Work Is Measured In Joules

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Why is estrus measured in joules?

• Thread starter Entanglement
• Start date Mar 20, 2014

My limited knowledge tells me that joules= newton.meters " Westward = F.d How can this equation be related to rut since this equation depends on force washed for a certain deportation which seems unrelated to rut, when something produces rut where is this such force and displacement ????

Answers and Replies

Oestrus is microscopic kinetic energy, stored in the motion of molecules. If you apply a net force F over a distance d to a mass m, it gains the kinetic energy: F * d = 1/2 * g * v²

Heat is measured in joules because James Prescott Joule experimentally showed that mechanical energy is converted to rut in 1843.

See http://en.wikipedia.org/wiki/James_Prescott_Joule

For more details study thermodynamics and statistical mechanics.

My express noesis tells me that joules= newton.meters " Westward = F.d How can this equation be related to heat since this equation depends on force done for a certain displacement which seems unrelated to estrus, when something produces heat where is this such forcefulness and displacement ????

Try playing with a bike tyre pump?

My limited knowledge tells me that joules= newton.meters " Due west = F.d How can this equation be related to heat since this equation depends on forcefulness done for a sure deportation which seems unrelated to heat, when something produces estrus where is this such force and deportation ????

Just to add together what others accept said, Joule's experiment involved applying a force through a distance to churn water in a tank and measuring the change in temperature. He institute that the change in temperature of a tank of water was proportional to the work washed. So it is quite natural for heat and work to be measured in Joules.

AM

Heat is microscopic kinetic free energy, stored in the movement of molecules. If you utilise a net force F over a distance d to a mass k, it gains the kinetic free energy: F * d = i/2 * grand * v²

This interpretation is a bit missleading. Microscopic kinetic energy(of molecul movements) corresponds with temperature rather then with heat. Heat doesn't equal temperature. Estrus is merely energy that a system gains or gives up. Strictly speaking, heat is transfer of energy. Thus rut is measured in the aforementioned units every bit energy, i.eastward. joules.

This interpretation is a chip misleading. Microscopic kinetic energy(of molecule movements) corresponds with temperature rather so with heat.

The microscopic kinetic energy of the molecules does not represent to temperature. The quantity that corresponds to temperature is ##\frac{\partial{East}}{\partial{S}}## where ##S## is the entropy; this is what determines which direction the heat is transferred between ii bodies every bit they move towards equilibrium.

The change in the microscopic kinetic energy of the molecules is exactly equal to the amount of heat transferred, which is why (to reply OP'due south question) we measure heat in Joules. If I have two bodies, one warmer than the other, and I bring them into contact with each other, estrus will flow from the warmer ane to the cooler one. When they reach equilibrium, I volition find that the full kinetic energy of the molecules in the ane has increased past exactly the same number of Joules equally the the total kinetic energy of the molecules in the other has decreased. Thus, information technology's very natural to think of the transfer of rut as the transfer of some number of Joules of energy.

At that place is an interesting history to this. The original idea was that Estrus was a substance (called Caloric) which would menstruation out of 'fire' and enter objects, every bit they got hotter. This idea was disproved by experiments which involved a lot of piece of work, tedious out a catechism, and lots of rut. Still, the first and cease (total) mass turned out to exist the same so it was concluded that heat was not an actual substance.

Even every bit recently as the 1960s, schools taught the 'Mechanical Equivalent of Oestrus', which is 4.2 Joules per Calorie, relating Work to Oestrus. (Same value as today but the name is not used whatsoever more.

This interpretation is a bit missleading. Microscopic kinetic free energy(of molecul movements) corresponds with temperature rather and then with oestrus. Rut doesn't equal temperature. Heat is merely free energy that a system gains or gives up. Strictly speaking, heat is transfer of free energy. Thus heat is measured in the same units as energy, i.e. joules.

That'south true, but for an platonic gas at constant book, all the oestrus added to the gas goes into kinetic free energy. If the book is allowed to modify, then heat added can become into expanding the volume in addition to increasing the kinetic energy.

The microscopic kinetic energy of the molecules does not stand for to temperature. The quantity that corresponds to temperature is ##\frac{\partial{East}}{\fractional{S}}## where ##S## is the entropy; this is what determines which direction the heat is transferred between two bodies as they motion towards equilibrium.

The change in the microscopic kinetic energy of the molecules is exactly equal to the corporeality of heat transferred, which is why (to reply OP's question) nosotros mensurate estrus in Joules. If I accept two bodies, one warmer than the other, and I bring them into contact with each other, heat will menses from the warmer one to the cooler one. When they reach equilibrium, I volition find that the full kinetic energy of the molecules in the 1 has increased past exactly the aforementioned number of Joules as the the full kinetic energy of the molecules in the other has decreased. Thus, it's very natural to think of the transfer of heat as the transfer of some number of Joules of energy.

For an ideal gas, the internal energy [itex]U[/itex] is merely an accounting of the kinetic energy of the molecules. And so a change in [itex]U[/itex] ways a change in kinetic energy.

But not all heat added goes into changing [itex]U[/itex]. By the equation

[itex]dU = dQ - dW[/itex]

or rearranged:

[itex]dQ = dU + dW[/itex]

nosotros can run across that only some of the heat added goes into kinetic energy, and some of it goes into performing piece of work (expanding the volume against a pressure).

I tin can't concord with the opinion that kinetic energy of a particle of gas does not correspond to temperature. Each caste of freedom has energy kT/2. Mean kinetic energy of molecules (regardless of number of atoms) ways three degrees of freedom. It ways that the mean kinetics energy of molecules is 3kT/ii. I don't know what your definition of the give-and-take 'to correspond' is, simply it is obvious there's a relationship betwixt mean kinetic energy of molecules and temperature. I judge, it was Feynman who said in his lectures that nosotros could possibly measure temperature in joules instead of kelvins since temperature is nothing else than hateful kinetic energy of the molecules.

However, everyone will hopefully agree that rut is only energy transferred from one object to another. We shouldn't misfile it with internal energy.

That's truthful, but for an platonic gas at abiding volume, all the heat added to the gas goes into kinetic energy. If the volume is allowed to change, and then heat added can go into expanding the book in addition to increasing the kinetic energy.

Isn't the expansion of the gas as a effect of weakening the intermolecular forces which results from the increase of K.E, it all happens due to the increment of K.Eastward, what do you mean by a part increases the Grand.E while the other part make the gas to aggrandize????

Isn't the expansion of the gas as a result of weakening the intermolecular forces which results from the increase of K.E, it all happens due to the increase of Grand.E, what exercise you mean by a office increases the K.East while the other part make the gas to aggrandize????

When a gas expands, it cools down--the temperature goes down, and and so does the average kinetic energy. And so if you do the following:

1. Heat upward a gas (this causes its average kinetic free energy to go up)
2. Allow the gas to expand (this causes its average kinetic to get down)

So after these ii steps, the net issue is to raise the temperature (and boilerplate kinetic energy) a little, and to increase the volume a picayune chip.

Isn't the expansion of the gas as a result of weakening the intermolecular forces which results from the increase of M.Due east, information technology all happens due to the increase of Thou.E, what do you mean by a part increases the K.E while the other part make the gas to expand????

No, the intermolecular forces are neglected in the ideal gas model.
And are negligible in real gases in "normal" conditions.
They play no office in the expansion of the gas.

When a gas expands, it cools downward--the temperature goes down, and then does the average kinetic energy. So if you do the following:

[*]Allow the gas to expand (this causes its average kinetic to go down)

[/LIST]

.

I idea increment in temp -------> increment in volume ------> increment in K.E how does the expansion let the K.E go down ??

Last edited: Mar 25, 2014

No, the intermolecular forces are neglected in the ideal gas model.

And are negligible in real gases in "normal" conditions.

They play no part in the expansion of the gas.

Then how are gases expanded ?????

I idea increase in temp -------> increment in K.E ------> increase in K.E how does the expansion let the K.Eastward go down ??

If you allow a gas expand against, say, a piston in a cylinder, the piston volition be moving abroad, against some restraining force and the momentum change as the molecules strike the piston will result in their speeds being decreased slightly. The loss of the KE of the molecules is fabricated up for past the gain of in energy of the piston and its restraint. This all refers to an platonic gas - no inter-molecular effects in the gas are involved.

And so how are gases expanded ?????

By increasing the volume available, for example. A gas will fill all the volume available to it. No external agent is necessary.
And this is because the inter-molecular forces are negligible in gases. The container is what keeps the gas from expanding and not inter-molecular forces. If the container cannot yield there will be no expansion, no affair how much you lot heat up the has and increase kinetic energy.

I don't really encounter how would think that expansion needs inter-molecular forces.
Perhaps y'all take in mind a condensed phase, like liquids and solids. Here the thermal expansion is due not to the "weakening of intermolecular forces" but to an increase in thermal energy combined with the non-harmonicity of the intermolecular potential.

I idea increment in temp -------> increment in book ------> increase in K.Eastward how does the expansion permit the K.East go downward ??

Well, imagine a molecule striking a wall. The molecule will bounciness off the wall. If the wall is moveable, then some of the kinetic free energy of the molecule will be transferred to the wall, causing the wall to move out, and therefore causing the volume to increment.

The energy [itex]\delta Eastward[/itex] expended on pushing the wall outward is given past:

[itex]\delta E = F \delta x[/itex]

where [itex]\delta x[/itex] is the altitude the wall has been pushed out, and [itex]F[/itex] is the force exerted on the wall. In the case of a gas, the force on the wall is but [itex]P \cdot A[/itex], where [itex]P[/itex] is the gas force per unit area, and [itex]A[/itex] is the area of the wall. Putting these facts together gives:

[itex]\delta East = P A \delta x = P \delta V[/itex]

The quantity [itex]A \delta 10[/itex] is the change in book of the box resulting from pushing the wall, of expanse [itex]A[/itex] out a distance [itex]\delta x[/itex].

Then when the box expands, the energy transferred from the gas to the wall is given by:
[itex]\delta East = P \delta V[/itex]. That is the free energy that is taken abroad from the kinetic energy of the gas molecules.

By increasing the volume available, for instance. A gas volition fill all the book available to it. No external agent is necessary.
And this is considering the inter-molecular forces are negligible in gases. The container is what keeps the gas from expanding and not inter-molecular forces. If the container cannot yield there will be no expansion, no affair how much you estrus upwardly the has and increase kinetic energy.

I don't actually see how would think that expansion needs inter-molecular forces.
Maybe y'all accept in mind a condensed phase, like liquids and solids. Here the thermal expansion is due non to the "weakening of intermolecular forces" but to an increase in thermal free energy combined with the non-harmonicity of the intermolecular potential.

But are intermolecular forces involved in the contraction of gases in example of cooling ???

I got your signal guys, on heating a gas the molecules will gain K.E, if the walls of the container are movable a function of the force will be consumed in doing piece of work to move these walls so the volume occupied by the gas increases, merely if the walls aren't movable the book will never aggrandize and all the oestrus volition just increase the K.E. Simply my question is what is the net change if the walls are movable, volition the K.E increase, decrease, or stay abiding????

Just are intermolecular forces involved in the contraction of gases in case of cooling ???

I got your bespeak guys, on heating a gas the molecules will proceeds One thousand.E, if the walls of the container are movable a part of the force will be consumed in doing work to move these walls so the book occupied by the gas increases, only if the walls aren't movable the volume volition never expand and all the heat will just increment the K.E. But my question is what is the net change if the walls are movable, will the K.Due east increase, decrease, or stay constant????

The intermolecular forces are relevant when the molecules are close enough together. In the limit, this results in a change of land (condensation), during which the temperature is unchanged and all the free energy change is taken up with the energy of the molecular attractions. (Merely this is non in the region where a gas tin can be considered as 'ideal').

You lot seem to desire to talk over the effects entirely in words, when the Gas Laws describe it all, perfectly in a line of Maths.

P₁V_ane/T_ane = P₂V₂/T₂

answers your question. You just plug in the values you want to change and the values y'all want to keep constant and information technology tells you what you want to know. The areas in the PV diagram show you the work done on or by the gas and energy is conserved. It is possible to add energy without increasing the temperature at all (isothermal), in which case, all the added free energy is used upwardly as work done on the moving cylinder.

The intermolecular forces are relevant when the molecules are close enough together. In the limit, this results in a change of land (condensation), during which the temperature is unchanged and all the energy alter is taken upwardly with the energy of the molecular attractions. (Merely this is not in the region where a gas can be considered every bit 'ideal').

You seem to want to discuss the furnishings entirely in words, when the Gas Laws depict it all, perfectly in a line of Maths.

P₁V_one/T₁ = P₂V₂/T_ii

answers your question. You simply plug in the values you want to change and the values y'all want to keep abiding and information technology tells you what you want to know. The areas in the PV diagram show you the work done on or past the gas and energy is conserved. It is possible to add together energy without increasing the temperature at all (isothermal), in which case, all the added free energy is used upwardly equally piece of work done on the moving cylinder.

I don't get how is the isothermal process done ???
I

I don't become how is the isothermal procedure done ???
I

Compress (or permit to expand) a gas in a state of affairs where the temperature cannot change and you get Boyle's Law behaviour. PV is constant.
I think you should read about the Gas Laws, starting at the very beginning. Many of your questions betoken that yous are not totally familiar with the basics of this stuff. These basics accept upward several weeks worth of lessons, unremarkably.

I retrieve yous should read near the Gas Laws, starting at the very beginning. Many of your questions point that you are non totally familiar with the basics of this stuff. These basics accept up several weeks worth of lessons, usually.

Thermodynamics was excluded from our syllabus last year due to the political situation in my state, in our syllabus this yr there is a chapter about cryogenics, then I'm studying low temperature physics with very picayune thermodynamics knowledge, I know that is stupid, only that's not my fault, the situation in my state is horrible, but I'll try to practice me best.

I don't become how is the isothermal procedure washed ???
I

If T is constant and V increases then P has to subtract. If P decreases, the walls practise not expand if the external pressure remains the same. And so if T is constant, the external pressure has to subtract if at that place is an expansion.

If you take your instance, if rut flows into a gas, T increases and P increases if V is constant. Withal, if the external pressure level is constant, then an increase in Five continues until P is equal to the external pressure. And then this describes a constant pressure level expansion.

AM

If T is abiding and V increases and then P has to decrease. If P decreases, the walls do not aggrandize if the external pressure level remains the aforementioned. So if T is constant, the external pressure level has to decrease if there is an expansion.

If you take your example, if heat flows into a gas, T increases and P increases if V is abiding. However, if the external pressure level is constant, so an increase in V continues until P is equal to the external pressure. So this describes a abiding pressure expansion.

AM

.......

Concluding edited: Mar 26, 2014

I think I'll accept to report the gas laws first, let's start with Boyle'due south law

It states that force per unit area is inversely proportional to volume, is that because the no of collisions of the molecules with the walls of the container due to Brownian move increase in a smaller volume ???

It states that pressure is inversely proportional to volume, is that because the no of collisions of the molecules with the walls of the container due to Brownian motion increase in a smaller volume ???

It'southward not that the motion of each molecule increases. If temperature is held contstant, the boilerplate velocity of the molecules will not change. It is that there are more molecules. So more than molecules hitting a given surface area of wall surface per unit time.

It'south not that the motion of each molecule increases. If temperature is held contstant, the average velocity of the molecules will not change. It is that in that location are more than molecules. So more molecules hitting a given surface area of wall surface per unit of measurement time.

Yes I got that, It just increases the chance of its collision with the walls, we can say it changes the no of collisions per unit area and per unit time but the boilerplate kinetic energy doesn't change since the Temperature is constant

when we are talking most the pressure of gases, what is really meant, its pressure due to gravity?? or its pressure due to its K.E and its collision with the environs ????

Shrink (or permit to expand) a gas in a situation where the temperature cannot modify and you become Boyle's Law behaviour. PV is constant.

.

If I compressed a gas in a container of movable walls which isn't thermally isolated from the surrounding, that is considered as a isothermal process ??

If I compressed a gas in a container of movable walls which isn't thermally isolated from the surrounding, that is considered every bit a isothermal procedure ??

It would exist far more efficient for you to read through and follow what a textbook has to say about all this. Await at This link and also Wikipedia volition tell yous all you lot need to know. Q and A is a very poor way of learning basic stuff, compared with going with the flow of the argument presented in a textbook - that has been tried and tested. Q and A is a bit like Brownian Move; every and so ofttimes you may randomly get in the right management but most of the time you can cease up astray.

If I compressed a gas in a container of movable walls which isn't thermally isolated from the surrounding, that is considered equally a isothermal process ??

Isothermal means that information technology is kept at a constant temperature. 1 way to exercise this is to take a very large reservoir (or air, or h2o, or whatsoever) that is kept at a abiding temperature, and do the compression so slowly that the gas in the container is always allowed to attain equilibrium with the reservoir.

Isothermal means that it is kept at a constant temperature. One manner to practice this is to take a very big reservoir (or air, or water, or whatever) that is kept at a constant temperature, and exercise the pinch so slowly that the gas in the container is always allowed to accomplish equilibrium with the reservoir.

And so during the slow compress the gas keeps gaining and losing heat to the surrounding medium at the same rate staying at an equilibrium position with surrounding, and then its temperature stays abiding

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Work Is Measured In Joules,

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