There is an example of exercise in a book "Thermodynamics, An Engineering Approach, the 8th edition" by Yunus A. Cengel and Michael A. Boles (example 7-9) stating that:

Air is compressed from an initial state of $100$ kPa and $17^\circ C$ to a final state of $600$ kPa and $57^\circ C$. Determine the entropy change of air during this compression process by using (a) property values from the air table and (b) average specific heats.

The answer of point (b) is that the below equation is used: $$s_2 - s_1 = c_{p,avg} \ln\left(\frac{T_2}{T_1}\right)- R\ln\left(\frac{P_2}{P_1}\right)$$

where $c_{p,avg}$ is specific heat for constant pressure at average temperature average of $37^\circ C$.

Why is $c_{p,avg}$ used in this example even though it is obvious that there is pressure change during process?

  • $\begingroup$ Did I answer your question? Do you need any further clarification? $\endgroup$
    – Bob D
    Apr 30, 2020 at 12:46

1 Answer 1


The example only gives you the initial and final equilibrium states without information on the process (path) connecting the states. But entropy is a state function that doesn’t depend on the path. That means you can use any convenient reversible path connecting the two states in order to calculate the entropy change. The path you choose need not bear any resemblance to the actual path. For an excellent description of this approach, check out the following: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

In this example it appears they chose a reversible constant pressure path to get to the final temperature (the first term on the right side of the equation) followed by a reversible constant temperature path to get to the final pressure (second term on the right side of the equation).

Hope this helps.

  • $\begingroup$ I am sorry for my late response. I've just checked this web again. Thanks a lot for your reply. I already have some idea about this. I just need to think the easiest path to imagine about this question based on diagram (maybe T-s diagram) $\endgroup$
    – elluthfi
    Apr 30, 2020 at 13:09

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