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My professor says the internal energy of a system can be calculated as $U=n T c_v$. Now I always thought that we could only calculate a change in internal energy $\Delta U = n c_v \Delta T$.

Following this idea, he says that a recipient that contains gas and loses a certain number of moles at constant temperautre undergoes a change in internal energy of $\Delta U = T c_v \Delta n$.

I've never seen any of this in any book. I believe he is assuming that the internal energy at 0 Kelvin is zero, which I believe is not true. However, in the case of the recipient losing gas,I wouldn't know how to calculate $\Delta U$.

Any thoughts? thanks!

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Suppose you assign an reference state for internal energy such that the internal energy per mole u of the gas is zero at $T=T_r$. This reference state is going to be held fixed for the material both entering or leaving the open system, as well as the material within the system at any time. Then, in the initial state of a constant-temperature system, $U_{init}=n_{init}C_v(T-T_r)$ and, in the final state, $U_{final}=n_{final}C_v(T-T_r)$. So, $$\Delta U=\Delta nC_v(T-T_r)$$Typically, in the overall energy balance on the system, including entering and exiting streams, $T_r$ will cancel out (just as it would for a closed system).

Please note, however, that if chemical reactions are occurring, the equation for the internal energy and the specification of the reference state has to be extended to take this into account (using the concept of "internal energy of formation").

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    $\begingroup$ No. Not if you include the exit stream in the balance. You have been learning about the version of the 1st law applicable to open systems, correct? That version of the first law includes the enthalpy of the exit stream. $\endgroup$ – Chet Miller Feb 27 '16 at 13:36
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    $\begingroup$ That's not a problem. You just use the amount that has left times its h. Doesn't your textbook cover situations like that? The amount entering does not have to match the amount leaving, except at steady state. $\endgroup$ – Chet Miller Feb 27 '16 at 14:17
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    $\begingroup$ Neglecting kinetic and potential energy effects, $$\frac{dU}{dt}=\dot{Q}-\dot{W}_s+\dot{m}_{in}h_{in}-\dot{m}_{out}h_{out}$$where $\dot{W}_s$ is the rate of doing shaft work. I'm sure you've seen this in some form before. $\endgroup$ – Chet Miller Feb 27 '16 at 15:03
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    $\begingroup$ Here is one of many links: learnengineering.org/2013/03/… $\endgroup$ – Chet Miller Feb 28 '16 at 0:19
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    $\begingroup$ The value of $h{out}$ depends on what is being done. If the tank is adiabatic, Q dot is zero, and, if the gas exits through a valve, the value of $h{out}$ is the same as h in the tank (no enthalpy change across the valve) at the moment that the differential mass is leaving. If the tank is somehow held isothermal, Q dot is not zero, but u and h in the tank don't change, and $h{out}$ is equal to the (constant) h in the tank throughout the change. $\endgroup$ – Chet Miller Feb 28 '16 at 16:09

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