Work Adiabat 1rst Law Calc VS Adiabat Calc

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SUMMARY

The discussion centers on the discrepancy between two methods of calculating work done during an adiabatic process in thermodynamics. The first method yields 2662 Joules, while the second method, based on the first law of thermodynamics, results in -1334 Joules. The confusion arises from the incorrect application of specific heat capacity (Cv) for a monotonic ideal gas, where the correct Cv is determined to be 3R due to the given gamma (γ) value of 4/3. This clarification resolves the inconsistency in the work calculations.

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Albertgauss
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Hello there,

I have been working on the efficiency of engine problems and have some trouble with this one. My question is very specific.

If you scroll down to the line encircled in red, you see that the Work they calculated there was 2662 Joules based on the formula derived for the work of an adiabat. However, if I calculate the Work of the adiabat a different way, by the 1rst law of Thero:

∆U = Q – W And the Q = 0 for an adiabat, so ∆U = – W

I get -1334 Joules, which does not agree with the 2662 Joules calculated in the attached file. What I did: Using the (3/2)nR(T2-T1) for ∆U where “n” is 1 mol, and R is the gas constant 8.3145, T2 at point A is 150 K and T1 at point C is 257 K, (Temps at A,B,C of the cycle are listed at the end of the document, I get

1.5*1*8.3145*(150 – 257 ) = -1334 Joules which is NOT the 2662 Joules they calculated.

Why does the Work calculated not agree between the two methods? My guess is that my mistake is that in my calculation where I got -1334 Joules, I did not put the Cv in correctly, but incorrectly assumed this was monotonic ideal gas for this problem where 1.5*R would be Cv. I did not see a Cv anywhere in the problem. I’m not sure, but I wanted a second opinion as to how to make the Work from ∆U = – W agree with the 2662 Joules using the derived expression of the Work they have in that adiabat.
 

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For an ideal gas, it is ##C_V=\frac{R}{\gamma-1}##, in your case it is clearly stated that ##\gamma=4/3## , hence ##C_V=3R##.
 
Last edited:
Excellent! I got it.
 
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