Let have $n$ mole of substance which is taken as an ideal gas. The internal energy is \begin{equation} U(n,T) = n u(T), \end{equation} where $u(T)$ is the molar internal energy which obeys \begin{equation} du = c_v dT. \end{equation} That implies $$ u(T) - u(T_0) = \int_{T_0}^T c_v d\tilde{T}. $$ Combining previous equation results into \begin{equation} U(n,T) = n \int_{T_0}^T c_v d\tilde{T} + nu(T_0) \end{equation} which is similar result mentions here.

My question and trouble is the difference of internal energy of two states. I always though that the internal energy is defined up to constant and the difference of two states does not depends on the constant.
However taken two states $(n',T')$ and $(n'',T'')$ and calculate the difference of internal energy between these two states results into \begin{equation} \Delta U = U(n',T') - U(n'',T'') = n' \int_{T_0}^{T'} c_v d\tilde{T} - n'' \int_{T_0}^{T''} c_v d\tilde{T} + n'u(T_0) - n'' u(T_0) \end{equation} If $n' = n''$ everything is ok, and the difference simplifies to \begin{equation} \Delta U = n'\int_{T''}^{T'} c_v d\tilde{T} \end{equation}

However, when $n' \neq n''$ then the difference depends on $T_0$ and it does not fit into my view that the internal energy is defined up to constant. Am I mising something?

  • 3
    $\begingroup$ If $n' \ne n''$ then the amount of substance in the system is not constant, so the "change in internal energy" includes the amount of energy transferred into and out of the system when the amount of substance changes. $\endgroup$
    – alephzero
    Jan 30, 2019 at 10:30
  • $\begingroup$ Thank you for your answer. Yes, I agree that the system is now not closed. My question was whether is correct that the difference depends on the constant $T_0$ (in my opinion it is wrong) in this case. Or in another words how to correct the formula to include these effects so the difference is independent on thechoice of $T_0$. $\endgroup$
    – Tomas S.
    Jan 30, 2019 at 12:24
  • $\begingroup$ Is $n$ the total number of particles in the ensemble (in which case $N$ is the more common symbol), or is $n$ the number density? $\endgroup$ Jan 30, 2019 at 13:48
  • $\begingroup$ the first one, $n$ is total number of moles, no density $\endgroup$
    – Tomas S.
    Jan 31, 2019 at 11:32

1 Answer 1


Short answer: in open systems, $\Delta U$ does depend on the reference state, but $\Delta u$ does not.

Because we can't peer into a system, $U$ and $u$ are not directly measurable; they can only be inferred using the First Law (conservation of energy). As has been pointed out in the comments, your question concerns a situation where the system's mass is changed, so conservation of energy can be written as

$$ \frac{\text{d}}{\text{d}t} U + \frac{\text{d}}{\text{d}t} KE + \frac{\text{d}}{\text{d}t}PE = \sum \dot{Q}_\text{in} + \sum \dot{W}_\text{in} + \sum \dot{m}_\text{in}(u+Pv) $$

(I have included sums to represent the general case where there may be multiple heat interactions, multiple work interactions, and multiple mass transfers between the system and its surroundings).

If we substitute $U = mu$ and apply the product rule, we get

$$ \underbrace{m\frac{\text{d}}{\text{d}t} u + u\frac{\text{d}}{\text{d}t} m}_{\frac{\text{d}}{\text{d}t} U} + \frac{\text{d}}{\text{d}t} KE + \frac{\text{d}}{\text{d}t}PE = \sum \dot{Q}_\text{in} + \sum \dot{W}_\text{in} + \sum \dot{m}_\text{in}(u+Pv) $$

Let's place all the terms containing $U$ or $u$ (i.e., all the terms that we could not measure directly) on one side and all the other terms (which we could measure directly) on the other. This gives

$$ \underbrace{m\frac{\text{d}}{\text{d}t} u + u\frac{\text{d}}{\text{d}t} m}_{\frac{\text{d}}{\text{d}t} U} - \sum \dot{m}_\text{in}u = \sum \dot{Q}_\text{in} + \sum \dot{W}_\text{in} + \sum \dot{m}_\text{in}Pv - \frac{\text{d}}{\text{d}t} KE - \frac{\text{d}}{\text{d}t}PE $$

What this shows is that, for open systems, $\frac{\text{d}}{\text{d}t} U$ (which can be integrated with respect to time to give $\Delta U$) can't actually be inferred from measurements; only the sum of this term plus another (negative) $u$-containing term can be defined. It is therefore ok for $\Delta U$ to be dependent on the reference state (/arbitrary additive constant) chosen because the extra term $ - \sum \dot{m}_\text{in}u$ is also dependent on the reference state, in such a way that their sum is reference-independent. The proof of this comes from conservation of mass: $$ \frac{\text{d}}{\text{d}t}m = \sum \dot{m}_\text{in} $$

Multiplying conservation of mass by $u$ and rearranging gives $$ u\frac{\text{d}}{\text{d}t}m - \sum \dot{m}_\text{in}u = 0 $$ which shows that the component of $\frac{\text{d}}{\text{d}t}U$ that depends on the absolute value of $u$ exactly cancels the extra term. Conclusion: $\Delta U$ can depend on the reference state (in open systems), but $\Delta u$ never does.

As a bonus, we can cancel the two identical terms in the energy equation to give

$$ m\frac{\text{d}}{\text{d}t} u = \sum \dot{Q}_\text{in} + \sum \dot{W}_\text{in} + \sum \dot{m}_\text{in}Pv - \frac{\text{d}}{\text{d}t} KE - \frac{\text{d}}{\text{d}t}PE $$

This equation shows that we can deduce $\frac{\text{d}}{\text{d}t} u$ from measurements. As we know from calculus, an indefinite integral of $\frac{\text{d}}{\text{d}t} u$ would introduce an arbitrary additive constant. This shows that $u$ is only defined up to an arbitrary additive constant.

  • $\begingroup$ Thank you! Your answer completely covered my question. $\endgroup$
    – Tomas S.
    Jan 31, 2019 at 11:32

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