Work Adiabat 1rst Law Calc VS Adiabat Calc

  • Thread starter Albertgauss
  • Start date
  • Tags
    Law Work
In summary, there is a discrepancy between the two methods of calculating the work for an adiabat, with one method resulting in 2662 Joules and the other resulting in -1334 Joules. The reason for this difference is likely due to a mistake in the second calculation not taking into account the specific heat capacity, which is 3 times the gas constant in this case.
  • #1
Albertgauss
Gold Member
292
37
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.
 

Attachments

  • Adiabat_Isotherm.pdf
    353.3 KB · Views: 231
Physics news on Phys.org
  • #2
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:
  • #3
Excellent! I got it.
 
  • Like
Likes Delta2

What is the difference between "Work Adiabat 1rst Law Calc" and "Adiabat Calc"?

Both "Work Adiabat 1rst Law Calc" and "Adiabat Calc" are calculations used in thermodynamics to study the behavior of a system. The main difference lies in their approach and the variables they consider. "Work Adiabat 1rst Law Calc" takes into account the work done on or by a system, while "Adiabat Calc" focuses on changes in temperature and pressure without any exchange of heat with the surroundings.

What is the first law of thermodynamics and how does it relate to "Work Adiabat 1rst Law Calc"?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted from one form to another. In the context of "Work Adiabat 1rst Law Calc", this means that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

How is the adiabatic process represented on a graph?

The adiabatic process is represented by a curve on a graph that shows the relationship between pressure and volume of a system. This curve is typically steeper than an isothermal process and can be identified by its equation, PV^(γ) = constant, where γ is the heat capacity ratio of the system.

What are some real-life applications of "Work Adiabat 1rst Law Calc" and "Adiabat Calc"?

These calculations are commonly used in the study of heat engines, such as car engines, to determine their efficiency and performance. They are also used in the design of refrigeration systems, gas turbines, and other industrial processes that involve the transfer of heat and work.

What are the limitations of using "Work Adiabat 1rst Law Calc" and "Adiabat Calc" in thermodynamic analysis?

One limitation is that these calculations assume ideal conditions, which may not always be the case in real-world scenarios. They also do not take into account factors such as friction, heat losses, and other non-idealities that can affect the accuracy of the results. Additionally, these calculations are based on the assumption of a closed system, and may not be applicable to open or isolated systems.

Similar threads

Replies
81
Views
3K
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
727
  • Introductory Physics Homework Help
Replies
19
Views
1K
Replies
22
Views
2K
Replies
56
Views
3K
Replies
3
Views
993
  • Introductory Physics Homework Help
Replies
8
Views
1K
Replies
1
Views
620
  • Engineering and Comp Sci Homework Help
Replies
11
Views
2K
Back
Top