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A.T.

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UltrafastPED

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

For more details study thermodynamics and statistical mechanics.

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CWatters

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Try playing with a bicycle tyre pump?

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Andrew Mason

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

AM

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This interpretation is a bit missleading. Microscopic kinetic energy(of molecul movements) corresponds with temperature rather then with heat. Heat doesn't equal temperature. Heat is simply energy that a system gains or gives up. Strictly speaking, heat is transfer of energy. Thus heat is measured in the same units as energy, i.e. joules.Fover a distancedto a massm, it gains the kinetic energy:F * d = 1/2 * m * v^{2}

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Nugatory

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The microscopic kinetic energy of the molecules does not correspond to temperature. The quantity that corresponds to temperature is ##\frac{\partial{E}}{\partial{S}}## where ##S## is the entropy; this is what determines which direction the heat is transferred between two bodies as they move towards equilibrium.This interpretation is a bit misleading. Microscopic kinetic energy(of molecule movements) corresponds with temperature rather then with heat.

The change in the microscopic kinetic energy of the molecules is exactly equal to the amount of heat transferred, which is why (to answer OP's 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, heat will flow from the warmer one to the cooler one. When they reach equilibrium, I will find that the total kinetic energy of the molecules in the one has increased by exactly the same number of Joules as the the total 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.

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sophiecentaur

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

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That's true, but for an ideal gas at constant volume, all the heat added to the gas goes into kinetic energy. If the volume is allowed to change, then heat added can go into expanding the volume in addition to increasing the kinetic energy.This interpretation is a bit missleading. Microscopic kinetic energy(of molecul movements) corresponds with temperature rather then with heat. Heat doesn't equal temperature. Heat is simply energy that a system gains or gives up. Strictly speaking, heat is transfer of energy. Thus heat is measured in the same units as energy, i.e. joules.

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For an ideal gas, the internal energy [itex]U[/itex] is simply an accounting of the kinetic energy of the molecules. So a change in [itex]U[/itex] means a change in kinetic energy.The microscopic kinetic energy of the molecules does not correspond to temperature. The quantity that corresponds to temperature is ##\frac{\partial{E}}{\partial{S}}## where ##S## is the entropy; this is what determines which direction the heat is transferred between two bodies as 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 answer OP's 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, heat will flow from the warmer one to the cooler one. When they reach equilibrium, I will find that the total kinetic energy of the molecules in the one has increased by exactly the same number of Joules as the the total 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.

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]

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

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However, everyone will hopefully agree that heat is just energy transferred from one object to another. We shouldn't confuse it with internal energy.

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That's true, but for an ideal gas at constant volume, all the heat added to the gas goes into kinetic energy. If the volume is allowed to change, then heat added can go into expanding the volume in addition to increasing the kinetic energy.

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 K.E, what do you mean by a part increases the K.E while the other part make the gas to expand????

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When a gas expands, it cools down--the temperature goes down, and so does the average kinetic energy. So if you do the following: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 K.E, what do you mean by a part increases the K.E while the other part make the gas to expand????

- Heat up a gas (this causes its average kinetic energy to go up)
- Allow the gas to expand (this causes its average kinetic to go down)

So after these two steps, the net effect is to raise the temperature (and average kinetic energy) a little, and to increase the volume a little bit.

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No, the intermolecular forces are neglected in the ideal gas model.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 K.E, what do you mean by a part increases the K.E while the other part make the gas to expand????

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

They play no part in the expansion of the gas.

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When a gas expands, it cools down--the temperature goes down, and so 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 thought increase in temp -------> increase in volume ------> increase in K.E how does the expansion let the K.E go down ??

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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 ?????

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sophiecentaur

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If you let a gas expand against, say, a piston in a cylinder, the piston will be moving away, 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 made up for by the gain of in energy of the piston and its restraint. This all refers to an ideal gas - no inter-molecular effects in the gas are involved.I thought increase in temp -------> increase in K.E ------> increase in K.E how does the expansion let the K.E go down ??

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By increasing the volume available, for example. A gas will fill all the volume available to it. No external agent is necessary.Then how are gases expanded ?????

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 matter how much you heat up the has and increase kinetic energy.

I don't really see how would think that expansion

Maybe you have 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.

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Well, imagine a molecule hitting a wall. The molecule will bounce off the wall. If the wall is moveable, then some of the kinetic energy of the molecule will be transferred to the wall, causing the wall to move out, and therefore causing the volume to increase.I thought increase in temp -------> increase in volume ------> increase in K.E how does the expansion let the K.E go down ??

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

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

where [itex]\delta x[/itex] is the distance 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 just [itex]P \cdot A[/itex], where [itex]P[/itex] is the gas pressure, and [itex]A[/itex] is the area of the wall. Putting these facts together gives:

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

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

So when the box expands, the energy transferred from the gas to the wall is given by:

[itex]\delta E = P \delta V[/itex]. That is the energy that is taken away from the kinetic energy of the gas molecules.

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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 matter how much you heat up the has and increase kinetic energy.

I don't really see how would think that expansionneedsinter-molecular forces.

Maybe you have 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.

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

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sophiecentaur

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

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

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

P

answers your question. You just plug in the values you want to change and the values you want to keep constant and it 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 energy is used up as work done on the moving cylinder.

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I don't get how is the isothermal process done ???The intermolecular forces are relevant when the molecules are close enough together. In the limit, this results in a change of state (condensation), during which the temperature is unchanged and all the energy change is taken up with the energy of the molecular attractions. (But this is not in the region where a gas can be considered as 'ideal').

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

P_{1}V_{1}/T_{1}= P_{2}V_{2}/T_{2}

answers your question. You just plug in the values you want to change and the values you want to keep constant and it 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 energy is used up as work done on the moving cylinder.

I

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sophiecentaur

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Compress (or allow to expand) a gas in a situation where the temperature cannot change and you get Boyle's Law behaviour. PV is constant.I don't get how is the isothermal process done ???

I

I think you should read about the Gas Laws, starting at the very beginning. Many of your questions indicate that you are not totally familiar with the basics of this stuff. These basics take up several weeks worth of lessons, usually.

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Thermodynamics was excluded from our syllabus last year due to the political situation in my country, in our syllabus this year there is a chapter about cryogenics, so I'm studying low temperature physics with very little thermodynamics knowledge, I know that is stupid, but that's not my fault, the situation in my country is horrible, but I'll try to do me best.I think you should read about the Gas Laws, starting at the very beginning. Many of your questions indicate that you are not totally familiar with the basics of this stuff. These basics take up several weeks worth of lessons, usually.

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