Dear Mr. Pandey,

let me try to help you out and clarify your doubts in a simple way.

A gas is constituted by many many molecules, each one having its own velocity. However, while studying a gas flowing in a given domain, we cannot follow each molecule so we introduce the concept of particle.

A particle is constituted by a bunch of many molecules (say millions) in the neighbourhood of a point, which however has dimensions much less than the domain to be studied. So the studied domain is composed of a very large number of particles. This is called the continuum assumption which is not valid, e.g. in rarefied flows.

More precisely, the particle must have a volume that is much larger than the cube of the gas mean free path and this volume must be much smaller than the volume of the studied domain. For air at STP, a cube having its side equal to a mean free path contains about 10,000 molecules.

The velocity of a particle of a single component gas is generally defined as the vectorial average of the several molecules contained in it. Of course, in a multi-component gas, a weighted vectorial average has to be performed. So, the velocity of a particle can be regarded as the velocity of the center of mass of the particle itself. Besides, the velocity of a particle depends on the adopted reference frame.

The kinetic energy of a gas particle is the one which is evaluated with the velocity of said particle.

But we have to remember that each molecule has its own velocity so, we may define the peculiar velocity of a molecule as its actual velocity minus the velocity of the particle it belongs to. The peculiar velocity of a molecule does not depend on the adopted reference frame because it is always measured with respect to the center of mass of the particle.

In this simple model, the internal energy of the particle is represented by the average kinetic energy of all the molecules contained in the particle evaluated with the peculiar velocities.

It is like if we split the total kinetic energy of the molecules in two parts, a macroscopic one (the particle kinetic energy) and a microscopic one (the particle internal energy).

Going to your example, in the gas expansion, the particle velocity increases ond so does the kinetic energy, while the internal energy (and so the gas temperature) decreases.

I hope to have been clear enough even not using equations.

Best regards and good luck for your work

Carlomagno

Dear Vikash,

in general U (internal) and K (kinetic) are not the same energy.

For ideal gases, where no attractive/repulsive forces are considered, we have U = K, but in the case of real gases U is not K. If you consider Van der Waals equation, U = - a/V + f(T) where f(T) is the kinetic term (it depends only on temperature T, as ideal case) and - a/V (V = Volume, a = VdW constant) is the potential energy due to forces between molecules (see E. Fermi, Thermodynamics, pag. 73-74 for a clear explanation).

Gianluca

Dear Gianluca,

Thank you for the reference. I have read it now. But it does not address my queries. Energy of an ideal gas is dependent on temperature and not on its volume. Fine, but what about the pressure dependence? If a gas expands, won't the pressure with which it exerts on the outer boundary also change as the volume expands? Further pressure and kinetic energy are related, the latter is dependent on temperature, and temperature decreases as the gas expands. So why not PdV + V dP? Why just dW = P dV?

Or KE of the gas has nothing to do with its Internal energy?

Maybe, to understand the issue you can start from the general equation for the total energy E= (K+U):

d/dt Int[V] (rho*E) dV + Int[S] n.v*(rho*E) dS = Int[S] n.(v.T) dS- Int[S] n.q dS

T=-pI + Tau

q=-k*Grad T

The RHS has two terms, the first one is the total mechanical work (irreversible + the reversible part) the second one is the heat

This equation is nothing else that the equation d Etot/dt = W -Q generally expressed in thermodynamics.

From this equation you can see the differences between K and U and understanding that they can transform each other.

Hi,

Internal energy U of a system is the energy possessed by the system

due to

• molecular motion
• molecular configuration

The internal kinetic energy UK of the molecules is due to the molecular motion and the internal potential energy UP is due to molecular configuration.

Therefore, U = UK + Up

But inter-molecular forces are negligible in case of IDEAL GASES making

Up = 0, which, btw means that, for ideal gas, U = UK and is a function of temperature only. (Remember, in case of real gases, it would depend upon both the temperature and the volume of the gas.)

Secondly,

• It is true that "dW = P dV + V dP"
• dW = P dV is used whenever the process is taking place at constant pressure or pressure changes are very very small.

By definition, a differential amount of work is defined as the dot product of a force external to an object and the object's incremental change in position: dW=F(dot)dx. The dot product leaves the component of the force that is "in line" with the direction of movement dx. It isn't hard to show that this leads to dW = (Fnormal/A)(Adx) = PdV. Note that this is the pressure external to the system upon which work is performed or extracted. That is, because work and heat in thermodynamics are quantities that must cross an external system boundary. In the case of thermodynamic changes that take place reversiblly, then the internal pressure of the system is only incrementally different from the external pressure. Thus, the system's internal pressure can be used as a proxy for the external pressure. But is is a fatal mistake to forget that work is performed on or by the system when that work appears or disappears in the environment, not in the system itself.

Hallow Vikash Pandey, I think my answer will remove your confusion.

For gases internal energy and kinetic energy are same. Because internal energy is the sum of chemical potential energy, electrical & gravitational potential energy and kinetic energy. For a gas with having no internal molecular bonding energy and negligible gravitational and electrical potential energy, the whole energy is kinetic.Also according to ideal gas postulates molecular KE 1/2mv2 = 3/2 KT and T is proportional to U.

Dear Mr. Pandey,

let me try to help you out and clarify your doubts in a simple way.

A gas is constituted by many many molecules, each one having its own velocity. However, while studying a gas flowing in a given domain, we cannot follow each molecule so we introduce the concept of particle.

A particle is constituted by a bunch of many molecules (say millions) in the neighbourhood of a point, which however has dimensions much less than the domain to be studied. So the studied domain is composed of a very large number of particles. This is called the continuum assumption which is not valid, e.g. in rarefied flows.

More precisely, the particle must have a volume that is much larger than the cube of the gas mean free path and this volume must be much smaller than the volume of the studied domain. For air at STP, a cube having its side equal to a mean free path contains about 10,000 molecules.

The velocity of a particle of a single component gas is generally defined as the vectorial average of the several molecules contained in it. Of course, in a multi-component gas, a weighted vectorial average has to be performed. So, the velocity of a particle can be regarded as the velocity of the center of mass of the particle itself. Besides, the velocity of a particle depends on the adopted reference frame.

The kinetic energy of a gas particle is the one which is evaluated with the velocity of said particle.

But we have to remember that each molecule has its own velocity so, we may define the peculiar velocity of a molecule as its actual velocity minus the velocity of the particle it belongs to. The peculiar velocity of a molecule does not depend on the adopted reference frame because it is always measured with respect to the center of mass of the particle.

In this simple model, the internal energy of the particle is represented by the average kinetic energy of all the molecules contained in the particle evaluated with the peculiar velocities.

It is like if we split the total kinetic energy of the molecules in two parts, a macroscopic one (the particle kinetic energy) and a microscopic one (the particle internal energy).

Going to your example, in the gas expansion, the particle velocity increases ond so does the kinetic energy, while the internal energy (and so the gas temperature) decreases.

I hope to have been clear enough even not using equations.

Best regards and good luck for your work

Carlomagno

For a typical real gas (no chemical reaction or nuclear decay involved), the internal energy comprises kinetic energy, associated to the motion of its molecules, and potential energy, due to spatial configuration/assemblage of its molecules.

While the kinetic energy is due to molecular agitation/vibration, associated with the fact that the gas absolute temperature is above 0 K (zero Kelvin), the potential energy refers mainly to electromagnetic attraction and repulsion forces between gas molecules. Gravitational forces between molecules are totally negligible.

For an ideal gas (your example), molecules are, by definition, non-interacting and the potential energy, due to their spatial configuration /assemblage is zero. Consequently its internal energy equals the kinetic energy.

Essentially I agree with Vigneshwaran Sankar.

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Internal energy of gas is simply a averaged random kinetic energy of all molecules by virtue of their vibration and motion of an electron. This random kinetic energy is only a function of temperature. Hear i have used world random because velocity (momentum) of all the molecule are not in unidirectional and hence this energy can not be compliantly concerted in to work. Hence it is categorize as low grad energy.

On the other hand Kinetic energy of gas represents energy of certain mass of the gas by virtue of its velocity in which all the molecules are moving uni direction. This it is possible to utilize motion of all the partial to produce work. hence kinetic energy of a certain mass compliantly can be converted in to work and categorized and high grad energy.

Hallow Vikash Pandey ,

While I added my answer, I had not seen Vigneshwaran Sankar's answer, but came to know when from R. P. Pecanha's agreement statement. I also agree with it and added a new relation KE 1/2mv2 = 3/2 KT and T is proportional to U.

And what about the expression for work done. Is it dw = pdv or pdv+vdp?

Of course, in my simple example, I do neglected otyher forms of internal energy such as rotational and vibrational modes to make the discussion much simpler.

As far as work is concerned, its expression is dW = pdV where W is the work performed by the system (w, lowercase, is specific, per unit mass), p is the pressure of the system and V is the system total volume (v, lowercase, is specific, per unit mass). You can simply derive it in a closed (constant mass) system with a moving piston and computing the quantity force times displacement.

If you develop the term for the mechanical work

W= Int[S] n.(v.T) dS = -Int[S] n.(vp) dS + Int[S] n.(v.Tau) dS

the first term in the RHS is expanded in two terms

-Int[V] Div(vp) dV = -Int[V] p Div(v) dV - Int[V] v.(Grad p) dV

Then consider that Div(v) is related to the production of volume in a gas system.

But as expansion occurs at the expense of internal energy of the gas in an adiabatic process, the pressure also changes. So why drop v dp in the expression?

Work done is calculated by the area under the PV indicator diagram.

For constant pressure process,

W = P*Volume change;

For an otherwise case,

W = integral (PdV) from V1 to V2;

(see the attached figure).

(If still not convinced, please post the context detail, where you encounter "dropping the VdP term in work done in an non-isobaric process".)

@ Vignesh

I have read all that in my bachelor's too. But I am trying to explore why V dP is not applicable. It satisfies the dimensionality as well, so where is the problem?

One can heat up the gas in an enclosed and constrained container such that dv = 0, but that does not mean gas has not work done at all. The added heat will increase the KE of the molecules and so the pressure which the molecules exert on the walls have increased too.

What is the meaning of d(pv)?

d(PV) = P1V1 - P2V2

or

d(PV) = VdP + PdV

In the first meaning for constant pressure d(PV) = PdV

if both of pressure and volume are not constants so; d(PV) = P1V1 - P2V2

If you go way back to Newton and consider a point particle

m*a = sum(Forces) -mg ...breaking the gravitational body for the sake of argument... I'm going to quit using the sum and take it as implicit in what is written below

Or

m(dv/dt) = F-mg

Now take the dot product of this with a change in displacement

m(dv/dt)dot(dx) = Fdotdx - mgdotdx

Rewrite as

m(dv)dot(dx/dt) = F(dot)dx- mg(dot)dx

This is

mv(dot)dv = F(dot)dx - mg(dot)dx

Or,

(1/2)mv^2 = INTEGRAL(F(dot)dx) - mgh (calling hthe vertical component of dx and assuming g is constant)

The lhs is identified as the kinetic energy of the particle, the first term on the rhs the work done on the particle by external forces and the second term the change in gravitational potential energy.

This simple analysis is the starting point for the idea of conservation of energy and the development of thermodynamics as a whole.

While it is true that the internal energy has a statistical component related to the kinetic energy of the atoms or molecules, that kinetic energy is lumpe into what is called the internal energy. The kinetic energy as it is understood for such a closed system is the actual translation of the control volume, not the statistical kinetic energy of the atoms/molecules within the control volume.

This development also explains why work is PdV and does not include VdP.

Well, I think that the answers could be also simplified and classified twofolds.

• Following what stressed by Prof. Carlomagno, if you are at the microscopic level (order of the m.f.p.) wherein the continuum hypotheses do not hold, there is no pointwise PDE to describe the thermodinamics (and other laws) and the energy of a system of particle is due to agitation of molecule. There is no other that the total energy (or internal energy) measured by convention in terms of the temperature. That is within the classical framework of thermodynamics Delta E = W - Q.
• In order to speak about a "deterministic" kinetic energy in a point, we need to invoke the existance of a velocity at a macroscopic level, a fact that is possible after introducing the continuum hypothesis. Now, the velocity field v is associated at each point to a small but of finite measure volume. In such description, we can discern formally between kinetic rho*(v.v)/2 and internal energy rho*e and we can express the mechanical work by means of the integral of v.T over the surface enclosing a generic volume. However, as I wrote before, also in the continuum assuption you can write a PDE conservation for the total energy, without taking care of the distinction between kinetic and internal energy. The PDE equations for these latter are, indeed, not in conservative form (that is can not express them only by means of what happen on the surface of the volume but you need production/destruction terms). Again, you can see this equation in terms of the equation for the power dE/dt = W* -Q*

Internal energy is a state function that defines the energy of a substance in the absence of effects due to capillarity action and external electric or magnetic fields.

Internal energy can be represented as a sum of terms that can be written as kinetic energy, potential energy and chemical energy of constituent particles of a thermodynamic system.

Internal energy is an extensive thermodynamic property which means that it's magnitude depends on the amount of substance present in a given state. Its value is expressed with reference to some standard given state and not usually expressed in absolute terms.

Mr Pandey, you say:

One can heat up the gas in an enclosed and constrained container such that dv = 0, but that does not mean gas has not work done at all. The added heat will increase the KE of the molecules and so the pressure which the molecules exert on the walls have increased too.

But in this case there is no work involved.

Please remember that in your physics course you learned that:

Work = Force times Displacement

If no displacement occurs, the work is identically null.

The heat you are adding increases only the system internal energy.

Dear Vikash, interesting questions. Even though they probably have been already answered, I would like to add a few comments.

The product P*V corresponds to the external virial of the system (Clausius, 1870), and in the case of ideal gases where intermolecular forces are neglected, P*V also corresponds to the total virial of the system. The virial of the system is basically a condition that guarantees that the system is at mechanical equilibrium. In other words, the virial represents the energy required to contain the system in a certain definite volume.

Now, a differential change in the virial can be expressed as: d(PV)=PdV+VdP. The first term corresponding to a change in volume at constant pressure (work), and the second as a change in pressure at constant volume.

Since the system needs to be contained, let us now consider two possible types of containers for the system: 1) A flexible container, and 2) A rigid container. If the container is flexible, the volume is defined by the equilibrium of mechanical forces, that is, internal pressure = external pressure. Therefore, assuming that the external pressure remains constant, then: d(PV)=P_ext*dV. This term corresponds to work. On the other hand, if the container is rigid, the volume contained is constant and therefore: d(PV)=VdP. In this case, the internal and external pressures might differ, and the maximum pressure difference will depend on the strength of the container.

Thus, if you are considering an adiabatic expansion, it implies that there is a flexible container with an internal pressure in equilibrium with the external pressure. For a constant external pressure, the chance of virial is the result exclusively of work. Also, since heat exchange is not taking place, the work of expansion is done by the molecules of gas, and therefore their internal energy is reduced. Since the gas is ideal, the only component of internal energy available is the kinetic energy of the molecules, thus the average kinetic energy of the molecules is reduced, and ultimately, the system temperature decreases.

Mr. Hernandez,

you say:

If the container is flexible, the volume is defined by the equilibrium of mechanical forces, that is, internal pressure = external pressure.

This is not true since you are forgetting the stress in the membrane.

Think of a fair balloon that, when you pinch it, kinds of explodes because the internal pressure is higher than the external one.

Dear Vikash,

to be very short :

internal energy includes all energy except that which is due to external interactions [ dixit Kevin Zhou ]

more details:

Internal energy U is all contained energy. And there are many ways to contain/store energy :

· In chemical bonds, when atoms join and form molecules

· In different phases of matter - ice at 0∘0∘ contains less internal energy than water at at 0∘0∘, because some energy has broken then bonds. Here a phase transition energy also called latent heat is the amount of energy needed to melt or freeze to do this phase change.

· As kinetic energy when particles move within a system

· As potential energy within the system when an object / objects is out of equilibrium (a book on a book-shelf wants to fall, or a compressed spring wants to jump out)

· As thermal energy, which is giving objects their temperature.

· Etc.

Finally, I found this from: https://physics.stackexchange.com/questions/287863/how-to-define-exactly-the-internal-energy

The first law tells you that you can change the internal energy of a system ΔU by either having work done by the system W or adding heat to the system Q

ΔU=Q−W

The system does not contain any heat. Heat and work are not state functions of a system. So you cannot say that this system has so much heat in it at one time and more heat in it at another time. However you will see δQ in textbooks which means a small amount of heat added not a change in the amount of heat in the system. Another common form which I cannot reproduce here is a little dd with a line through it.

The internal energy of a system is the sum of the kinetic energies and the potential energies.

If a system does no work and you add heat to it, the internal energy of the system increases. If your system is an ideal gas then this increase in internal energy is an increase in the kinetic energy of the atoms of the gas. For most systems heat entering a system will affect both the kinetic energy and the potential energy of the system.

Temperature can be defined for a system in equilibrium and so you can define a difference in temperature.

Heat and work are energies in transit, i.e. energies that are exchanged between the system and its environment.

I totally agree with you Giovanni that these energies are in transit (e.g. heat is the enrgy transferred from A to B due to a temperature gradient ... but when inside the system boundaries we add them to the internal energy balance (stored heat or work).

We can make an analogy with local populations living in a country (internal energy) and foreigners (migrants + travellers transiting)....

Dear collegues I am not sure we are sharing a common nomenclature...

Kamal Mohammedi, you wrote " Internal energy U is all contained energy" and that would define nothing else that what is also called (at least in a nomenclature common for continuous systems) Total Energy of the system. But I understand that books of Gasdynamics and books of Thermodinamics can adopt a nomenclature with some differences.

However, the way in which the "total energy" is exchanged between a system and the environment can be described by using a continuous model or a particle model. But I think that is only a way in which we use mathematics for describing what is always a unique physical phoenomenon.

Provided that the continuous model applies, heat and work (reversible + irreversible) are exchanged between environment and systems as fluxes over the surface that encloses a system. That is described in general by starting from the transport theorem. Of course, for an Eulerian description we get also the flux of total energy due to the convection of macroscopic mass flux.

Sorry if I am more inclined to see this issue in a more fluid dynamics view but I think that different mathematical descriptions have to describe the same physics within the hypothesis introduced by the model.

Hey, Thanks everyone for answering. Now if you can focus on what should be the correct expression for work done since in most thermodynamic systems of daily observation both pressure and volume changes with time.

W = PdV + Vdp ?

W = PdV ?

W = Vdp ?

In the case of a reversible expansion/contraction, dW=PdV

Dear Filippo,

Wether using macroscopic or microscopic approach, I am sure we are talking about the same internal energy (which is a macroscopic quantity which includes microscopic kinetic energy due to the motion of the system's particles and the potential energy associated with the microscopic forces, including the chemical bonds ). But the Total energy is not the internal energy. There is no simple universal relation between these quantities of microscopic energy and the quantities of energy gained or lost by the system in work, heat, or matter transfer.In statistical mechanics, the internal energy corresponds to the ensemble average of the sum of the microscopic kinetic and potential energies of the system. Of course we need to define our system (closed, open, isolated).

To focus on Vikash work... what kind of processes you are interested in? isobaric? isochoric? isothermal?...

@ Kamal,

Could be isothermal or adiabatic. And in rare cases it is reversible, mostly not. It is not isochoric for sure.

If the process isn't reversible, then work has nothing to do with the internal system pressure. Work and heat manifest themselves as changes in the environment. For example, imagine an ideal gas contained in a cylinder with locks on the piston. Outside there is a vacuum. The locks are released, the piston moves until it encounters another set of locks. Even though the gas expanded, no work was done because the piston didn't "push" against anything. Another example, a high pressure gas is contained in the piston/cylinder apparatus. The outside pressure is 1 bar. The latches are released and the gas expands. Even though the gas pressure is much greater than 1 bar at all times, and changes continuously as the piston moves, the work is related to the pressure of 1 bar and the distance moved by the piston. Only for a reversible process can the external and internal pressures be used interchangeably.

Dear Kamal,

I follow the hystorical book of Zucrow & Hoffman, Gas Dynamics Vol I. According to 1-11(b), I also define the total stored energy of a system as sum of internal, kinetic, potential and all other forms of energy.

Focusing on the Vikash question, Zucrow defined the work in Eq.(1.44) assuming no friction in the system.

Hallow Vikash Panday.

I was very much busy, I could not track your subsequent questions, Now I am giving a justified Answer.

According to you in tn the expression W =Pdv +VdP

the term VdP violets the definition of work, that is volume remains constant, no overall displacement occurs. The change of pressure, according to the kinetic theory gases is the result of increase of molecular kinetic energy. So this term must not be accounted as work. The actual term of work is PdV. VdP term will act as internal energy restoring term, in other words it will prevent a large fall of internal energy.

Mr. Hanley, you say:

If the process isn't reversible, then work has nothing to do with the internal system pressure. Work and heat manifest themselves as changes in the environment. For example, imagine an ideal gas contained in a cylinder with locks on the piston. Outside there is a vacuum. The locks are released, the piston moves until it encounters another set of locks. Even though the gas expanded, no work was done because the piston didn't "push" against anything.

But, what accelerates the piston and wins its inertia?

Where the piston kinetic energy comes from ?

Another example, a high pressure gas is contained in the piston/cylinder apparatus. The outside pressure is 1 bar. The latches are released and the gas expands. Even though the gas pressure is much greater than 1 bar at all times, and changes continuously as the piston moves, the work is related to the pressure of 1 bar and the distance moved by the piston.

But again, what accelerates the piston ?

If the two forces are not equal, so the piston, must obviously accelerate.

Since:

Work = Force times Displacement

The force is the one exerted by the system on the inner surface of the piston. The displacement is the piston displacement.

Mr. Hanley,

think it over.

Regards

Mr Carlomagno,

The usual assumption in any introductory discussion is to treat the piston as massless and frictionless. And if the piston is massless, it has no kinetic energy (and there is also no change in potential either). And if it is frictionless, then there is no loss here either.

Including details of the construction of the apparatus, like the density of the metal, the length of the piston stroke, friction with the container walls, etc., serve only to add a layer of complexity that serves very little purpose in the discussion of internal energy, work, and heat here. The gas is, after all, ideal. If we are going to broaden the discussion by adding what is, in all likelihood, a minor perturbation on the ultimate answer, then we might as well include the other small perturbations, like the fact that no gas behaves perfectly ideally, and that there will be contributions to the internal energy beyond that of the average kinetic energy of the gas molecules, or whether the piston/cylinder apparatus is horizontal or vertical and whether or not it is located in earth's gravitational field, the field of another planet, or it is located in empty space. Text here is limited and it isn't possible to include a discussion of every possible scenario. In any event, it is the mass of the piston and the friction due to movement that are important when the piston's movement is anything other than infinitesimally slow,. In this case the work performed is certainly not the internal gas pressure times the piston's surface area. Those two sets of forces approach one another to a high degree only when the movement of the piston is infinitesimally slow.

Think it over.

Regards

Mr. Hanley,

being the piston massless, its acceleration will go infinity.

What happens ?

Regards

Nor is the work, in the other case, with a very high pressure gas on the one side, and atmospheric pressure on the other - even including any other complexities like the weight of the piston and friction with the walls - related to the the volume change and the pressure inside the container. It is related to the volume change, the pressure outside the apparatus, and any losses you want to include due to the kinetic energy of the piston, friction, change of potential, etc., etc., etc.

Now, you are complicating the model including friction, change of potential, etc., etc., etc.

Let us stick to the simple model of ideal gas, no friction, no variation of potential energy, no etc., but massless piston.

Please tell me what happens.

Regards

So what if the acceleration is infinite? How does that in any way change the essence of the discussion? Make the piston's mass very small, but not zero. Now the acceleration is large but finite. Are you quibbling over the fact that the gas does some work on the piston? Granted. Please answer MY question. Pint*dV is certainly not the same as Pext*dV, where Pint is the internal system pressure and Pext is the external pressure (let's say atmospheric), in the cases I have laid out. They clearly don't have to be equal. In the case of the work of pushing back the atmosphere due to the gas expansion, one is related to the work done by the system and the other isn't. Which is more closely related to the work performed by the expansion in the case of movement of the piston at finite speed? Is it the integral of the Pint*dV or is it the integral of Pext*dV?

Dear Mr. Hanley,

I prefer to quit this discussion.

It seems to me that in Baton Rouge nature behaves differently than in Naples. So, I will not add any further comment. I taught elementary thermodynamics a long time ago, but now I am retired.

Regards.

Mr. Carlomagno,

I'm sorry that you have decided to quit the discussion because it is an important one that leads to a lot of confusion. The development of the concepts embodied in the energy conservation equations of thermodynamics are based on Newton's laws, and here it is the forces external to a system that act upon it. That leads naturally to the concept of work in terms of the external forces. Reversibility offers the great advantage of allowing us to use internal system variables to get at quantities, like work, that are based on forces acting on the system externally. The laws of nature are everywhere the same. Their understanding and application is what varies from place to place.

See

https://books.google.com/books?id=igG7BwAAQBAJ&pg=PA95&lpg=PA95&dq=work+external+forces+internal+forces+thermodynamics&source=bl&ots=Akl5lGmUtx&sig=ff_bSy9szRfR3KCBclGOWs9aM28&hl=en&sa=X&ved=0ahUKEwjJ0Nbzoa3YAhVExWMKHd6bDpo4FBDoAQgmMAA#v=onepage&q=work%20external%20forces%20internal%20forces%20thermodynamics&f=false

Dear Brian,

I am following your discussion but I am not sure about the systems you are considering...I think we are talking about an Eulerian volume with moving boundary, not a material volume.

Considering an ideal gas (no viscosity) you can write the work done by the isotropic stress on one surface of the piston as -Int [S] n.v p dS. Formally, that means you can consider the internal surface of the piston as well as the external surface of the piston and elaborate two different expressions for the work. They are boundaries of two different systems of fluid, right? Formally, you can set arbitrarily the movement of the piston at a chosen velocity (movement by some force) and elaborating the work transmitted to the two systems, the one subject to expansion and the one subject to compression, right?

Now, the issue about the fact that the piston has a mass or is massless seems somehow marginal. It would make sense if you want to study the flow with a control volume enclosing the piston. In such case, the difference of works from the two parts an the (mass*acceleration) of the piston has to be taken in to account to define the time evolution of the total energy of this system.

Filippo,

The discussion began with whether the internal energy of a closed system containing an ideal gas is proportional to the average kinetic energy of the atoms/molecules contained therein, with no other contributions, and then, whether the expansion of such an ideal gas had two terms related to the work, PdV and VdP. The answer to the internal energy/kinetic energy question for an ideal gas is that the internal energy is directly proportional to the average kinetic energy of the atoms/molecules and that there is no other contribution. The answer to the PdV + VdP question for work, based on many responses, is that work is only related to PdV and not to VdP. I then pointed out that for expansions at finte rates for that system the work would be equal to the integral of the external force(s) acting times the incremental change in position of the piston or system boundary. Work in the case of expansions at finite rates is not equal to the integral of the internal pressure times the differential change in position of the piston or system boundary. The idea of thermodynamic reversibility was introduced early on in thermodynamics so that the movement of the piston or system boundary can be considered infinitesimally slow and therefore the internal pressure must be only infinitesimally different external pressure. Thus, the one can be substituted for the other. The work done BY the system on the surroundings can be approximated as the integral of the internal pressure times the differential displacement of the system boundary. The distinction between finite expansion rates and infinitesimal expansion rates is crucial. This is all laid out in the link to the book I provided above.

One can always in theory choose the material characteristics of the piston and whether the piston is included or excluded from the definition of the control volume. In the case where the piston is included in the definition of the control volume and in the case where the piston has a finite mass, then its change in kinetic energy due to the finite velocity of the piston and in its potential energy due to its change of position within the gravitational field amount to conversions of the internal energy of the system into kinetic and potential energies of the system. The work performed on the atmosphere by the piston is still the integral of the external pressure times the differential change in volume.

In the case where the piston is considered a part of the environment rather than a part of the system, then the external force would be that due to the weight of the external atmosphere plus the weight of the piston. The work performed in this case would involve those two terms - external pressure and change in volume and piston weight times change in position. and here there would be a negligible change in the kinetic energy of the "system" composed of the atmosphere and the piston. The gas now would experience a change in internal energy alone (to a first approximation), due to the work it performs on the combined system of piston weight plus atmosphere.

In either case, energy is conserved and the issue of the weight of the piston and its contribution to the discussion is kind of ancillary. I hope that long-winded response is related to your post.

Also note that my thought experiments have the piston stop at the end. There is thus no change in kinetic energy whether you consider the piston as part of the system or part of the environment.

Ok, but do you agree that given a gaseous system with its boundaries, the mechanical work that is exchanged at the surface only in reversible way is :

Int [S] n.v p dS

being p the pressure of the gas over the surface?

That is in accord also with the Batchelor book. The confusion I still see is in the nomenclature of the contribution of the work to the energy of the system. Zucrow defines the work in the contribution to the stored total energy (see Eq.1.46) while Batchelor talks about internal energy of the system (see Eq.1.5.2).

I think that the original question has source in this issue...

Yes, I do. I don't have easy access to the books you mention.

Why you do not understand that, if you consider the external pressure acting on the system, you are considering the piston as a part of it and so you must consider as well its kinetic energy, elastic energy and so on ?.

Even if you consider a quasi-static process, the external work goes partially in the elastic energy of the piston (when in the most general case is deformable) and the rest of it in internal energy of the gas.

But now I really quit !!.

The piston is stationary at the start of the stroke and it stops at the end of the stroke - held in place by latches both at the start and at the end (read the statement of the problem) - so its change in kinetic energy is zero...whether the piston has mass or not. Its kinetic energy is a non-sequitur. The kinetic energy that the piston had is either converted to heat and lost if the cylinder is conducting or it is returned to the internal energy of the gas if the cylinder is insulating (with, perhaps some small change in its potential if the expansion occurs in a vertical cylinder in a gravitational field). In any event, it certainly does not appear in the work required to push back the atmosphere nor does the kinetic energy of the piston during the expansion allow one to calculate the work to push back the atmosphere based on the internal pressure rather than the external pressure. Is your argument that one must consider the kinetic energy of the piston during the expansion in some integrated form as it moves between the latches in order to complete the energy balance? Please reference any peer-reviewed text or article or book where this is done. I gave you a reference above that is in complete agreement with what I have contended.

I really understand Giovanni's frustrations! He has willing to help and tried to answer the question based on concept that should be known by everybody included in this discussion. I would like to recommend Vikash to read a nice and short introduction in the matter in R. Panton: Incompressible fluid flow. After that he will understand what is Giovanni talking about, and he will find answers to his questions by himself.

Sorry I don’t know this book but what would be the contribution in case of incompressibile flows where the pressure has no termodynamcs meaning and the divergence-free constraint inplies no volume production of the fluid System??

Hi! I have gone through some standard text books and online discussion forums. I must admit I had confusions, and it seems they were from my high school days. Also, I guess we have not been able to converge in this discussion partly because we could all have such misunderstandings and unfortunately we have taken them for granted as they seem trivial. I would like to add some inputs based on my latest studies to the discussion:

1) Internal energy of a gas manifests itself from two factors: KE of its constituent atoms/molecules and spatial/temporal arrangements (configurations), i.e., the potential energy.

2) In the case of an ideal gas, IE is due to its KE alone. So, any heat added to such a system will increase the velocity of the atoms, and so the pressure which the atoms exert on the container walls will increase too.

3) In the classical well-known formula W = P dV. The P here is the EXTERNAL pressure that is applied through the piston to the gas (if its a compression process), and it is further assumed that the acting P is maintained, i..e, a constant.

It is NOT the pressure of the gas in the system.

Similarly for the expansion process the pressure felt by the outward moving piston is P and again assumed a constant.

Such an assumption of constant P is okay provided the process is quasi-static (very slow process).

https://www.khanacademy.org/science/chemistry/thermodynamics-chemistry/internal-energy-sal/a/pressure-volume-work

http://physics.bu.edu/~duffy/py105/Firstlaw.html

4) However in my problem under investigation, 3) does not hold. And I have limited information. I have a gas trapped in a small enclosure (mm-scale) and then the enclosure ruptures or expands or changes in volume, and it is not quasi-static since it occurs in microsecond scale. So in such a process both pressure and volume of the system changes. What should be the expression for the work done in such a process?

5) At least in fluid mechanics it seems VdP is used to calculate work.

https://www.quora.com/When-Integration-Pdv-Vdp-is-used-to-calculate-the-work-done-in-a-process

Another example where VdP is considered is here, https://physics.stackexchange.com/questions/52001/what-is-vdp-work-and-when-do-i-use-it

Could also be possible in a thermodynamic open system?

@Vikash

In fluid dynamics, the work (for time unit) is defined by means of the stress tensor as I addressed above (I attach again )

___

the term for the mechanical work is

W= Int[S] n.(v.T) dS = -Int[S] n.(vp) dS + Int[S] n.(v.Tau) dS

the first term in the RHS (due to the isotropic part) is expanded in two terms

-Int[V] Div(vp) dV = -Int[V] p Div(v) dV - Int[V] v.(Grad p) dV

Then consider that Div(v) is related to the production of volume in a gas system.

___

Of course, for an ideal gas the viscosity is disregarded and there is no work from the deviatoric part of the stress tensor.

If I understand your problem, you are studing something resembling the shock tube, isn't that?

its Bubble Dynamics but it has shock wave aspect too. This has been studied before but with a lot of assumptions and so my questions posted in this forum are to better understand them and further adopt an appropriate approach to study the problem.

Do you model the front of the bubble in 1D assumption or is multidimensional? If you define the gas system corresponding to the bubble, you have an Eulerian control volume with moving boundaries and you can compute the reversible work by its definition

As far as I know bubble case is not reversible thermodynamic process. they often end with violent implosion. My question is simple, a bubble of gas when changes from volume v1 to v2. what should be the expression of work done? Often I have seen PdV in scientific literature but I am not completely satisfied with that since it is reasonable to assume that in a realistic case pressure changes alongside volume as well.

Have a look to the definition of work for a bubble here

https://authors.library.caltech.edu/25017/5/BUBBOOK.pdf

I will try to add my contribution for this dense discussion, focusing only on Vikash's last concern. Up to my understanding:

- The work done on the outer boundary is; p_out . integral(dv). Note that outer pressure is constant.

- The work done by the system (inner fluid) is; integral(p_in . dv). Note that inner pressure increases with volume.

- Thus, as the volume increases, pressure difference increases, and hence the difference between work done by the system and work done on the boundary also increases. The difference between the two works is equal to the strength of the bubble itself which should increases with volume as well.

- As you said: "in a realistic case pressure changes alongside volume as well". YES; that it is why you need to find a relation between p-in & v to be able to integrate and get the work done by the system (inner fluid).

I'm sure that this answer contains bunch of assumptions and simplifications.

Regarding your concern about the work formula. I think the confusion starts from the point you had assumed in your question that work is equal to PV. This is not true, because work is a process function, so it can't be defined by any single state! Wort is defined as the dot product of force and displacement (note that displacement is a process). When you convert this definition into formulas, you can find that;

- in closed system (as described in details above by Mr. Hanley) it leads you to integral(p.dv), while

- in open system (like compressors and turbines) it leads you to integral(v.dp).

That is implied by the definition of work, if you consider the external environment at constant pressure p_out, then Int[S] n.(vp) dS = p_out*Int[S] n.v dS, being Int[S] n.v dS the volume production

Dear Vikash

I believe this discussion would profit if we first agree on the definition of thermodynamic system (matter, a gas in this case), boundaries (imaginary, fixed or movable) and surroundings (matter, everything else except the gas). But I’ll move a little away from it.

Now that your practical problem has been stated in more detail, I believe it is similar to cavitation phenomenon: moving liquid particles, that is finite portions of matter, reach a region where the pressure is lower than the liquid vapor pressure at the flow temperature. A sudden liquid vaporization occurs locally and bubbles (or cavities) are formed whitin the liquid mass. This is now a very complex, accelerated (that is, transient) viscous two phase flow: a surrounding liquid practically incompressible and a highly compressible vapor bubble-phase. Bubbles are carried with the flow just to reach another nearby region where the pressure is higher than the liquid vapor pressure at the flow temperature. Then bubbles suddenly implode and again the flow is single phase. Bubble implosion is the damaging effect to equipments: local presssure near 1000 atm and temperatures of 800 ºC have been measured. So, just may be existing mathematical models, possibly based on the Navier- Stokes equation, or even empirical correlations for cavitation, migth help you out.

Dear R. P. Peçanha,

I agree with you. But also see an alternative benefit of this discussion, there are so many misconceptions among us regarding simple thermodynamics.

Yes, it is cavitation bubble dynamics. It is adiabatic, not reversible in most cases, moving or even oscillatory boundary. It may also change from a closed system to an open system (the moment when violent implosion occurs). There is a nonlinear ODE - Rayleigh-Plesset Equation, a variant of NSE used to study bubble dynamics, but the equation does not reveal the underlying physics. This is what I am trying to understand so that an alternative equation could be derived (not empirical) which gives a better insight into the physics. So, I am trying to consider as many possible mechanism as possible, with fewer assumptions.

Why don’t you start from the NS system without using the assumption of the RP model?

@ Denaro

I tried there, nothing much to explore other than solving the problem numerically. I am more into analytical solving.

Anyway, the final verdict is that W = PdV is not an expression which always holds, rather V dP could also be a possibility?

I am not an expert of cavitation propblems, I don't know what you tried to do, I was thinking an approach that, while retaining the 1D assumption of the RP equations, works with the compressible 1D NS equations. This resembles the way in which we solve analytically the equations for the viscous shock layer.

Finally, I still do not like the compact expression W=PdV for the work in a fluid system. As a fluid dynamic teacher, I will always introduce the work (reversible + irreversible) from the general integral expression I wrote above.

I am not sure if I understand the integral expression. Can you please give the latex expression or something more viewing-friendly? Also what is S and n?

Have a look to this page

Dear All,

I may add some confusion to this discussion.

a) Internal energy

First of all internal energy is not an absolute quantity. The only relevant quantity is the change in internal energy. Furthermore the quantity of internal energy depends on the processes that are assumed to be relevant. For example in a gas without macroscopic movement and with negligible molecular vibration and with negligible chemical reactions and with negligible ionisation and and with negligible electronic excitation and with negligible nuclear reactions... and with negligible long distance molecular interaction obviously, the internal energy is dominated by the sum of the translational and rotational kinetic energy of individual molecules. This will also be an ideal gas. As temperature is increased various processes can be "frozen" or active. For example at very low temperature, only translational energy will be significant. Rotation will be frozen due to quantum mechanical effects. At extremely low temperature also even translation could become frozen. As temperature increases one can reach conditions at which vibration of multi-atomic molecules becomes significant. At that moment the potential energy between atoms within a molecule become essential. For low amplitude vibration the oscillation are harmonic and one can demonstrate that potential and kinetic energy involved in these molecular vibration are equal. From that point on obviously the internal energy is NOT ANYMORE only the sum the kinetic energy of individual molecules. Increasing the temperature gradually (step by step) all the processes described above will become significant.

b) Thermodynamic equilibrium

Thermodynamic equilibrium is a condition in which nothing changes rom a macroscopic observation point of view, what ever the time scale we consider. Obviously, this is a useless concept. We are only interested in changes. However we consider that the theoretical concept of "thermodynamic equilibrium" can be a fair approximation when the time scale considered to wash out a perturbation in the system is short compared to the time scale considered in our applications. In fluid dynamics we define material elements that are small enough to allow assuming macroscopic quantities to be uniform in the element, but large enough so that the statistic fluctuations in amount of molecules, kinetic energy...show negligible fluctuations within our measurement time. We call this the continuum assumption and the local thermodynamic equilibrium assumption. Obviously, there are many applications in which this does not apply. But there are fortunately also a huge number of applications for which this is a erasable application.

c) Ideal gas

By definition in thermodynamic equilibrium the INTERNAL ENERGY U is an equilibrium state variable, which means that it does not depend on the path (process) that has been followed to reach this equilibrium state. Basically we cannot prove this, and as every statement that we cannot prove, but appear to be useful we say that this is a basic (fundamental) law of the equilibrium science called "thermodynamic". Another law states that TEMPERATURE is also an equilibrium state variable. Density rho and pressure p are other state variables. For a uniform gas the "state" is uniquely determined by specifying two state variables. Hence U is a function of p and T. Therefore U=U(p,T). One defines an ideal gas as a gas for which U is only a function of the temperature T, in other words U=U(T).

d) Work

When considering a system allowed to receive heat Q from its environment and allowed to deliver work W to its environment infinitesimal processes (infinitesimally slow) such that the departure from equilibrium of the environment is negligible and they are "reversible" the mechanical work performed by a fluid on the environment will be p_ex dV, where p_ex is the pressure in de environment assumed to remain constant and dV is the increase in system volume V. The force of the environment on the system for a locally flat (contact) surface element dA has a magnitude -p_ex dA and is directed in the direction normal to the surface element dA from the system to the environment. Due to Newton's law we know that the reaction for of the system on the environment is p_ex dA. For a displacement dx of the contact surface between the system and its environment (dx is normal to the surface element and positive in the direction towards the exterior, toward the environment), dV=dAdx. By definition the work performed by this displacement is dW=(p_exdA)dx=p_ex dV. This will remain valid for all processes that do not involve departure from equilibrium in the environment even when there is departure from equilibrium within the system considered.

The idea that dW could be equal to pdV+Vdp is simply NONSENSE and more importantly even the formula dW=pdV should be used with care. When p is defined as the pressure in the system, we have a problem with non-equilibrium processes, because p is not well defined!

Yours sincerely,

Mico Hirschberg

Please excuse me for a confusion between local quantities such as mass density rho and global density U. U depends of course also on the volume of the system U=(p,T,V). For an ideal gas U=(T,V).

Dear Avraham,

I like your comments. From your statement "... the internal energy is dominated by the sum of the translational and rotational kinetic energy of individual molecules. " we perhaps could convey that in Gas/Fluid Dynamics we are used to define as "internal energy" only this part. And all other types of contribution you mentioned are summed explicitly in what is denoted at "total stored Energy". Maybe this convention is not usual in Thermodynamics.

I also agree about your point about work. This is on the line of my previous definition where n.v is nothing else that the displacement dx (for time unit) in the direction normal to the elementary surface dS.

Regards

We seem to agree. Of course when considering air as an ideal gas with constant specific heat (calorically perfect gas) we only take the translational and rotational kinetic energy into account. I however have been studying heat transfer in partially ionized gasses for re-entry problems in aerospatial applications. In such cases, electronic excitation and ionization becomes essential for monoatomic gasses (argon). For air one also have to take dissociation of molecules into account. Obviously, studying combustion involves chemical reactions. Hence there are common engineering applications of fluid dynamics for which the internal energy is not only kinetic energy.

Hallo,

Vicash Panday And other followers, I can not accept the the Term VdP work as it does not justify the definition of work as the product of force and displacement in the direction of force. Any mathematical model what ever one uses should be feasible with practical changes the particles undergo. What is cavitation bubble model and what is its contribution towards work and what is your research on it I do not know. I only understand a work term should satisfy the first basic definition.

I suggesto to have also a look to the book of Kundu Cohen (Fluid Mechanics), specifically pages 12-13 and 104-109. That would clarify both issues on kinetic/internal energy and work for the case of a fluid system (highlighiting differences from a thermodynamic system)

@ Krushna

Let there be a "real" thermodynamic system whose initial pressure and volume be p1 and v1, and final pressure and volume be p2 and v2. p1 and p2 are not the same, and, v1 and v2 are not the same. What is the work done?

I can give my opinion. According to what I wrote above, you have

Int [t1,t2] ( Int [S] Vn*p dS ) dt

and as an approximated estimation you can write

0.5*(t2-t2)*[ Int [S(t1)] Vn*p dS + Int [S(t2)] Vn*p dS]

and if you assume that Vn*p is approximatively constant over the surfaces

0.5*(t2-t2)*[Vn1*p1 |S(t1)| + Vn2*p2 |S(t2)|]

Of course, I considered only the reversible part of the work, in a real flow system you have to add the irreversible part due to the deviatoric stress.

Because the work W is not a state variable, the change in work from state (p1,V1) to (p2,V2) will depend on the path that is followed. Basically you have to integrate pdV along the chosen path. Elementary examples can help you to realise that integral of pdV does depend on the path. For example you can start by increasing the pressure from p1 to p2 at constant volume (no work as dV=0) and then increase the volume at constant pressure p2. This implies W=p2 (V2-V1). Another path would be increase the volume from V1 to V2 at constant pressure p1 (W=p1(V2-V1) and later increase the pressure at constant volume V2 from p1 to p2. Hence for this path W=p1(V2-V1). Als long as p1 is not equal to p2, these two work will be different.

Hello Vikash Panday

Simultaneous ptessure and volume change occutes in isothermal or adiabatic or politropic process where W= mRTln(V2/V1) or (P2V2 _ P1V1)/(n_ 1) but not simple PdV or VdP form.

Dear all

I just hope not to start another side discussion with the following.

Avraham initial detailed argument in “(b) Thermodynamic equilibrium”, resorted to “small enough” and “large enough” amount of material elements, as stated, a very common reference in fluid dynamics textbooks.

Browsing the Internet some fifteen years ago, I came across a sort of memory book by Clifford Truesdell (1919 - 2000), considered the “father” of modern continuum mechanics, where the two above, shall I say “wordings”, are somehow demolished. I quote Truesdell (1984): “to speak of an element of volume in a gas as “a region large enough to contain many molecules but small enough to be used as an element of integration” is not only loose but also needless and bootless”. That’s it.

Truesdell, C., An Idiot’s Fugitive Essays on Science – Methods, Criticism, Training, Circumstances, Springer – Verlag, New York (1984), pg. 23.

I also call your attention to an earlier paper of mine on this subject (online free access):

Peçanha RP (2015) Fluid Particles: A Review. J Chem Eng Process Technol 6:238.

doi: 10.4172/2157-7048.1000238

I will check that.

My mentor at TU Delft, prof. H.J.Merk, used to say " I lost one year of my life reading Thruesdell". Therefore I never read Thruesdell. I howver did enjoy very much reading Landau and Lifschitz. As an engineer, I did found their approach to fluid mechanics quite nice.

Dear Avraham

As a chemical engineer, aiming to solve daily practical problems, I would, probably, agree with some of the views of your mentor at TU Delft, regarding continuum mechanics, not Truesdell himself. Read Truesdell prose (e. g., the book I mentioned): straight forward, clear and unequivocal. Perhaps, sometimes, friendly with less used English terms.

Continuum mechanics is such a generalized approach to materials’ mechanical behavior that even in its early stages the mathematics required (basically Linear Algebra and Multivariable Calculus) makes it foggy and distant from the original physical phenomena under study. Most of the time, you’ll have to stick to mathematical rules to operate with the variables involved, not questioning their immediate physical meaning. Conservation laws (mass, momentum and energy), mappings between vector spaces and tensor quantities abound. Symbols of physical quantities/parameters with multiples subscripts and superscripts do resemble hieroglyphs. However, a theory aimed to treat gases, liquids, solids as well as multiphase systems (Mixture Theory) with the same basic equations, had to be complex.

Dear friends,

as an engineer, I must define the equilibrium condition of my system in an operational way and I do it with the following sequence.

First of all I have to fix my degree of approximation in solving the problem I am facing (1%, 0.1% and so on).

Then, I have to fix the observation time of my system (1second, 1 hour, 1day, 1 year), i.e. the characteristic time of my phenomenon.

Finally, I have to check the percentage variation of all the extensive quantities of my system (mass, momentum, energy, entropy etc.) during the observation time. These variations may be due to either fluxes at the system boundaries (e.g., flows), or to production within the system (e.g., chemical reactions, entropy production, etc.).

If all these variations during my observation time are less than my degree of approximation, I can consider my system in equilibrium.

Let us do a naïve example.

Consider a fair balloon full of helium. Being the helium a gas with a monoatomic small molecule, you have some helium leakage through the balloon membrane and spout, no matter how tight you close it. Then, if your degree of approximation is, say 1%, and your observation time is, say one hour, you may consider the system in equilibrium with respect to the mass fluxes. This is not true if, with the same degree of approximation, your observation time is one month. You will find on the floor the balloon which escaped to the ceiling one month before. The same reasoning could be applied to energy fluxes if the room temperature changes and so on.

So, from the operational point of view, the condition of equilibrium of a system is not absolute or objective, but it depends on your degree of approximation and observation time.

Regards to all.

Of course, for systems not in static equilibrium, you have to introduce the evolutionary equilibrium concept, but this is another story.

I agree about all was written but it seems turning around the original question: how to compute the work for a general real fluid system deforming its shape at the boundary (such as that of a bubble expanding or collapsing) not for the standard hystorical thermodynamics system.

I wrote my idea, is that operatively acceptable?

Hi everybody,

Thanks Giovanni for his perfect answer. I think his answer is great and enough. Simply, Internal energy comes from several types of energy resources which one of them is particle's kinetic energy (molecular and submoleculars). Beside the kinetic, potential and others can be factored in the internal energy.

Regards

SAMD

https://physics.stackexchange.com/questions/287863/how-to-define-exactly-the-internal-energy

Dear Ahmed Sir, Out of your two questions

Please, remember that Q is not the heat made by system, but is the heat transferred to the system.

Regards

Dear Vikash,

For an ideal monatomic gas, the internal energy is simply equal to the kinetic energy of thermal motion, but a real gas is not.

By definition, pdV denotes that p is constant, V is changed, δW=pdV denotes work done between a system and its surroundings, Vdp denotes that V is constant, p is changed, Vdp can be caused by heat exchange or by internal energy conversion but no work done between a system and its surroundings, so we have

d(pV)=pdV+Vdp.

δW=pdV.

δW ≠ pdV+Vdp.

Best

Interesting question

Internal energy represents total energies inside substance. It is purely molecular energy and might exist in different forms such as sensible, latent, chemical, and nuclear energy.

For gases, part of the internal energy associated with the motion of particles which takes the form of kinetic energy and classify as sensible energy. The motion and configuration of the substance depend on temperature. The average velocity of the molecules is proportional to the temperature, hence higher temperature lead to higher kinetic energy as well as internal energy.

In brief, the kinetic energy is part of internal energy but not constitute the whole value. For the second part of your question why the term of Vdp omits in work equation. That occurs because pressure has constant value, so dp is equal to zero.

Dear Mr. Almutairi,

In your second sentence, you should change your word particle with molecules. Please see my previous answer.

Then, in your simple model, the average velocity of the molecules is proportional to the square root of temperature.

Regards

Dear Researcher, any one will understand molecule from the word particle as in chemistry tiny particles represent molecules.

In fluid dynamics, molecule has a meaning and particle another one. Please see my previous answer. Thanks

this problem has been solved. Thanks to all. I wish I could delete this thread but I cannot. No need to answer it any further. This is resolved now.

I agreed with Giovanni Maria Carlomagno

Dear Giovanni Marie Carlomagno, I read your previous answer about heat made in a system. It is possible to form or release heat in system. For example a system subjected to oxidation releases heat. Again you say heat is transferred to the system, but can not be removed or transferred from the system.

Please think it in cool mind.