The first law states that $ \Delta U = Q - W $ where W is the work the system does on the outside. If there is no heat added, the change in energy is equal to the work done. If the work is negative, i.e. the outside is doing work on us, we get $ \Delta U = W $. But in mechanics, $W = \Delta $ KE = $ - \Delta $ PE, not the total internal energy / mechanical energy, as in thermodynamics. So is there a difference between the two internal energies? How can both equations be true?
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1$\begingroup$ Does this answer your question? Why we dont consider momentum of gas when tackling any specific thermodynamics problems? $\endgroup$– ChemomechanicsSep 21 at 16:09
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$\begingroup$ @Chemomechanics It seems to me that this question is more about the sign convention for work. Therefore, the linked question is irrelevant. $\endgroup$– GiorgioP-DoomsdayClockIsAt-90Sep 22 at 6:10
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$\begingroup$ Possible duplicates: physics.stackexchange.com/q/572771/2451 and links therein. $\endgroup$– Qmechanic ♦Sep 22 at 7:25
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3 Answers
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So is there a difference between the two internal energies? How can both equations be true?
$\Delta U$ in the first law equation applies to the internal energy of the system, that is, the KE and PE at the molecular level. The mechanical energy of the system, which I like to call the "external energy of the system" because it depends on a frame of reference external to the system, is the KE and PE of system as a whole. Both the internal and mechanical energy of the system are included in the general form of the first law, as already pointed out.
The diagram below is my attempt to help illustrate the difference between internal energy and mechanical energy.
Hope it helps.
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The form of the 1st law of thermodynamics you gave applies to cases where there are negligible changes in kinetic energy and potential energy. The more general form of the 1st law is $$\Delta U+\Delta (KE)+\Delta (PE)=Q-W$$ How does that work for you?
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$\begingroup$ But isn't change in u(change in internal energy) already defined as the change in mechanical energy of the system $\endgroup$ Sep 21 at 15:16
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$\begingroup$ No. It's defined as the change in sum of random molecular kinetic energies plus change in molecular potential energies of interaction. $\endgroup$ Sep 21 at 17:42
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The most complete form of the first principle of thermodynamics for a closed system should be written as
$\Delta E^{tot} = W^{ext} + Q^{ext}$,
being:
- $E^{tot}$ the total energy of the system, that can be written as the sum of the kinetic energy $K$ and the internal energy $U$, $E^{tot} = K + U$;
- $W^{ext}$ the work done on the system by external forces;
- $Q^{ext}$ the heat transfer "pointing outwards", i.e. reducing the energy of the system.
You can put together:
- the theorem of kinetic energy, $\dot{K} = P^{tot} = P^{ext} + P^{int}$; this is a result of classical mechanics, providing that the time derivative of the kinetic energy of a system equals the power of all the forces (both external and internal) acting on the system;
- the first principle of thermodynamics, $\dot{E}^{tot} = P^{ext} + \dot{Q}^{ext}$
to find an equation for the internal energy, $U = E^{tot} - K$,
$\dot{U} = P^{int} - \dot{Q}^{ext}$.
For a brief introduction to the principles of thermodynamics, you could take a look at these notes: https://basics.altervista.org/test/Physics/TD/td_principles.html
