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I am currently studying a chapter on entropy of ideal gases from a textbook by Yunus Cengel - Fundamentals of Thermodynamics. There are two equations T ds that can be used to determine the entropy of an ideal gas using approximate or exact analysis. In one example, the parameters given are as follows:

P1 = 100 kPa, T1 = 17 C P2 = 600 kPa, T2 = 57 C

But here's my problem - How do I solve the problem? Do I use first T ds equations which states: $$ T ds = dU + P dv $$ (which I cannot since dU = cv dT and this is not an isochoric process since there is compression involved)

Or do I use constant pressure specific heats which comes under the second Tds relation which states: $$ T ds = dH - v dP $$ (where dH = Cp dT, but this process is not an isobaric process). Note that the textbook uses the second equation which requires which requires constant pressure specific heat.

Given all this, my question is, why do we use constant pressure specific heats when we know for a fact that this is not an isobaric process. Secondly, even if did use constant pressure specific heats (second T ds equation), is an isobaric process evaluated between infinitesimal intervals, such that although the initial and final pressure values are different, the pressure at any given point during a quasi-equilibrium process is the same, in that these states can be retraced in an internal reversible process? Otherwise, I don't know why we can use constant-pressure specific heats.

Thank you.

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The general equation is $dH=C_P\,dT+V(1-\alpha T)\,dP$, where $\alpha$ is the thermal expansion coefficient. (For derivation and discussion, see The cruelest equation in introductory thermodynamics.)

But for an ideal gas, $\alpha=1/T$, so it doesn’t matter that $dP\neq 0$ because its coefficient is zero. $H =C_PT +H_0$ is just a relation than holds for the ideal gas, whether a process is occurring or not. It’s just bad luck for the new thermodynamics practitioner that the only material constant that doesn’t cancel out of a simple equation for a simple state of matter unfortunately happens to have “constant-pressure” in its name.

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  • $\begingroup$ my browser (Brave, a chrome clone) does not like your website, it always gives a warning "The connection to john.maloney.org is not secure" $\endgroup$
    – hyportnex
    Oct 14, 2023 at 19:58
  • $\begingroup$ anyhow, as a thermo amateur, based on your cruel example and several others, too, I am getting to the point that the ideal gas law should be the very last thing an introductory thermodynamics course should be teaching... $\endgroup$
    – hyportnex
    Oct 14, 2023 at 20:03
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But here's my problem - How do I solve the problem?

I'm assuming you want to determine the change in entropy in going from state 1 to state 2.

Since the change in entropy of a system depends only on its initial and final equilibrium states, you can choose any convenient reversible path between the initial state 1 and final state 2 and apply the definition of entropy change for a reversible transfer of heat:

$$\Delta s=\int_1^2\frac{\delta q_{rev}}{T}$$

For your example you can choose a combination of a reversible constant volume (isochoric} heat addition process to the final pressure, where you can calculate the change in temperature and use the specific heat at constant volume, followed by a reversible constant pressure (isobaric) expansion process to the final equilibrium temperature, where you can use the specific heat at constant pressure. Apply the above equation to each process and add the entropy changes for the total entropy change.

Hope this helps.

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  • $\begingroup$ Thank you, Bob D. I understood the first two paragraphs of disregarding the actual process, but instead we identify a reversible process path between the two states to determine entropy. But, can you explain more about the final paragraph where you mention about dissecting the actual process into reversible isochoric and isobaric processes, instead of just using reversible isobaric process between the states 1 and 2? $\endgroup$ Oct 16, 2023 at 7:37
  • $\begingroup$ @PrajwalKori how can it be isobaric if p2 = 6p1? $\endgroup$
    – Bob D
    Oct 16, 2023 at 9:46

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