Assuming a closed system, show how the first law of thermodynamics. $$\mathrm dU = \mathrm dQ + \mathrm dW$$


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

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

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).

How do I make my assumptions?

Do I assume holding one part constant to solve the equation in one way - that is via an adiabatic process, isothermal process - do the same for another path and compare?

Working for above proof


2 Answers 2


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 $\mathrm dQ=0$ (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,

$$\mathrm dQ=T\mathrm dS=C_v\mathrm dT+P\mathrm dV$$

But, for an ideal gas,

\begin{align} &\mathrm dT =\frac{P\mathrm dV+V\mathrm dP}{R}\\ &C_v+R = C_p \end{align} Therefore,

\begin{align}\mathrm dQ&=C_v\frac{P\mathrm dV+V\mathrm dP}{R}+P\mathrm dV \\ &=\frac{C_v}{R}V\mathrm dP+\frac{(C_v+R)}{R}P\mathrm dV \\ \implies\mathrm dQ&=\frac{C_v}{R}V\mathrm dP+\frac{C_p}{R}P\mathrm dV \end{align} This equation applies only to an ideal gas and a reversible path.


I think it might be wrong because when you were differentiating $PV=RT$, you have assumed pressure to be constant and in your derivation, in some places, you have assumed changing pressure.

You have to know what parameter is changing when you differentiate an equation. When you assumed $P$ to be constant you got $P\Delta V = R\Delta T$. If $P$ weren't constant you would have had a completely new formula that would complicate things even more.

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

  • $\begingroup$ 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
    Commented Feb 16, 2017 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.