For more information, please see full course syllabus of Physical Chemistry
For more information, please see full course syllabus of Physical Chemistry
Discussion
Download Lecture Slides
Table of Contents
Transcription
Related Books
Entropy Changes for an Ideal Gas
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
- Entropy Changes for an Ideal Gas
- General Equation
- The Fundamental Theorem of Thermodynamics
- Recall the Basic Total Differential Expression for S = S (T,V)
- For a Finite Change in State
- If Cv is Constant Over the Particular Temperature Range
- Change in Entropy of an Ideal Gas as a Function of Temperature & Pressure
- Change in Entropy of an Ideal Gas as a Function of Temperature & Pressure
- Recall the Basic Total Differential expression for S = S (T, P)
- For a Finite Change
- Example 1: Calculate the ∆S of Transformation
- Intro 0:00
- Entropy Changes for an Ideal Gas 1:10
- General Equation
- The Fundamental Theorem of Thermodynamics
- Recall the Basic Total Differential Expression for S = S (T,V)
- For a Finite Change in State
- If Cv is Constant Over the Particular Temperature Range
- Change in Entropy of an Ideal Gas as a Function of Temperature & Pressure 11:35
- Change in Entropy of an Ideal Gas as a Function of Temperature & Pressure
- Recall the Basic Total Differential expression for S = S (T, P)
- For a Finite Change
- Example 1: Calculate the ∆S of Transformation 22:02
Physical Chemistry Online Course
Transcription: Entropy Changes for an Ideal Gas
Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.0000
Today, we are going to discuss entropy changes for an ideal gas.0005
The equations that we developed so far, as we said earlier they apply to any every situation solid, liquid, gas, real gas and ideal gas.0010
For all practical purposes we do not really need to take those equations and somehow fiddle around with them0019
in order to apply them to the case of an ideal gas because they apply as is.0026
However, it is nice to do this because most of the time when we are dealing with gases we are doing really specialized work.0031
We always going to treat gas as an ideal gas so it is nice to have some equations specifically for ideal gases0037
that you can just turn to when you know you are dealing with an ideal gas.0044
You do not have to if you do not want to.0049
Again, all the other equations are absolutely valid.0050
If you want to do the derivations for yourself, it is always nice mathematically.0053
Personally, it gives you a nice sense of accomplishment.0058
I think it would be nice to go ahead and actually discuss it specifically for an ideal gas because it does tend to be the system that we talk about the most.0061
Let us jump right on in.0069
What is nice about the equation for ideal gas is that they tend to be really simple.0076
Let us go ahead and start off first of all we will do a little bit of a mathematical derivation, very quick one though.0081
In general, you remember that we had for energy with respect to temperature and volume,0086
the differential change in energy was equal to CV=CV DT + Du DV under constant T DV.0096
But because we are dealing with an ideal gas, what was nice about it, you also remember hopefully.0111
For an ideal gas, this term the DU DV T that is equal to 0, that was Joule’s law.0117
For an ideal gas, this change in energy with respect to a change in volume under constant temperature actually equal 0.0124
This was Joule’s law.0133
When you change the volume of an ideal gas, you do not change the energy of the system.0138
Basically, what you have when this term goes to 0, what you are left with is just that.0144
We have DU = CV DT.0148
I will go ahead and put that aside for a second.0152
Let us deal with fundamental theorem of thermodynamics.0155
The fundamental theorem of thermodynamics, it says the following.0158
It said that DS = 1/ T DU + P/ T DV.0168
We had DU under conditions of an ideal gas so we will just go ahead and put this into here.0177
What we end up with is DS = 1/ T CV DT + P / T DV.0186
I will go ahead and write this as DS = CV/ T DT + P / T DV.0199
This particular differential, it looks familiar.0210
It does not really look altogether that different than what we just did with the other questions, it is the same equation.0212
Notice, this is a function of temperature and volume but notice we have pressure here.0219
The ideal gas law, again, we are dealing with an ideal gas so it make things simpler.0225
We have an ideal gas we have a relationship between pressure, volume, and temperature.0229
Because this is temperature and volume every single variable in here of this side has to be expressed in terms of temperature and volume.0235
As pressure, we have to express it in terms of temperature and volume.0243
We can do that it is right here P = nRT / V.0246
I take this and I put in there.0253
Here is what I get so DS = CV/ T + nRT / V / T DV.0256
The T cancels the T and I'm left with the following equation DS = CV/ T + nR/ V DV.0278
When we consider entropy as a function of,0293
I’m sorry I forgot my differential here DT.0297
When we consider entropy as a function of temperature and volume for an ideal gas this is what0299
the general equation takes the form as it is CD / T DT that part is the same + nR/ V DV.0306
In other words the number of moles × the gas constant divided by the volume of the system.0314
This is the differential coefficient for the volume component of the entropy.0318
Let us go ahead and call this equation 1 and for an ideal gas this is the equation that you want to memorize if you want to.0326
It is pretty easily derived so it is not a problem but there it is.0333
We recall the basic total differential expression for S = the function of T and V.0339
What you have is DS = DS DT constant V DT + DS DV under constant T DV.0361
I will call this one equation 2.0380
When we compare equations 1 and 2, the DT DV DT DV that means this and this and this and this are identified with each other.0381
We already know the DS DT under constant V is just a constant volume capacity divided by temperature.0391
Nothing is new there.0397
However, the change in entropy with respect to volume under conditions of constant temperature for ideal gas is equal to nR/ V.0399
That is really nice.0408
What we have is DV T =nR / V.0409
Notice, this is always positive.0423
Therefore, because it is always positive that means an increase in the volume implies an increase in the entropy of the system.0432
In other words, if I have a smaller volume, if I have a particular ideal gas and if I make the volume bigger now the gas has more room to bounce around in.0446
It becomes more chaotic, it becomes more disorder.0455
It is entropy rises.0459
T is positive S has to be positive.0461
This is positive because this is positive.0465
An increase in the volume implies an increase in entropy, we knew that already but it is always nice to have this corroboration,0470
that is what is nice about this.0476
For a finite change in state, we just integrate the differential expression.0479
Integration of equation 1 which I have to go ahead and write again, it is not a problem.0496
DS = CV/ T DT + nR/ V DV.0506
When I integrate that, I integrate this, I integrate this, I integrate this, and I get the following.0514
I get δ S = the integral from temperature 1 to temperature 2 of CV/ T DT + the integral from volume 1 to volume 2 of nR/ V DV.0523
If the constant volume heat capacity happens to be constant over the temperature range of T1 and T2.0548
In general, we tend to use constant values for heat capacities.0556
Heat capacity is actually a function of temperature so as the temperature of the system rises, if heat capacity changes0562
but in general over a reasonably good temperature range, it tends to be pretty constant.0567
If CV is constant which in most cases it will be, constant over the particular temperature range, we can pull this out of the integral.0574
nR our constant so we could pull those out of the integral then we get, when we pull these out of the integral.0598
I will go ahead and write that down, it is not a problem.0616
We get DS = CV/ the integral from T1 to T2 of DT/ T + nR × the integral from V1 to V2 of DV/ V.0619
These are very simple integrations that we can do and we end up with the following equations.0634
CV × log of T2/ T1 + nR × log of V2/ V1.0639
This is the equation that we want, the final version for a finite change from state 1 to state 2.0651
The change in entropy of the system, if I change its temperature and I change its volume or either one, it is a whole lot of constant,0659
I will just knockout, I do not need it.0666
It is going to be the constant volume heat capacity × the log of T2/ T1 + n × R × nat log of T2/ T1.0668
For an ideal gas this will give me the entropy change, nice and simple.0676
For a temperature change and volume change or just a temperature change or just the volume change, that is all, very straightforward.0682
This was temperature and volume, let us deal with temperature and pressure like we always do.0691
For the δ S of an ideal gas, as a function of temperature and pressure here is what we have got.0700
In general, we will go ahead and do the derivation.0716
We have DH= CP DT + DH DP under constant T DP this was the general expression for the change in enthalpy of the system.0722
The enthalpy is just a different way of looking at energy that includes the pressure, volume, work, that was the definition of enthalpy.0737
For ideal gas though, this term is 0.0747
I would actually write it down.0755
For an ideal gas, this term DH DP T = 0.0760
Therefore, we just have DH = CP DT.0771
Another form of the fundamental thermodynamics that deals with enthalpy instead of just energy is the following DS = 1/ T DH - D/ T DP.0780
We have seen this before, this is just the fundamental equation of thermodynamics except now in terms of energy DU0801
we have used enthalpy because under conditions of constant pressure.0806
That is all that is going on here.0810
We have DH = this, so we go ahead and put this there and we end up with DS = CP/ T DT - V/ T DP.0812
Notice, this differential expression is in terms of temperature and pressure which means that every single variable0832
on the right has to be expressed in terms of temperature and pressure.0839
This is volume not a problem, we are dealing with an ideal gas PV= nRT.0842
I will just go and ahead express volume in terms of temperature and pressure nRT / P.0847
I take this and I put in there I get the following.0853
I get the DS = CP/ T DT - nRT/ P / T DP.0859
The T cancel and I'm left with DS = CP / T DT –nR / P.0879
Let us correct here R / P DP this is the other equation for the differential change in entropy with respect to temperature and pressure.0891
I will go ahead and call this equation 3.0906
Recall the basic total differential expression for S as a function of T and P it is DS = DS DT P DT + DS DP T DP.0911
T and P these 2 I call this equation 4.0955
If I compare these 2 that means I have identified this with this which we know already.0964
In this case, I’m identifying this for the case of an ideal gas with this right here.0971
What I have got is a change in entropy with respect to a change in pressure under conditions of constant temperature for an ideal gas = -nR/ P.0976
In other words, if I have an ideal gas and if I hold a temperature constant and I change the pressure, the entropy change is this right here.0994
Notice that it is negative.1004
What this says that if I increase the pressure of the system and keep the temperature constant, the entropy has to go down,1005
that is what is interesting here which says under isothermal conditions, you know what isothermal means, it means T constant.1013
If I increase the pressure of the system that implies that the entropy of the system has to go down.1046
That is what this says.1056
This change is negative, this is the entropy change per unit change in pressure.1059
If I change the pressure that entropy change is negative which means it goes down.1066
Rising pressure of the system means a decrease in entropy of the system.1072
Notice this is isothermal that is what makes this important.1077
I will get to that in just a minute, I will say actually a couple of things.1083
Let us first talk about a finite change.1087
For a finite change, we just integrate the DS expression.1091
We have the integral of DS = the integral of CP/ TD T from T1 to T2 - the integral from P1 to P2 and nR/ P DP.1100
Again, if the constant pressure heat capacity is constant / the particular temperature range,1125
this of course is constant we can pull those out of the integral.1131
When we pull these out from under the integral sign we end up with the following.1135
Let me actually right them out.1139
I get CP × the integral from T1 to T2 of DT/ T - nR × the integral from P1 to P2 of DP/ P.1141
I get the change in entropy of the system = CP × LN of T2 / T1 - nR × LN of P2 / P1 this is the entropy change for an ideal gas1155
whenever I change the temperature and the pressure or either one.1181
Let us go back and talk about what I was going to talk about up here regarding the negative.1191
Notice that the change in entropy with a change in pressure under constant temperature is actually negative.1197
It is -nR/ P so once again an increase in the pressure of the system implies a decrease in the entropy of the system.1211
Here is why it makes sense, you probably think yourself and say if I increase the pressure of the system1227
does it mean that the volume is actually going to get bigger?1232
If the volume gets bigger does not the entropy increase?1237
Yes, but we are holding the temperature constant, an ideal gas PV = nRT.1239
If temperature is constant that means this n, this R, this T, they are all constants.1248
That means that the right side of the equation is constant.1252
If I increase the pressure of the system the volume of the system have to go down in order to retain the fundamental equality.1255
When a volume of the system goes down, we already know that when volume goes up entropy goes up1262
which means that when volume goes down entropy goes down which makes sense physically.1267
If the volume increases now there is less room for the molecules to bounce around, there is less disorder in the system, the entropy goes down.1271
Pressure actually alternately comes down to a volume.1281
Let us go ahead and finish this off with a little bit of example.1291
Again, let us finish with a theoretical discussion, we would be doing a lot of example problem.1295
Example problems are going to come in a couple of lessons and there are going to be several of them.1303
We are going to do in bulk, all at once.1308
We have a chance to reuse these equations over and over again in the context of problems.1310
Do not worry there are plenty of examples to come, just like there were for energy.1315
2 mol of oxygen gas we will go ahead and treat it as an ideal gas is taken from 25°C.1326
I’m going to do this one in red, I think.1332
It is taken from 25°C and 1 atm pressure to 100°C and 5 atm pressure.1334
Here you are changing the temperature and we are changing the pressure.1341
Calculate δ S for this transformation, the molar constant pressure heat capacity = 7R / 2.1344
This is nice and simple.1353
Since we are dealing with constant pressure and it is an ideal gas, I can go ahead and just fall back on the equation.1354
If I do not know it, I can derive it, it is not a problem.1363
I have CP × LN of T2 / T1 -n × R × LN of P2 / P1.1366
Just plug my values in δ S.1380
The biggest problem you run into is arithmetic.1383
We have 7/2 of 2 mol, the molar CP is 7 R/ 2, molar CP that is CP/n = 7/2 R.1387
The CP = 7/2 Rn that is what it means, molar means divided by the number of moles.1406
For actual heat capacity you have just move the n over so you have 7/2 Rn.1413
We have 7/2 × 8.314 J/ mol-°K × n which is we have 2 mol of the oxygen gas.1418
mol and mol cancel × the log 25° to 100° so 100°C is 373°K, we have to use Kelvin temperature divided by 25 which is 298 - the number of moles which is 2 × R which is 8.314.1434
I’m going to keep writing the units, it is not a problem 0.314 × log of 5 atm/ 1 atm.1459
Hopefully, I have done the arithmetic correctly and I hope you will go and check this out.1472
I know I tend to make arithmetic mistakes but this is what is important, this right here, not the actual arithmetic.1476
You end up with - 13.7 J/-°K.1482
In this particular case, notice the change in entropy is negative.1492
The affect of temperature on the entropy is going to be positive.1497
The effect of pressure on the entropy is going to be negative.1501
In this particular case, the temperature from 25 to 100°C is overcome by the increasing pressure from 1 atm to 5 atm.1503
In this particular case, the effect of pressure, the effect of P exceeds the effect of T which is why it ended up with a negative entropy change.1516
There you go.1537
Thank you so much for joining us here at www.educator.com.1539
We will see you next time, bye.1541
Start Learning Now
Our free lessons will get you started (Adobe Flash® required).
Sign up for Educator.comGet immediate access to our entire library.
Membership Overview