Sign In | Subscribe
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Physical Chemistry
  • Discussion

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books

Bookmark and Share

Start Learning Now

Our free lessons will get you started (Adobe Flash® required).
Get immediate access to our entire library.

Sign up for Educator.com

Membership Overview

  • Unlimited access to our entire library of courses.
  • Search and jump to exactly what you want to learn.
  • *Ask questions and get answers from the community and our teachers!
  • Practice questions with step-by-step solutions.
  • Download lesson files for programming and software training practice.
  • Track your course viewing progress.
  • Download lecture slides for taking notes.
  • Learn at your own pace... anytime, anywhere!

Three Miscellaneous Example Problems

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Example I 0:41
    • Part A: Calculating ∆H
    • Part B: Calculating ∆S
  • Example II 24:39
    • Part A: Final Temperature of the System
    • Part B: Calculating ∆S
  • Example III 46:49

Transcription: Three Miscellaneous Example Problems

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

Today, I want to throw in these 3 miscellaneous problems.0004

It is not like they have another place in the previous problem sets.0009

In fact, they actually technically belong to the entropy section but I want talk about few other things before I brought these back.0014

They have to do with a little bit of what we discussed previously.0023

They have to do with heat transfer, changes in phase, things like that.0027

In some sense, it is a little bit of everything.0031

The third example problem is strictly an entropy problem but I just thought I collect them here like this.0033

Let us get started.0040

The first example is on a fully insulated flask 40 g of liquid water at 25°C is added to 25 g of ice at -7°C,0044

What is the final state of the system given the following data and pressure is held constant during this process.0056

The constant pressure heat capacity for liquid water 4.18, constant pressure heat capacity for solid water, ice is 2.09 and the δ H of fusion is 334 J/ g.0063

In addition to finding the final state calculate δ H and δ S for this transformation.0079

The final state of system, what they are asking for is we have 40 g of liquid water and we have 25 g of ice.0084

Some of the ice is going to melt and it is going to melt it because the liquid water is going to end up cooling.0094

We want to find out what is the final temperature of the system?0103

How much ice is there and how much liquid water is there?0110

That is what they mean by what is the final state of the system.0112

Let us just jump on in and see what we can do.0116

A lot of this has to do with a lot of heat capacity that we have talked about earlier but 0119

it actually harkens back to what it is that you learn in general chemistry.0124

Remember Q = MC δ T and the amount of heat that something loses is the amount of heat that something else gains.0128

In this particular case, the amount of energy lost upon cooling of the water is going to be gained by the ice and that is going to end up melting.0136

We are going to reach some final equilibrium temperature that is going to be some mixture of water and ice.0144

Maybe it is going to be all ice or maybe it is going to be all water.0149

We do not know, that is what we need to find out.0152

Let us see what we have.0157

Once again, we know that liquid water is going to cool.0160

We know that the ice is going to warm up and it is going to melt.0162

The first thing I'm going to do is find out how much energy is required to bring the 25 g of ice from 7°C up to 0°C.0168

The reason that I happen to pick the ice first just to see it, is just that I notice that since the ice was at -7, it is not that far from 0.0180

I want to see how much energy and it is only 25 g, where is the water itself you have actually 40 g and it is at 25°C.0191

The heat capacity of liquid water is higher than the heat capacity of the ice.0200

Chances are just qualitatively looking at the problem, 0205

I'm probably going to be able to raise all of the ice to 0 without dropping the liquid water temperature too far.0211

That is what my intuition tells me just based on the data that I have.0219

Let us see if that is actually true.0223

It does not really matter where you start, whether it is with the ice or with the water.0225

You just have to start somewhere and inch your way, converge on some final temperature and final state.0229

Remember from general chemistry we had this Q = MC δ T.0237

The amount of energy lost or gained by something = the mass of something× the specific heat × the change in temperature.0245

This δ G gives us the change in temperature of that particular substance.0254

To find out how much energy is required to bring 25 g of ice to 0°C, here is what we do.0262

Our Q is going to equal the mass which is 25 g of ice and the heat capacity of the ice which is 2.09 J/ g/°C.0272

°C and °K, I do not give it to you in J/ GK.0287

The difference of degree for °C and °K is the same so it does not matter whether I use °C or °K.0294

I'm going from -7 to 0°C.0302

It is going to be δ T of 7°.0305

I’m just going to write it as 7°C.0310

When I go ahead and multiply and cancel units, I'm going to end up with 365.75 J.0313

This is how much energy is required to bring the 7 g of ice from -7° all the way up to 0°C.0323

Energy has to come from somewhere.0331

The energy is going to come from the water actually cooling down.0333

Let us see what this affect has on the liquid water because that is where the energy is being pulled from.0338

This is going to be Q = MC δ T but this time we are going to be looking for our δ T for water.0349

Our Q is going to be 365.75 J and I’m going to skip the units, I hope you do not mind.0357

We have 40 g of water and a specific heat of water is 4.18 J/ g/°C.0364

And we want to know what δ T is.0374

When we only do this math, nice and simple arithmetic, we get 2.19°C.0375

That means the water cools from 25°C it drops down by 2.19°C.0382

That amount of energy that is lost by the water in cooling, that energy goes towards bringing the ice from -7 to 0.0390

That is what is happening here, that is what I have done.0398

This is an exchange of energy.0401

Let us see where it is that we are.0404

The water is now at 22.81°C.0417

It started at 25, it dropped by 2.19, now it is at 22.81°C.0426

We have 40 g of liquid water at 22.81°C and we still have a 25 g of ice.0433

Except now, the ice is at 0°C.0455

We bought the solid ice from -7 to 0 but we have melted the ice yet.0459

This is very important.0464

Just because something comes to 0°C does not mean that it starts to melt or the other way around.0467

It does not mean that the liquid water starts to freeze.0472

That is the δ H of fusion, there is a certain amount of energy that is required to make the conversion from one phase to the other.0475

We do not have 25 g of ice, we do not have 25 g of water to 0, we have 25 g of ice at 0°C.0483

Let us see what we can do about the water at 22.81°C.0499

I have got to write it a little bit better.0503

Water at 22.81°C, in order to take that to water at 0°C, we need a certain amount of energy.0516

Let us calculate that energy.0534

Once again, Q = MC δ T.0536

Q = I have 40 g of water, 4.18 is its heat capacity J/ g/°C.0539

I want my difference in temperature to be 22.81.0553

I want to take it from 22.81 down to 0°C.0556

When I do this calculation, it tells me that I require 3113.8 J.0561

Water is going to lose 3813.8 J in going from 22.1° down to 0, that heat is going to go into the ice at 0°C 0572

and convert it from solid ice at 0 to liquid water at 0.0586

It is going to cause the ice to melt, this is where we use the δ H of fusion.0591

Let us see how much ice actually melts.0596

Once again, let me go blue because I like the blue.0600

We are going to take this 3813.8 J and we are going to multiply it by the reciprocal of the δ H of fusion.0614

1 g and 334 J, 334 J/ g is the δ H of fusion, that is how much energy it takes to either melt ice at 0°C and convert it to liquid water at 0,0634

Or it is the amount of energy that is going to be lost by liquid water at 0 and then convert that to ice.0650

When I do this, I end up with 11.42 g.0661

What is this 11.42 g?0674

11.42 g is the amount of ice at 0°C converted to liquid water.0677

Let us recap what is that we have done in terms of a nice picture.0702

We had liquid water at 25°C, I will just go ahead and put a 0 mark right there.0705

This is 0°C and then we have solid water at -7°C.0713

This H₂O liquid and this is H₂O solid.0720

A certain amount of heat was lost by the water in cooling and drop it down to 22.81°.0730

This amount of heat was enough to bring the ice of 20°C but still ice.0738

Now from the 22.81 there is a certain amount of heat that is going to be lost upon the water actually cooling down to 0°C.0746

Water at 0°C.0755

This amount of heat that is lost by the water in cooling, that amount of heat, 0758

since the ice is now already at 0°C it is going to convert a certain amount of ice into liquid water.0762

That is the 11.42 g, that is our final state.0771

Here is where we are now.0775

Let us see, I have 40 g of H₂O liquid at 0°C, that is the initial 40 that drop down to 0.0777

I have 11.42 g of H₂O liquid at 0°C is the water that came from the ice that melted.0796

That gives me a total of 51.42 g of liquid water0811

I'm left with 25 g of ice initially -11.42 that had melted, that leaves me with 13.58 g of ice or solid water at 0°C.0822

This is our final state.0844

Our final state, once everything comes to equilibrium I have 51.42 g of liquid water at 0°C and I have 13.58 g of ice at 0°C.0846

That is what is going on, that is our final state.0861

Let us go ahead and do the δ H and δ S.0865

Any time you see the word insulated that means no heat is allowed to come in or go out.0870

It means that Q = 0.0879

Since, the pressure is constant, since P is constant we know that under conditions of constant pressure the δ H = Q so δ H= 0.0888

There is no δ H, there is no enthalpy transaction in this particular process.0908

Let us go ahead and talk about the δ S.0915

The δ S for this process is going to be composed of three parts.0919

It is going to be the δ S total.0923

It is going to be equal to the change in entropy of the first phase, 0926

that means 40 g of H₂O liquid at 25°C taken to 40 g of H₂O liquid at 0°C.0931

The first part of the total entropy is going to be the entropy change in taking the 40 g of liquid water at 25 0950

and converting it to 40 g of liquid water at 0.0956

To that we are going to add δ S2, to that is going to be the change in entropy associated with taking 25 g of H₂O solid at -7°C and converting it to 25 g of H₂O solid at 0°C.0961

We have to account for each and every single transformation, that is what we are doing.0986

To that we are going to add the δ S3, 0991

We have 11.42 g of H₂O solid at 0°C converted to 11.42 g of H₂O liquid at 0°C.1006

There are three things that happen here.1034

Liquid water went from 25 to 0, there is a certain entropy change associated with that, that is δ S1.1036

25 g of ice went from -7° to 0 that is δ S2.1043

Out of that 25 g, 11.42 g of that was converted to liquid water and there is a change in entropy associate with that.1056

The total change in entropy, I just have to add them all up.1067

Let us see what we can do.1070

Let us do δ S1, let us always go back to our basic equations, always start with your basic fundamental equation.1074

From there, derive what you need.1088

DS = for conditions of constant temperature and pressure CP/ T DT – V Α DP and 1092

that is our basic equation for calculating changes in entropy when the temperature of the pressure are constant.1107

In this particular case, this is great.1113

They tell us that the pressure is constant and that mean DP = 0.1116

This is equal to 0 because the pressure is constant.1121

This is not hard, most of these problems are actually very simple as long as you to start with the basic equation.1126

You want to be consistent, always start at the same place and then take it from there.1131

You might go this way or this way, but always start at the same place.1136

That means, once I integrate this I have δ S1 = the integral from T1 to T2 of CP/ T DT.1141

The CP is constant and this is liquid water that we are talking about so it was going to be 4.18 × the integral of T1 to T2 of DT/ T = 4.18 × the log of T2/ T1.1156

δ S1 = 4.18 × the nat log of the final temperature was 0°C, 273.1177

The initial temperature was 25°C, 298.1186

We have to work in °K and when we do that I get -0.366 J/ g/°K.1190

This is 4.18 J/ g/°K.1208

It is J/ g, I have 40 g and I still have to multiply this by the 40.1211

This is the part that you have to be very careful of.1217

It is really important to keep track of your units.1220

We have this many J/ g/°K.1224

We have 40 g so let me write that down.1228

We have 40 g of water so 40 g × -0.3,1233

There are numbers everywhere, 366 J/ g/°K or J/ g/°C.1245

And we end up with the δ S1 =- 14.65 J/°K.1253

I'm hoping that you confirm my arithmetic.1262

Our δ S1 = -1465 J/°K.1265

Let us do δ S2, it is going to be the same thing.1271

I mean, except now the heat capacity is going to be 2.09 instead of 4.18 because now we are talking about the ice going from -7 up to 0.1275

We have 2.09 × the nat log, final temperature is 273, initial temperature was 266.1287

We end up with 0.054 J/ g/ °K.1299

We have 25 g of ice so we have 25 g × 0.054 J/ g °K.1307

We are left with δ S2 of +1.36 J/ °K, that is δ S2.1318

Let me go ahead and do the next one on the next page.1331

We have δ S3, this one had to do with the melting of the ice, the converting of the 11.42 g of ice to 11.42 g of liquid water.1334

We are going to use the δ H of fusion.1351

The δ S3 remember that is the δ H of fusion divided by the temperature, that was the entropy change when there is a phase change.1355

During the phase change, the temperature is constant.1366

Because the temperature is constant we can go ahead and use the δ H.1368

This is how to calculate the entropy for phase changes.1373

The δ H of fusion is 334 J/ g and it is happening at 0℃ which is 273 °K.1378

We end up with 1.22 J/ g °K.1381

We have 11.42 g of ice as being converted 11.42 × 1.22 J/ g °K.1398

We end up with 13.97 J/ °K, that is δ S3.1409

We are going from ice to liquid water.1419

Entropy is rising, it is positive.1422

We just put them all together.1426

Our δ S total for this particular process is - 14.65 + 1.36 + 13.97 and we get a δ S total equal to 0.68 J/ °K.1430

A very practical problem because you are putting something at a certain temperature, something else at another temperature, 1457

and there is going to be some final state that is going to be involved here.1463

Let us see what we have got now.1472

Let us go to example 2.1476

This is going to be the same sort of thing except that now we would be dealing with steam and liquid water.1481

It is going to be handled precisely the same way.1486

We are going to combine our intuition based on heat capacities which is going to absorb more heat.1489

Without change in temperature, it is going to be liquid water.1496

You are going to see how much you have of each and decide where you want to start the problem.1498

25 g of steam at 100°C and 310 g of liquid water at 25°C are combined in a fully insulated flask.1503

We are fully insulated so again Q = 0 so δ H is going to equal 0.1514

Given the following data, pressure is constant, we have the heat capacity for liquid water.1521

We have the heat capacity for the gas.1527

Notice 1.86 is not very high, it is a bit surprising given that for water.1529

The δ H of vaporization is 2257 J / g.1535

The δ H of vaporization, this is the amount of heat necessary to take 1 g of liquid water from the liquid phase to the gaseous phase, still at 100°C.1540

Or it is the amount of heat that is lost by 1 g of steam in transforming from 1 g of steam to 1 g of liquid water at 100°C.1553

They want us to know, what is the final temperature of the system and what phase or phases are present?1566

Same thing as before, what is the final state?1570

It is just a different way of asking it.1572

They want you to calculate δ S for this transformation.1573

We already know δ H is going to be 0 because we are at constant pressure and this is a fully insulated flask.1576

Let us see what we can do.1583

Let us deal with the 25 g of steam first.1587

I have got Q = MC δ T.1591

I got 25 g of steam at 1.86 J/ g/°C × 30°C.1595

What I'm calculating here is the amount of energy that is going to be required 1606

in order to take the steam at 130°C and bring it down to steam at 100°C.1609

This is still steam, it has not converted yet.1617

This multiplication, I end up with 1395 J.1620

This is the total amount of energy has lost by steam in going from H₂O gas at 130°C to H₂O gas at 100°C.1626

Let us see what this heat lost, this 1395 J.1647

If that heat is lost, that heat is going to end up going into the water so the water temperature is going to rise.1652

Let us see what is the rise in water temperature is.1659

1395 J = we have 310 g of water, its heat capacity is 4.18 J/ g/°C.1664

And we are going to multiply that by δ T.1676

When I calculate δ T, I get δ T = 1.08°C.1680

The water rises by only 1.08°C.1686

We have 310 g of H₂O liquid at 25°C and we just took it to 310 g H₂O liquid at 26.08°C.1693

Now the water is at 26.08°C.1714

Let us see what is next, let us find out how much energy is lost now that the steam is at 0°C.1719

We have 25 g of steam at 100°C and we have water at 26.08°C.1732

They still need to find some in between temperature that is going to be the final temperature, the equilibrium state.1742

The steam has to cool a little bit further but now it is steam at 100°C.1749

It cannot drop any temperature now it is going to undergo a phase change once it hits 100.1756

Now we are going to have to use the δ H of vaporization.1761

We have 25 g of steam and the δ H of vaporization is 2257 J/ g.1767

We have 56,425 J.1777

This 25 g of steam at 100°C, when it converts to liquid water at 100°C it gives up 56,425 J of energy.1783

It is going to give that energy to the liquid water, the 310 g.1796

This amount of energy is going to raise the temperature of water yet some more from the 26.08.1801

Let us find out what the new temperature is.1807

Let us just say what this is here, write it down.1813

For H₂O gas at 100°C converted to H₂O liquid at 100°C.1819

This is the amount of heat lost by the 25 g of steam.1835

Let us see what this does to the water.1840

Now, we have 56,425 J = 310 g 4.18 J/ g/°C × δ T.1842

Our δ T is going to be 43.54°C.1858

Let us do this on the next page.1870

We are at 26.08° and we just raised the temperature at another 43.54°C.1876

Now, my temperature is 69.62°C.1889

310 g of liquid water is at 69.62°C.1904

We also have 25 g of liquid water at 100°C.1918

They are both liquid water but at very different temperatures.1934

This 25 g came from the condensation of the steam.1936

This 310 g at 69.62 just came from the normal liquid water that is rising the temperature.1941

We have water at 100°C, we have water at 69.62°C.1948

This water is going to cool, this water is going to warm.1966

They are going to reach some final temperature.1969

We need to find what that is.1973

The heat lost by the water at 100°C as it cools, is the heat gained by the water at 69.62 as it warms up.1977

That is the whole idea, heat out heat in.1987

We have Q lost = - Q gained.1992

This negative sign is very important.1999

The heats are equal.2002

When I add them, they add to 0.2003

This mathematic this ends up being a negative sign, very important.2005

The heat that is lost by one substance, in this case, water at a certain temperature, is negative of the heat gain by the other.2009

One is gaining and one is losing.2016

Let us write it out.2020

The heat lost that is going to be the MC δ T of the water at 100 and the heat gain is going to be the MC δ T of the water at 69.62.2023

Let us write that out, MC δ T.2034

I’m going to write M1 C1 δ T = M2 C2 δ T to let you know we are talking about two different masses.2041

In this particular case, it is not going to be two different heat capacities because now they are both water but the masses ore different.2052

We have got 25 g of water, the heat capacity is 4.18 and the change in temperature is temperature final – initial.2059

This initial temperature is 100°C for the 25 g = negative.2075

I have 310 g of water, its heat capacity is 4.18 and now it is going to be temperature final.2083

Its initial temperature is 69.62, this is the equation.2091

I'm looking for a final temperature.2097

We have done this from general chemistry.2100

Let us see what the arithmetic looks like.2105

The 4.18 go away and I'm going to be left with 25 TF - 2500 = -310 TF + 21,396.2.2109

I should have 335 TF = 23,896.2.2132

I have a temperature final of 71.33°C, this is what I wanted, the final temperature.2144

We have 335 g of H₂O liquid, the 310 g + 25 g of liquid water at 71.33°C.2157

This is our final state, that is what happened here.2177

The steam cooled, the water rose in temperature.2181

The steam converted to liquid water at 100 and that conversion, the heat that was lost by that conversion, 2185

one of the 310 g of water lifting the temperature to the 69.62.2192

We have water at 100 and water at 69.62.2198

We are going to find some medium temperature.2201

I’m going to write it in between them.2203

Again, 25 g and 310 g there is a big difference.2204

There is only one from 69 to 71.2208

This one is from 100 down to 71, that is the whole idea.2211

Let us go ahead and calculate the δ S for this process.2218

δ S this time is going to be 4 different things that happened.2223

δ S total = we have δ S1.2227

This is the conversion of 25 g of gaseous water at 130° taken to 25 g of H₂O gas at 100 °.2232

To that we are going to add the δ S2, the second process we are going to take this 25 g of H₂O gas at 100°C.2253

And we are going to convert it to 25 g of H₂O liquid at 100°C.2266

And we have another process, δ S3 is going to be taking 25 g of H₂O liquid at 100°C and converting it to 25 g of H₂O liquid at 71.33°C.2277

Then the final entropy change is going to be 310 g of H₂O liquid at 25°C and 310 g of liquid water at 71.33°C.2305

25 g of steam at 130 went to 25 g of steam to 100, there is an entropy change associated with that.2337

25 g of steam at 100 went to 25 g of liquid water at 100, there is entropy change associated with that.2344

+ 25 g of H₂O liquid at 100 and now this 25 g of H₂O liquid at 71.33, there is an entropy change associated with the temperature change.2351

And this one we have 310 g of H₂O liquid at 25 going to 310 g of H₂O at 71.33.2360

There are 4 parts to this entropy.2368

Let us go ahead and calculate them.2370

Let us see here, what should I do.2374

The general equation once again, DS = CP/ T DT – V Α DP, this goes to 0 because pressure is constant.2377

Therefore, our general equation for change in entropy upon a temperature change 2390

is going to be the integral of CP/ T DT from temperature 1 to temperature 2.2394

And for a phase change, our entropy is going to be δ H of vaporization ÷ T.2405

For 3 of these, our temperature changes we are going to use this one.2412

One of these is going to be a phase change, we are going to use this one.2416

Let us do it.2419

The δ S1 = 1.86 we are taking it from 403 which is 130 to 373 which is 100.2424

Which is going to be DT/ T and we are going to get 1.86 × the nat log of 373/ 403.2438

We end up with -0.144 J/ g/ °K.2451

We have 25 g of steam so 25 × 0.144 J/ g/ °K gives us - 3.60 J/ °K this is δ S1.2459

δ S2 is going to equal the,2482

This was the phase change 1.2485

We have the δ H of vaporization/ T which id equal 2257 J/ g ÷ 373 °K because the transformation took place at 100°C.2487

We get 6.05 J/ g/ °K.2504

We are changing 25 g of it so g/ °K and we end up with -151.27 J/ °K.2511

It is negative, there is a negative sign here but it is negative because it is dropping.2532

I'm actually going from a gaseous phase to a solid phase, entropy decreases.2537

You can use your qualitative understanding.2544

As things go from solid, liquid, to gas, the entropy is going to be positive.2547

As things go from gas to liquid to solid, the entropy is going to be negative.2551

These are just the absolute values, these are the actual numbers.2555

The magnitudes of the energy changes of things like that.2559

But I put negative sign here, the entropy is negative because we are going from a gaseous phase to a liquid phase, in more ordered phase.2562

Our entropy change is negative.2569

By all means, use the qualitative.2571

If there is ever a place where you want a qualitative analysis, it is in thermodynamics.2574

δ S3, it is going to equal 4.18 × the integral 373 to 344.33 DT/ T = 4.18 × the nat log of the final temperature which is 344.33 °K / 373.2580

We end up with -0.334 J/ g °K.2610

Once again, we have 25 g of this -0.334 J/ g °K and you get -8.36 J/ °K, this is δ S3.2617

In this particular case, the negative that shows up automatically based on the mathematics because we are actually doing integration.2639

As integration ends up being a logarithm.2645

And when you have a logarithm in the quotient and the quotient is less than one, you will get a negative number.2649

This negative number is already taken care of.2655

We do not have to worry about it being qualitative.2657

We have got one more δ S here.2663

We have our δ S4 and that is going to equal the 4.18 because now are talking about water × the nat log .2665

And we went from 298 °, we went to the 344.33 °, and tat is going to equal 0.604.2678

This is positive because we are rising in temperature.2691

Temperature increases, entropy increases.2694

J/ g °K and this time we have 310 g of the water that we raised from 25 to 78 or whatever it was.2699

× 0.604 J/ g °K and end up with 187.3 J/ °K.2710

Our δ S total = -3.60 - 151.27 - 8.36 + 187.3.2726

Our δ S total for this particular transformation = 24.02 J/ °K, a net increase.2746

There you go, a long problem, a little bit tedious but nothing altogether difficult in terms of what is actually happening.2758

Just take it one piece at a time.2767

Any time you are dealing with any sort of a mixture of two things, one is at a certain temperature and one is at another temperature, 2771

the heat is lost by one thing is going to be the heat that is gained by the other thing.2782

Particularly, when we are talking about something that is going to be in a completely insulated flask which is often how we run these things.2785

We do not have to worry about any heat being transferred in or out of the combined system.2791

There you go.2797

Let us round this out with one final example and it does have to do with entropy.2801

Let us see what it says.2807

Given the following expression DS = CP/ T DT – V A DP.2812

We already know what this expression is, this is one of general expressions for finding the change in entropy.2817

This should be S.2822

Finding the entropy under conditions of constant temperature and constant pressure.2826

Given the following expression, calculate the decrease in temperature that occurs when 1 mol of water 2832

at 25°C and 1100 atm is brought adiabatically and reversibly to a pressure of 1 atm.2837

Assume that K, the coefficient of compressibility = 0.2847

That is extra information, do not worry about that.2851

You do not need to do anything with it.2852

You have the following data for water.2855

The molar volume of water is 18 cm³/ mol, constant pressure heat capacity this time is expressed in J/ mol °K,2858

75.3 J/ °K and Α the coefficient of thermal expansion is 2.07 × 10⁻⁴/ atm.2867

Let us see what they are asking us here.2880

I have 1 mol of water it is at 25°C but its pressure is 1100°,2882

Under conditions of adiabatically and reversibly, I drop the pressure to 1 atm.2892

Adiabatic means there is no change in the Q = 0.2899

In other words, there is no heat is allowed to flow.2904

I know that when I drop the pressure from 1100 to 1 atm, my intuition tells me that the temperature is going to drop.2907

The temperature is going to drop precisely because no heat is allowed to flow, this is the adiabatic part.2915

No heat is allowed to flow into the system, in order to keep the temperature up.2921

The temperature is definitely going to drop.2926

Remember that an adiabatic process is when you increase the pressure, it is the hugest, 2928

the largest drop in temperature happens in an adiabatic reversible process.2935

We want to know what that pressure drop is.2941

In other words, starting at 25°C if I do not allow any heat to flow, if I conduct this reversibly and very slowly, 2943

adiabatically what is my final temperature going to be when my pressure in the system is now 1 atm, that is what this is asking.2952

Let us do it, adiabatic and reversible, that means that DQ reversible = 0.2960

Adiabatic Q is 0, we now we just put a reversible because it is telling us that it is actually reversible.2985

DS by definition = DQ reversible/ T.2993

If DQ reversible is 0 it implies that DS = 0.2998

There is no entropy change for this process.3005

We have an equation 0 = CP/ T DT – V Α DP.3007

Move is over, I get CP/ T DT = V Α DP.3021

Integrate this expression, the heat capacity is going to be constant so I end up with the following.3031

CP × the integral from T1 to T2 of DT/ T = V Α × the integral from P1 to P2 of DP.3038

I’m just taking the expression and found out that DS = 0.3055

I set it equal to 0, I rearranged it and I integrate this expression.3058

My final equation is going to be.3062

I get CP × the nat log of the final temperature/ the initial temperature = V Α δ P.3070

I have everything that I need, now I need to put it in terms of units that match and everything should be fine.3084

Let us go ahead and deal with the molar volume first.3091

I have 18 cm³/ mol = 18 × 10⁻³ dm³/ mol.3094

A cubic decimeter is a liter.3108

If there is 1 unit that you should know, 1 liter is 1 cubic decimeter.3111

We get 18 × 10⁻³ L/ mol, that just means that 1 mol of liquid water occupies 18 × 10⁻³ L.3115

Volume is it L and pressure is at atmosphere.3131

We are going to end up with something in L atm and we need to convert it to J.3134

Let us go ahead and take care of the V DP first.3140

I have got V × δ P.3145

The volume is 18 × 10⁻³ L/ mol.3150

The δ P was final - initial so it is going to be 1 atm - 1100 atm.3167

V δ P when I actually multiply this out I get - 19.782 L atm/ mol.3177

In order to convert it, I multiply by 8.314 J ÷ 0.08206 L/ atm.3191

And I end up with - 2004 J/ mol.3203

V × DP = V × DP = -2004.3212

I just have to convert to J because I’m working J, because heat capacity is in Joules.3221

I need to make sure that my units match.3226

Let us go to the next page.3233

I have my equation CP LN TF/ TI = VΑ DP.3236

I have got 75.3 × the nat log of the final temperature.3252

That is what I want, I want the final temperature.3261

They wanted to know what the temperature drop was.3262

TF I started off at 298 = Α V DP.3265

Α is 2.07 × 10⁻⁴ and V DP was – 2004.3275

And so I get 75.3 × LN of TF/ 298 = - 0.415.3287

I go ahead and I will do the division and exponentiation.3301

Let me write that out LN of TF/ 298 = -0.00551.3306

When I exponentiate, e to this power, e to that power.3317

TF/ 298 = 0.9945 and I get the final temperature of 296.36.3326

And therefore, δ T = final temperature 298.3345

I think it is a little bit of an arithmetic problem here.3364

The final temperature, final – initial will give me something positive, it is going to be something negative.3366

I apologize if I calculated some of it a little bit of arithmetic issue here.3372

I’m going to go ahead and leave this.3376

The process is correct but I seem to have done the multiplication wrong.3378

It is from what I see on the table.3382

This is correct and now if you just go ahead and multiply this × that.3384

I'm sorry, I apologize.3394

I wrote 273 on the paper, it should be 298.3395

This was correct.3400

We have 296.36 and our δ T = 296.36 – 298.3402

I'm sorry I thought it was 0 so I had 273 on my paper.3416

Our δ T -1.64°C or °K, I will just go ahead and write °C and there we go, this is our temperature drop.3420

Just use what is given to you in the problem.3437

You have a general equation because it is adiabatic and reversible, there is no change in entropy.3439

DS= 0, rearrange the equation and just put the numbers in.3444

The only thing you have to watch out for in this problem is the units.3449

Cm³/ mol you have to change that to L because the pressures and atmospheres.3455

That is just my particular preference.3462

You can work in cm³ and work in Pa, it does not really matter.3464

Just what you are comfortable with.3468

I’m comfortable with L atm, convert it to J.3470

Once I’m in Joules, I’m going to go ahead and run in the problem.3473

There you have it, a change in temperature is -1.64°C .3477

Thank you so much for joining us here at www.educator.com.3482

We will see you next time, bye.3485