrong with my "short" approach?

    Thanks!

    P.S.: I somehow suspect that this exercise has been created in a "reverse" way which is ambiguous… I mean going back from the Cp/Cv equation gives me one condition for all variables, whilst going back from my short approach gives two conditions…

    P.P.S.: I checked this in Maple: If I fix all state variables (pi and Vi), Q1 = 501 J and Q2 must be 1500 J. So the ratio between Q1 and Q2 was perfectly fine, but my approach fixes them to absolute values…
    Is this correct?
     
    Last edited: Jan 7, 2015
  11. Jan 7, 2015 #10
    Well, if I use the equations separately to evaluate the situation, I get:

    ##\frac{C_v}{R}=\frac{182}{(0.001)(3.34\times 10^5)}=0.54491##

    ##\frac{C_p}{R}=\frac{546}{(0.002)(3.\times 10^5)}=0.91##

    Neither of these is consistent with ##C_v = 1.5 R## and ##C_p = 2.5 R##. So, there is something wrong with the problem statement. If ##Q_1## were 500 J, and ##Q_2 = 1500 J##, then everything would be consistent.

    Chet
     
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