# Work in the First Law of Thermodynamics

The First Law can be stated as $\Delta U=Q-W$, where $W$ is the work done by the system. My question is what kind of work $W$ includes. $W$ certainly includes $PV$ work, i.e. expansion and compression of gas. Does it also include work on the system as a whole? For example, if the system is a billiard ball on a frictionless surface, and I push the billiard ball, doing positive work on it, increasing the kinetic energy of the ball. However, both $Q$ and $\Delta U$ are 0, so it seems this kind of work (me pushing the entire system) is not included in $W$. Am I correct? If so, what references spells this out? I was not able to find any references that specify the exclusion of this type of work.

• Pushing the ball increases its kinetic energy, so $\Delta U$ is certainly nonzero. In this case, $Q = 0$, so the system's change in energy is due entirely to the work done on the ball (or equivalently, the negative work done by the ball on the environment). Jun 29, 2015 at 4:16

Question: Does Internal Energy ($\Delta U=Q-W$) also include work $W$ on the system as a whole?

Answer: If work is done on the "whole system", that work does not increase the Internal Energy of the system.

Wikipedia Definition: * Internal Energy: In thermodynamics, the internal energy of a system is the energy contained within the system, excluding the kinetic energy of motion of the system as a whole and the potential energy of the system as a whole due to external force fields. It keeps account of the gains and losses of energy of the system that are due to changes in its internal state.

Elaboration of Definition: Internal Energy, refers to all kinetic and potential energies internal to the boundaries of a system. The purpose of identifying and quantifying the internal energy of a system is to allow for the quantification of energy passing in/out of the system.

• Potential Energies such as 1) $E=mc^{2}$ mass-energy of atoms, 2) the potential energy of inter-atomic and inter-molecular bonds, and 3) gravitational potential energies are disregarded/ignored if those potentials are not released or changed to another energy form during a process.
• In thermodynamic problems, the kinetic energy within a system may be coherent (mass in motion) or incoherent (thermal energy of many small particles randomly moving within the system).
• When macroscopic Kinetic Energy, such as rolling balls, is considered, we typically categorize such as Newtonian mechanics problems. Nevertheless, a single particle system is the degenerate case of a multi-particle system, and can thus fit into the thermodynamic model.
• Reservoirs of macro-Kinetic Energy (i.e., mass with velocity) must be explicitly included as a term in the Internal Energy Thermodynamic Equation when that coherent Kinetic Energy penetrates the system boundary during a process.

Regarding System Boundaries and Energy Types:

• Boundary conditions of a system may be assigned anywhere/arbitrarily. But, the choice of boundary conditions is natural/useful, when assigned to capture changes in energy from crossing the boundary.
• Thermal energy crossing system boundaries is called Heat
• Coherent/measurable macroscopic motion of a mass is called Kinetic energy.
• Moving a mass over a distance against a Force is called Work.
• Macroscopic Kinetic Energy can be converted into work by the impulse of a collision, the compression of a gas, deformation of metal, lifting a mass, stretching a rubber band, or sliding a boulder.
• Force opposed by an equal and opposite field, but with displacement, is called work.
• Force opposed only by the inertia of mass results in acceleration.
• Force acting against a field produces displacement until the reactionary force is equalized.
• Kinetic Energy arises from a reservoir of Potential Energy.

Ambiguities in the Question: There are several possible factors confusing/obscuring this question. Those factors include:
a) Definitions of terms,
b) Choice of system boundary,
c) Detailed consideration of the forces acting within a real physical system.

1) Examine the question using different system boundary conditions:

a) Billiard ball and pool table as the system, and pool cue as the environment,
b) Billiard ball as "whole system", with pool table and cue stick as the environment.

System A: (not the frame asked about in the question) - a billiard ball on a frictionless surface (the system boundary includes both billiard ball and the pool table - both are inside the system).

• When a cue stick impacts the billiard ball, the cue stick penetrates the boundary of the system and imparts kinetic energy to the system from the environment.
• The Internal Energy of this system has increased. The billiard ball is the only particle in this thermodynamic system. In this case, the particle is a macroscopic mass whose Kinetic Energy is increased due to the impulse of force on the particle ($W= \int Fdx$), which in a frictionless/lossless system converts completely into the Kinetic Energy of the mass/particle. Thus, the Internal Energy of the system has increased.
• This is a normal thermodynamics problem (although simplified greatly by eliminating all thermal losses during the impact of cue stick and ball). Energy has passed through the boundary between the environment and the system. Energy is lost from the environment and gained by the system, thereby increasing the system's Internal Energy. The particle inside the system has more kinetic energy after having had work done on it by the collision/impact. This is not new, interesting, or enlightening (nor the question being asked) - we knew the ball would have more energy if we hit it.
• Considering/examining this system does not answer the question about giving kinetic energy to the system as a whole, (i.e. we didn't give energy to the whole system - we penetrated the system and gave energy to a particle inside the system).

System B: (the actual system in the question) - a billiard ball on a frictionless surface. The billiard ball is the "whole system" - the frictionless table and the pool cue are the environment.

Note: this system allows us to consider the actual question as framed.

• The pool cue does work on the billiard ball, giving it kinetic energy since the impulse of momentum from the cue stick was opposed only by the inertia of the billiard ball mass.
• (Note: if the pool table has friction, it exerts a force on the billiard ball opposite to the force of the pool cue. But, friction or frictionless, this factor is a distraction/ irrelevant to the illustration of the principle. Both the friction and pool cue impact are forces acting from the environment on the system.)
• The actual relevant simplification, of perfect elasticity and rigidity of the billiard ball was assumed, but not specified in the question.
• The pool cue exerts a force on the billiard ball and accelerates the "whole system". In this idealized example, the molecules are infinitely rigid (without deformation), or perfectly elastic (lossless deformation and rebound); thus, no energy is lost to thermal motion inside the billiard ball.
• The mass of the billiard ball acquires the entire force impulse delivered by the pool cue. Thus, there is no change in the state variables (P, V, T, S) of the billiard ball. Thus, the internal energy of the system has not changed.
• (Note: I think this is the system intended by the original question - i.e., whether the energy transferred to the whole system is to be included as an energy component when computing the Internal Energy.)
• Again, the answer is no. Doing work on the whole system, and thereby giving it kinetic energy is not included in the computation of the system's Internal Energy.

2) Examine the question as c) an ideal lossless system and d) a real physical system:

• After having analyzed the ideal system, we must now look at a real system to understand the subtleties of such a question.
• With a billiard ball composed of real molecules, the impact of a pool cue increases the internal energy of the billiard ball molecules by losses due to the inelasticity of the material.
• Coherent Kinetic Energy is converted into random Kinetic Energy (i.e., Thermal Energy, aka "Internal Energy", the behavior of which Thermodynamics concerns itself).
• Due to its imperfect elasticity, a percentage of the momentum applied to the billiard ball by the pool cue force impulse is retained by the billiard ball molecules as random motion. The full impulse of momentum delivered to the billiard ball is not converted into coherent Kinetic Energy of the billiard ball's mass.
• Imperfect atomic and molecular bond elasticity means that the molecular bonds composing the billiard ball deform after impact, and are converted (by various mechanisms) into random molecular motion upon recoil.
• This increase in molecular kinetic energy corresponds to an increase in Temperature, which is a state variable proportional to the increase in Internal Energy: $U=\frac{3}{2}nRT$

System C: Idealized System - The molecules composing the billiard ball are perfectly elastic (i.e. no loss of impact energy to thermal energy after deformation and recoil).

• The impulse of force provided to the billiard ball by the cue stick is converted completely into the billiard ball momentum.
• (Note: The frictionless surface of the pool table is irrelevant because that surface is outside of the system.)
• If the initial thermal spectrum of velocities between molecules within the ball does not change after the collision, and there is no change in volume of the system (indicating an increase in potential energy by the amount of $P\Delta V$), then the internal energy of the system has not changed.

System D: Physically Realistic System - The molecules composing the billiard ball are not perfectly elastic, which means, some of the momentum imparted to the billiard ball by the force impulse from the cue stick is retained by the billiard ball molecules as thermal energy.

• Thus, the average velocity of the molecules of the system increases; the billiard ball temperature rises due to the increase in its increase in thermal energy (i.e. Internal Energy).
• Billiard balls are considered to be highly elastic, which means, only a small portion of the impulse-momentum is retained by the billiard ball molecules.
• Therefore, in a real system, a change in the kinetic energy of the billiard ball system will change the internal energy of the system by only a small amount.
• Thus, an answer to the question: The majority of the impulse given to a highly elastic system will not be added to the system's internal energy.
• The kinetic energy of the coherent velocity of the billiard ball is not considered to be part of its internal energy.

Summary:

• Accelerating the "whole" system does not increase the internal energy of an ideal, perfectly elastic system.
• Important elements of thermodynamic problems: 1) define the system boundaries; 2) Note the flow of heat and work across the system boundaries; 3) Identify the forces operating within the system; 4) note the portion of energy divided as coherent energy, incoherent energy, and potential energy after the system comes to equilibrium.
• No real physical system is composed of perfectly elastic materials. Thus, when a system is accelerated, there will be deformations of bonds, and the recoil will not fully transmit the accelerating force to the whole system. Some of the momentum transmitted to the system will be lost to random thermal motion/thermal energy. This manifests an increase in the temperature of the system and an increased Internal Energy.

All of these scenarios are offered as comparisons to give a sense of parallax, perspective, and proof by example, regarding the question of whether to include the kinetic energy of the whole system in the computation of the system's Internal Energy.

The question resolves by clearly defining the "system boundary conditions", and noting there has been no change in the state variables after the billiard ball acquired its impulse of momentum (at least in the lossless-perfectly-elastic system). Even in the "real process, with both friction and inelastic deformation, the state variables of the billiard ball do not change in the amount associated with the absorption and conversion of the full momentum impulse into Internal Energy of the system.

The Big Picture: Kinetic Energy (momentum change by force impulse) imparted to the whole system does not increase the Internal Energy of the system, except for the amount of energy converted to thermal energy due to inelastic molecular collisions.

The First Law formula you cite is for work done BY a system. If the system is a billiard ball and you hit it with a cue, you do work ON the system. The formula in that case would be delta U = Q + W. The work increases the internal energy of the billiard ball.

• Unless you define work done on the system as negative. Jun 29, 2015 at 6:46
• @Steeven: Yes, you are correct. Jun 29, 2015 at 6:51

When a billiard ball is pushed by a cue, the K.E. of the ball increases as it gains a velocity from rest ( supposing it is at rest). Well, then the internal energy of the ball increases. Thus the work done is done by you on the ball and definitely non-zero.

First law as you have stated is applicable to only systems at rest (w.r.t. you). For the example that you have given the first law must be generalized to:

$Q=W+\Delta U+\Delta (Kinetic~energy)+\Delta (Potential~energy)+...$

You are correct. If you have a bulb with gas at a given temperature, putting on a train and letting it depart to another city does not increase its temperature. Uniform movement is not a force!. But if you apply for instance an electric field, gas particles if ionized to certain degree will accelerate and after thermalization through collisions with the boundaries (the glass flask for instance) you would have a system with increased energy.

For any increase in internal energy you need to exert a force onto the system and then letting it achieve the equilibrium state. Then you should be able to find the increase in internal energy of the system.