Energy forms of a system

Question

Energy exists in many forms and their sum is the total energy of the system. Thermodynamics deals with the change of the total energy. The book I am studying from (Thermodynamics: An engineering approach) divides the total energy into two groups: macroscopic or microscopic. The macroscopic forms of energy are those a system possesses as a whole with respect to some outside reference frame. The microscopic forms of energy are those related to the molecular structure of a system and the degree of the molecular activity, and they are independent of outside reference frames.

How are we not accounting for certain forms of energy twice when we consider the system macroscopically and microscopically?

For example, if the phase of the system is a solid or liquid and we fix a reference frame, find its center of mass, and measure the change in its position with respect to a fixed coordinate frame over time. Macroscopically, we can observe (even qualitatively without taking measurements) and tell if the system is at rest or moving, and we can calculate the kinetic energy the system possesses and determine changes in kinetic energy. I am not sure how we calculate the kinetic energy the system possesses if the system is in the gas phase, since most gases are colorless and do not have a definite volume and shape. I understand that changes in velocity will be caused by a force acting on the system, and that if we reduce the system to a point (center of mass) we can analyze the system and determine the external force acting on it.

If we take the same system in the gas phase (I chose gas phase since in the solid phase and liquid phase there are intermolecular forces) and view it on a microscopic scale, given a molecule in space, there are other molecules in space colliding with it and imparting a force on it, causing it to move through space with a constant/changing velocity and thus they posses kinetic energy. If we sum the kinetic energies and divide by the number of molecules, we obtain an average kinetic energy. Is this the same as the kinetic energy calculated above?. How is the system at rest (macroscopically) and moving at the same time (and possesing different kinetic energies with time as a result of the collisions)?

Answer 1

How are we not accounting for certain forms of energy twice when we consider the system macroscopically and microscopically?

We avoid "double counting" by having two versions of the first law of thermodynamics. The first and most frequently used version only deals with changes in the internal (microscopic) energy of a closed system. The second general version adds the macroscopic energy that deals with the motion and position of the system as a whole with respect to an external (to the system) frame of reference.

I am not sure how we calculate the kinetic energy the system possesses if the system is in the gas phase, since most gases are colorless and do not have a definite volume and shape.

If the system is a gas it is generally a closed system, meaning there is a physical boundary between the system and the surroundings that does not permit the exchange of mass with the surroundings. This boundary defines the volume of the gas. Its macroscopic kinetic and potential energy is then that that of the motion and position of mass in this volume with respect to an external frame of reference.

Is this the same as the kinetic energy calculated above?

All the kinetic energy associated with the random molecular motions of the gas molecules are part of the internal energy covered by the first version of the first law described above. It is separate and apart from the kinetic energy of the gas as a whole, which is included in the general form of the first law. The changes in microscopic and macroscopic kinetic energy are separately calculated and added together to get the total energy of the system.

How is the system at rest (macroscopically) and moving at the same time (and possessing different kinetic energies with time as a result of the collisions)?

The figure below illustrates how the microscopic and macroscopic kinetic energies are separately accounted for in the general first law equation. Perhaps it will answer your question.

Hope this helps.

Answer 2

It's certainly possible to count twice if you're not careful. If you have a box of monatomic gas, and somehow know the velocity of every single molecule and calculate the kinetic energy of each one and divide by the number of molecules, that is absolutely going to be the average kinetic energy of each molecule. But if you calculate the average velocity of the gas molecules, or the velocity of the center of mass of the gas molecules, it will be zero (the box and parcel of gas contained has no overall translational motion), and also velocity being a vector has direction. On average there are just as many gas molecules moving up as there are down, left as there are right, etc. It's no different than any other "internal" motion vs. overall motion. If am sitting in a chair wiggling my thumbs, my center of mass is completely stationary, so I have no "macroscopic" kinetic energy, but my thumbs sure are moving so I do possess kinetic energy due to smaller scale motions that don't cause overall translation.

Answer 3

Why are we not counting twice? By definition: "macroscopic kinetic energy" is defined with the average velocity, while the "macroscopic internal energy" is defined as the kinetic energy of the fluctuations around the average energy.

Let's take the energy of a set of particles contained in a volume $\Delta V$, and let's evaluate the kinetic energy of all the particles,

$$\Delta K = \sum_{i=1}^{\Delta N} \frac{1}{2} m |\mathbf{v}_i|^2 = \sum_{i=1}^N \frac{1}{2} m \mathbf{v}_i \cdot \mathbf{v}_i \ , $$

and evaluate it's average $\langle K \rangle$ writing the velocity of the particles as $\mathbf{v}_i = \langle \mathbf{v} \rangle + \delta \mathbf{v}_i$, so that $\langle \delta \mathbf{v}_i \rangle = \mathbf{0}$,

$$\begin{aligned} \langle K \rangle & = \langle \sum_{i=1}^{N} \frac{1}{2} m \mathbf{v}_i \cdot \mathbf{v}_i \rangle = \\ & = \langle \sum_{i=1}^{N} \frac{1}{2} m (\langle \mathbf{v} \rangle + \delta \mathbf{v}_i) \cdot (\langle \mathbf{v} \rangle + \delta \mathbf{v}_i) \rangle = \\ & = N \frac{1}{2} m |\langle \mathbf{v} \rangle|^2 + \sum_{i=1}^{N} \frac{1}{2} m |\delta \mathbf{v}_i|^2 = \\ & = K + U \ , \end{aligned}$$

having defined $K$ as the "macroscopic kinetic energy" and $U$ as the "macroscopic internal energy", following the definitions given in the first lines of this answer.