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For class we have been asked to show how the First Law of thermodynamics.

$$\mathrm dU = \mathrm dQ + \mathrm dW$$

can be shown to be

$$\mathrm dQ = \frac{C_V}{R}V\,\mathrm dP + \frac{C_p}{R}P\,\mathrm dV$$

assuming a closed system.

I have an answer, but am hesitant to say it is a final answer or even the correct one at that.

I will admit that while doing the problem I had trouble following what I was doing. Hence my posting the question. It seemed to me that because we know how state functions act and change with certain processes, that much of the problem is a more of a "plug and chug" approach.

But, I am concerned with the partial derivatives when rearranging this equation. How do certain parts cancel out or how does one approach the problem without them? Is it safe to assume that since you know how the state variables will act in a certain process, that you can use that knowledge to provide a better or more exact answer to the problem?

Attached you will find a picture of the problem that I have done to the best of my ability (which is limited I must say).

If anyone can provide an explanation as to why certain things are done the way they are, or correct my dissection of this problem, I would be grateful for your time in the matter.

What I am looking for in particular, is how do I make my assumptions? Do I assume holding one constant to solve the equation in one way, i.e. adiabatic process, isothermal process, etc. Then do the same for another path and compare?

Working for above proof

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  • $\begingroup$ You can edit your post, that's better than posting a new comment. :) $\endgroup$ – orthocresol Feb 15 '17 at 23:24
  • $\begingroup$ You have 4 subqueries in your post. You might want to focus on fewer of them or break you post into separate ones. As it stands, it's very broad and might be closed by the community. $\endgroup$ – Todd Minehardt Feb 16 '17 at 0:08
  • $\begingroup$ and what might i break it up into? any recommendations? $\endgroup$ – Dylan Russell Feb 16 '17 at 1:10
  • $\begingroup$ I for one didn't think, but know it was very broad. The answer in general would be very broad as it encapsulates the understanding of the subject in its entirety surrounding the laws of thermodynamics. $\endgroup$ – Dylan Russell Feb 16 '17 at 1:18
  • $\begingroup$ The relationship that you are trying to show is valid only for the case of an ideal gas undergoing a reversible path. See my Answer below. $\endgroup$ – Chet Miller Feb 17 '17 at 20:01
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I think it might be wrong because when you were differentiating PV=RT, you assumed pressure to be constant and in your derivation in some places you have assumed Pressure to be changing. You have to know what parameter is changing when you differentiate a equation. When you assumed P to be constant you got P∆V = R∆T. If P weren't constant you would have got a completely new formula that would complicate things even more.

I would suggest to determine what kind of process is occurring and then manipulating the first law equation accordingly.

Please edit your question and write what you have assumed and done in each step of your derivation.

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  • $\begingroup$ thank you for the comment. I will rework the problem with all my assumptions. Would it be safe to say that I need to work this problem with multiple paths in mind? $\endgroup$ – Dylan Russell Feb 16 '17 at 16:29
  • $\begingroup$ I would do a derivation for a single process first then make a generalization for that particular process Because, for example, work done in adiabatic process wont be the same to the work in isothermal process. But the first law equation is general, a factual equation, for the processes that we encounter in our questions. $\endgroup$ – Mitchell Feb 16 '17 at 16:38
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This equation can not be applied in general because it can't describe the amount of heat transferred to an incompressible solid or liquid at constant pressure. For these cases, it would predict that dQ=0 (which is not necessarily correct). If it is not generic, it also cannot describe real gases beyond the ideal gas region. Therefore, the equation must apply exclusively to an ideal gas. The presence of the ideal gas constant R in the equation is a dead giveaway.

For a reversible path, $$dQ=TdS=C_vdT+PdV$$But, for an ideal gas, $$dT=\frac{PdV+VdP}{R}$$Therefore, $$dQ==C_v\frac{PdV+VdP}{R}+PdV=\frac{C_v}{R}VdP+\frac{(C_v+R)}{R}PdV$$But, for an ideal gas, $$C_v+R=C_p$$Therefore, $$dQ=\frac{C_v}{R}VdP+\frac{C_p}{R}PdV$$This equation applies only to an ideal gas and a reversible path.

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  • $\begingroup$ thank you for this information. I have since worked out the problem and had the same answer. I forgot to add that detail that it was an ideal gas and assumed a closed system. $\endgroup$ – Dylan Russell Feb 20 '17 at 21:05

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