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Joule's Experiment

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• Intro 0:00
• Joule's Experiment 0:09
• Joule's Experiment
• Interpretation of the Result 4:42
• The Gas Expands Against No External Pressure
• Temperature of the Surrounding Does Not Change
• System & Surrounding
• Joule's Law
• More on Joule's Experiment
• Later Experiment
• Dealing with the 2nd Law & Its Mathematical Consequences

Transcription: Joule's Experiment

Hello, welcome back to www.educator.com and welcome back to Physical Chemistry.0000

Today, we are going to talk about Joules experiment.0004

Let us just jump right on in.0007

Recall in the last lesson what we did is we said that energy was a function of both temperature and volume.0010

We created this total differential expression we had which was du = du dt sub VDT + du DV sub TDV.0021

In that lesson, we associated this particular thing with the constant volume with a heat capacity under constant volume.0041

Now, the question is, we took care of this, we associated with some easily measurable quantity.0052

That is the question here.0061

We are just going to deal with the other partial derivative in this total differential expression.0063

What can we do with that?0066

As it turns out, the answer is not much.0068

A little bit disappointing but we will be able to say a couple of things about it.0073

But the truth is we cannot really do much about it right now.0077

Let us go ahead and draw out what is that he did.0090

He came up with something very ingenious.0097

This is going to be water and of course the water has a thermometer in it.0100

Its measures change in temperature, things like that.0106

There is a little stir in here to make sure that the water is stirring up well.0110

Let me go ahead and draw this apparatus in here.0115

This is A and this is B and I will go ahead and shade this one in.0135

Basically, the idea is this.0145

We want to find out, remember du DV, we want to find out if there is a change in volume, what is going to be a change in energy?0148

Because that is what this derivative says.0160

It is the rate of change of energy with respect to volume per unit change in volume, how much does the energy change?0161

He took his apparatus which consisted of two balls separated by a stop cock.0176

He evacuated this chamber altogether, this B, and A is filled with his gas.0182

Basically, what we do is we take this thing and we drop it into this tub of water and we allow this to come to thermal equilibrium.0188

The system to come to thermal equilibrium with its surroundings.0198

When that happens, what we do is we open up the stop cock because of the evacuated his gas is going to expand.0202

It does experience a change of volume.0209

And then upon this change of volume, we allow the system to become the thermal equilibrium again.0212

What we do is we measure the temperature change of the water.0221

If that is what we are doing, we are seeing if there is going to be some energy change based on a change in volume.0225

We are trying to measure this, what is the relationship between a change in volume and a change in temperature which is a change in energy?0232

Here is what actually happened.0242

Here is the result, no change in temperature.0248

Is it surprising, the expectation was that you open this up, the system would undergo volume change,0261

there would be a temperature drop or temperature rise and then heat would either flow in or out and that heat flow in or out will be measured by this.0268

That would give us a change in energy vs. A change in volume.0277

Here is the interpretation of the results.0283

We have the interpretation, the gas expands against no external pressure.0290

Remember, we said that B is completely evacuated, there is a vacuum in there.0300

There is no external pressure against which this gas would expand is pushing.0305

It expands against no external pressure.0310

In a situation where that is the case, it is called a free expansion.0317

If a gas expands against 0 external pressure, it is called a free expansion.0322

No work is done, P external is 0.0335

Work = P external × the change in volume.0340

If there is a change in volume if the external pressure is 0 the work is 0.0343

No work is done.0349

In other words, DW is equal to 0.0355

Du = DQ – dw, DW= 0 so du = dq.0361

Therefore, this particular case, the change in energy is just a change in heat.0372

So far, so good.0377

Since T of the surroundings which is the water, does not change, the temperature of water does not change so no heat is transferred.0382

Dq = 0, dq = Du = 0.0407

The energy change is 0.0416

The system and the surroundings are in thermal equilibrium.0426

The change in temperature is equal to 0.0443

The system DT, the temperature of the water does not change which means the temperature of the system does not change.0448

That is this DT.0457

Du=DUDT sub V × DT +du dv sub T × DV or it becomes du is equal is 0.0463

Therefore, this goes to 0.0486

What we are all left with the terms of that equation, this total differential equation is du DV sub T × Dv is equal to 0.0488

Because we know from the previous part of this that the du is equal to 0.0501

For this equation that DT is equal to 0, that goes to 0.0507

All we are left with is, du = du dv sub t dv.0510

We know that = 0.0516

But you know the change in volume does not equal 0.0520

There is a change in volume, the system change this volume.0524

If dv is not equal to 0, basic property of mathematics A × B.0527

If B is in 0 and A has to be 0, therefore, du Dv sub T is equal to 0.0532

That is the interpretation.0544

That is what this is, we just found that this derivative is not some property, it is actually equal to 0.0553

Nothing is happening.0560

In other words, the rate of change of energy with volume is equal to 0.0564

The change in energy is not associated with a change in volume.0577

The volume can change, the energy of the system does not change.0584

That is what this is saying.0587

The previous lesson, we found out that if the temperature changes, the energy changes.0589

Now, we are working with this one.0595

This experiment says that this is actually equal to 0.0597

Energy does not depend on volume.0601

Energy only depends on temperature.0603

Since the rate of change of energy with volume = 0, that is that thing, energy is independent of volume.0605

Our initial U is equal to TV becomes U is just a function of T, not a function of V.0630

That is what this is.0642

That is Joules experiment says, this is Joules law.0644

This is what Joules law says, that the energy is a function of temperature not a function of temperature and volume.0651

That is what Joules law is.0657

Here is what is interesting, Joules law is not exactly true.0661

That is pretty close to being true.0674

As it turns out, the large heat capacity of the water and the small heat capacity of the gas in that apparatus,0678

it caused the temperature effect caused the effect to be too small to be observed.0714

As it turns out, there was a temperature difference.0738

If the difference was so tiny, that the difference in the capacity of a water and small heat capacity of that gas itself, it is as if the water did notice.0742

It is like dropping a teaspoon of water into the ocean.0751

The ocean is not going to notice.0754

Later experiments, with much more sophisticated sense instrumentation showed that du DV under constant temperature for a real gas does not equal 0.0761

It is very small though.0792

For all practical purposes, we can take DU DV T equal 0, even under for real gases.0794

For ideal gases it is always true.0799

For real gas it is not, it is very small though.0801

For an ideal gas, for the most part we would be dealing with ideal gases unless told otherwise.0808

From ideal gas, du/ dv sub T = 0, definitely something that you want to remember.0819

When we start doing problems, we will go over all this again.0828

When we get to the second law of thermodynamics, when we deal with the second law and0838

its mathematical consequences, then we will identify this derivative, this du/ dv sub T with an easily measurable quantity.0850

Remember, that other derivative, that DUDT in the last lesson, we associate it with a very easily measurable quantity, the heat capacity of the system.0888

In this particular case is 0.0898

Later on, when we deal with the second law, we will actually be able to identify this with an actual measurable quantity.0900

Even though, we now have a U, the energy is just a function of T, to be as precise as possible0917

both qualitatively and mathematically, for the purposes of derivation and working with the mathematics of equations,0942

we will continue to write that the differential change in energy is equal to the heat capacity × DT.0962

We will replace the derivative of what we know what to be the heat capacity + du/ DV sub TDV.0974

We are dealing with an ideal guess, we can just go ahead and take this equal to 0 and it just drops out, it just vanishes.0983

For mathematical purposes and to be as precise as possible, we will go ahead and keep it this way.0989

That is it, that is what the Joules experiment was.0996

We just want to associate and we want to find out what this partial derivative represent and can we measure it.0998

What it is associated with.1005

Thank you so much for joining us here at www.educator.com.1007

We will see you next time, bye.1009