WEBVTT chemistry/physical-chemistry/hovasapian
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Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.
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Today, we are going to continue our entropy example problems.
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Let us jump right on in.
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For the first problem we are going to look at here is the following.
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We have 1 mol of an ideal gas initially at 25°C and 1 atm pressure and it is taken to 50°C and 0.55 atm,
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during this transformation 260 J of work are done on the surroundings.
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They give us the work that is done during this particular expansion.
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They want us to calculate the heat δ U, δ H, and δ S for this particular process.
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Let us see what we can do.
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We are dealing with an ideal gas so it is going to change a couple of things.
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We have 25°C to 50°C.
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Let us go ahead and start with, in this particular problem when you are finding the Q, in this case they gave you W,
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but in subsequent problems we have to find the work, the change in energy, the change in enthalpy and the change in entropy.
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There is no systematic way as far as choosing one before the other.
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Sometimes it is easier to find the energy and then the heat and work.
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Sometimes it is better to do the work than heat so do not think that you have to go down the line in order, it is not like that.
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It is what you have at your disposal and whatever strikes your fancy.
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Let us start off with this one.
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Let me go ahead and start with energy.
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When we do that let me stick with black.
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Let us do DU = CV DT again, it is always nice to start with the equations that you know to write them down and take it from there.
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It is a great review to constantly write down the equations that you need.
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DDT + DU DV T DV is basic equation as far as the energy of a system in terms of temperature and volume.
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Now this is an ideal gas so this is automatically goes to 0.
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All we are dealing with is DU=CV DT and when we integrate this we get δ U =CV × δ T.
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This is the equation that we are going to use.
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It looks like we have everything.
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δ U = CV δ T, CV is 3/2 RN δ T.
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We can go ahead and put some numbers, we have 3/2, R is 8.314 J/mol-°K and we have 1 mol and we have δ T.
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δ T is going to be 50 -25 and again Kelvin and Celsius, the δ is the same because the increment is the same so we have just 25°K.
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°K vs °K and mol vs. mol, energy is going to be in Joules.
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δ U for this process is 312 J and again I'm hoping that you will actually confirm my arithmetic, I’m notorious for arithmetic mistakes.
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The energy is taken care of we have 312 J.
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We know that δ U, let me go ahead and use the differential expression because it is probably the best to begin with.
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DU = DQ - DW that is the fundamental relationship, that is the first law of thermodynamics.
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This implies in for a finite change that δ U =Q – W.
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Again, there is no δ Q δ W, these are Q and W are path functions, this is a state function which is why we have a δ here.
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When you rearrange this, I get Q = δ U + W.
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Therefore, Q I have δ U, I have 312 J, I’m going to go ahead and skip the unit.
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I hope you do not mind just work with numbers.
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They tell me that 250 J of work are done on the surroundings.
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Work is set positive if it is being done on the surroundings.
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Therefore, we have the 260 J.
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For Q, we end up with 572 J.
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For this particular expansion, 572 J of heat transpire.
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Let us go ahead and take a look at δ H.
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I’m going to go ahead to do this on the next page, it is not a problem I have few pages here.
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Let us go ahead and write our basic relationship that DH = CP DT + DH DP DP.
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Again because this is an ideal gas this goes to 0.
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Basically, all we have is DH =CP DT, nice and straightforward.
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When we integrate this function, I will integrate this differential equation we end up with CP δ T.
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This is really nice and straightforward.
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The only thing we need is they gave us CV, we have to find relationship, we have to find CP.
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We know the relationship, we know that for an ideal gas CP - CV = Rn.
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Therefore, CP = Rn + CV.
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Therefore, CP =Rn + 3/2 Rn CP = Rn × 1 + 3/2.
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Therefore, our constant pressure heat capacity = 5/2 Rn.
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Working with an ideal gas is really easy.
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We have our constant pressure heat capacity now we can just go ahead and put in here.
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Let us go ahead and write δ H =CP δ T CP is 5/2 Rn and we have δ T.
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Let us put in some numbers δ H =5/2 × 8.314.
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Again, I hope you forgive me, I’m going to skip the units.
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1 mol and δ T is 25°K.
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When we work out this arithmetic, we end up with 570.
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No, I’m sorry that was going to be our δ H is going to be 520 J.
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Let us see what we can do here.
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Let us go ahead and go to example 2.
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Example 2, says 1 mol of an ideal gas with the CV = 3 Rn/ 2 was initially in the following state 350°C and 1.5 atm.
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At constant volume the gas is heated to 450°K, calculate Q, W, δ U, δ H and δ S.
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They also want us to compare δ S with the value of Q/T.
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We have 1 mol of an ideal gas, they gave us the constant volume heat capacity 350°K , so we are doing this process at constant volume.
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These problems are really great because they allow us to actually review our first law issues, the heat, work, and energy, enthalpy, things like that.
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We just add that final state property which is entropy.
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Let us see what we have, constant volume is going to be important.
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These constraints are what changes the equations that we deal with.
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Constant volume that implies the DV = 0.
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DW =the external pressure × DV.
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If DV is 0 that means DW = 0 that means that the work = 0.
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Constant volume nothing is changing so no work is being done so work is 0.
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Let us go ahead and see what we can do about DU.
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Again DU = CV DT + DU DV at constant temperature DV and again because we are dealing with an ideal gas this term goes to 0.
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It is constant volume so DV goes to 0.
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Even if this were an ideal gas, in this particular case this would go to 0.
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In either case, this term goes to 0.
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What we are left with is DU = CV DT and for a finite difference when we integrate this differential equation we get CV δ T.
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Let us put in some numbers, we have 3/2, let us go ahead and write it all in, 3/2 Rn δ T and it is going to be 3/2 × 8.314 × 1 mol and
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the change in temperature 350°K to 450°K, we have 100°K temperature change.
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Our energy for this process is going to be 1247 J.
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That is nice.
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We have W and we have δ U so let us go ahead and find Q.
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δ U= Q - W and let me just go ahead and just do it this way Q =δ U+ W =δ U which is 1247 + W which is 0.
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Our Q for this process is also 1247 J.
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That takes care of the heat, now let us go ahead and take care of the enthalpy.
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DH =the constant pressure heat capacity × the differential change in temperature or constant pressure heat capacity × temperature +
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this DH DP the change in enthalpy per unit change in pressure at constant temperature × the change in pressure.
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But again we are dealing with an ideal gas so this goes to 0.
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We have DH =CP DT and for integration we have DH = CP δ T.
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Nice and straightforward, nothing strange is happening here.
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Again, since we have the 3/2 Rn the constant volume heat capacity was 3/2 Rn because this is an ideal gas our constant pressure heat capacity is going to be 5/2 Rn.
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It is going to be like that.
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DH = 5/2 Rn δ T = 5/2 × 8.314 × 1 × 100 we get a δ H for this of 2079 J.
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That takes care of our δ H.
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Let us go ahead and see what we can do about δ S.
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This is what we have been discussing, this particular unit is entropy.
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We have DS =CV / T DT + nR/ V DV for an ideal gas this is the expression for the entropy or the differential change in entropy of the system.
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This is happening at constant volume, therefore, this term goes to 0.
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What we have is just DS =CV/ T DT.
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We are going to integrate this and what we are going to get is δ S =the integral from temperature 1
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to temperature 2 of CV/ T DT which = CV × the log of T2/T1.
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CV is a constant so it comes out of the integral which we are left with is the integral of DT/ T, the integral of DT/ T is LN T2/T1.
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Now I will just put the numbers in.
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In this particular case, it is going to be 3/2 Rn × the log of temperature 2/ temperature 1.
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Let us go ahead and DS = 3/2 R is 8.314, the number of moles is 1 log, and we have 450°K/ 350°K.
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Real quickly here, when you are dealing with temperatures involving the temperature on top of another temperature, you have to work in °K.
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I know that in previous problems, in this particular problem or either ones where we did Celsius temperature, when we did δ T we just use the 50°C - 25°C.
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The difference in the δ T is 25°C or 25°K.
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There we do not have to convert to °K because the increment of Celsius and Kelvin is actually the same.
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However, for here we cannot go 25°C/ 50°C.
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The °C do not cancel and the reason is because in the Kelvin temperature is going to be 25 + 273/ 50, + 273 the 273 do not cancel.
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When you do this, the temperature has to be in °K.
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In general, we are just avoiding problem, just do everything in °K.
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In that way, you will never go wrong.
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I just want to let you know that if the temperature in this particular problem we are given in Celsius,
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you would not be able to put Celsius on top of Celsius you will get the wrong answer.
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When we run this, let me go ahead and go back to black here.
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δ S should be 3.13 J/°K.
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Since we are dealing with 1 mol, we can just go ahead and put 3.13 J/°K mol or mol°K.
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It does not matter how you do it because we are talking about 1 mol.
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It is okay, let us go ahead and put them all there.
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It also asks us to compare δ S with Q/ T.
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I do have another page here and that is good.
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Let us go ahead to the next page.
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The δ S we just calculated so δ S = 3.13 J/°K mol.
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The problem with Q/ T is T is not fixed.
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T goes from 350°K to 450°K so technically we cannot really solve Q/ T so we should leave it alone.
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However, I’m going to do something different here.
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I’m going to take T average.
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I’m going to take the average between 350°K to 450°K instead of just 400°K just for illustrative purposes.
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1247 J that was the heat for this particular transaction and I'm going to divided by the 400°K.
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When I do that I will get 3.12 J/°K.
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Notice that they are almost the same, 3.13 is the actual change in entropy calculated.
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The Q/ the T average is 3.12.
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Let me write down this particular note again, we cannot really do Q/ T because T is not fixed.
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I chose to simply average the temperatures for illustrative purposes.
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If you are ever asked to do something like this on an exam, it is perfectly fine for you to say this cannot be done.
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Or if you do it, use T average and make sure you specify that you are actually using an average temperature here.
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Let us go ahead and do example 3, we have 1 mol of an ideal gas with this particular constant volume heat capacity initially in the following state 350°K, 1.5 atm.
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This is the same as the previous problem, the initial state is, at constant pressure the gas is heated to 450°K.
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The previous problem starting with this particular state we took a 450°K under constant volume.
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We are doing under constant pressure.
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At constant pressure, this is going to be important part here.
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I think I’m going to go in red here just for a change of pace, make it a little brighter.
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Calculate Q, W, δ U, δ H, and δ S and compare δ S with Q/ T.
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At constant pressure and we are dealing with an ideal gas.
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At constant pressure we know that δ H = the heat transpired, that is the whole idea behind the constant pressure process.
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Let us go ahead and take a look at the DH.
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The DH = CP DT this is one of our equations from the first law + DH DP constant T DP.
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This is an ideal gas so this term goes to 0 or its constant pressure so this term goes to 0.
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In either case, it goes to 0.
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What we have is DH = CP DT and when we integrate that we get δ H =CP δ T.
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When we do the mathematics I will get δ H =5/2 Rn δ T = 5/2 × 8.314 × 1 mol and the δ T is again 100°K.
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We get a δ H of 2079 J.
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At constant pressure we know the δ H = QP.
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Therefore, we automatically know the Q.
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The Q under constant pressure, the Q for this particular process is also = 2079 J.
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It takes care of the heat.
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Let us see what we can do about δ U.
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I will do it in this page it is not a problem.
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I know the δ H =δ U+ δ PV because the definition of enthalpy H =U + PV just apply the δ operator to all of that and you have this.
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Well δ H =δ U + P δ V because P is constant.
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Since there is no change in P, you can just go ahead and pull that out.
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That means δ U =δ H - P δ V or dealing with an ideal gas so PV =nRT.
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V = nRT/ P therefore δ V δ U =δ H – P.
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δ V is just V2 – V1 so we have nR T2 / P because it is constant pressure –nR T1/ P.
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The P cancel and I will get δ U =δ H - nR T2 - T1.
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I’m just fiddling around with the basic equation that is all it comes down to.
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We have δ U =δ H which we calculated which was 2079 J – n 1 mol R is 8.314 and δ T which is T2 - T1 is 100.
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When we end up doing that, we end up with δ U =1247 J.
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That is the energy of the system spans by 1247 J.
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δ U =Q – W, therefore W = Q - δ U.
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Q is 2079 J and δ U is 1247 J.
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Therefore, we get that the work = 832 J and I'm hoping that you are performing all of my arithmetic.
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It takes care of the work.
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We have the work, we have the energy, we have the enthalpy, we have that, let us go ahead and find our δ S.
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Let me see if I have an extra page.
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Yes, I do.
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Let me go ahead and go to the next page.
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In terms of pressure we have DS.
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We are express things in terms of temperature and volume, temperature and pressure.
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In this case because pressure is involved we are going to use this version CP/ T DT - nR/ P DP.
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The pressure is constant so this term goes to 0.
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What we have is just DS =CP / T DT and when we integrate the get δ S = it is going to be the integral from T1 to T2 of CP/ T DT.
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CP is a constant so what comes out from under the integral and which are left with is DS = CP × the log of T2/ T1.
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When we go ahead and put numbers in here, we go down here a little lower we get 5/2 × 8.314 × 1 mol
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× the log of 450°K/ 350°K and we get δ S = 5.22 J/°K.
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Our comparisons so we have δ S = 5.22 J/°K that is our δ S.
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Again, the T is not fixed so we technically cannot do this.
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I decided to just go ahead and do a T average.
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Our Q here is 2079 J and we have an average of 400°K.
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We get 5.2, it looks like a 6.
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We have 5.20 J/°K that is our Q/ T.
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We are just running through our basic equations that is what we want do.
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We have derived all of these equations and we have set aside specific ones that are important to learn, for example this one regarding entropy.
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We are just gaining some practice and of course reviewing all of our concept from energy that is why we are comfortable with this.
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At this stage of the game, we want to be able to understand the thermodynamics but mostly we just want to be able to be comfortable
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with the mathematics and be able to solve some of the problems that are thrown at us.
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Let us see what is next, example number 4 I think.
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Example number 2, example number 3, start with the same state.
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Example number 4, the next example 5 and the next couple of examples in the next lesson that we do,
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they are all going to be based on the same initial state this 350°K, 1.5 atm, an ideal gas with a constant volume heat capacity of 3/2 Rn.
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What we are going to be doing was we are going to be transforming it from one state to another state under different circumstances.
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For example 2, we made a transformation from 350 to 450 under constant volume.
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In the example after that example 3, we do the transformation of 350 to 450 at constant pressure.
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It is going to be the same starting state, we would be doing something else to it.
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That is all we are doing, we are just trying to figure out how to deal with these thermodynamic variables under different conditions.
00:28:58.100 --> 00:29:07.300
1 mol of an ideal gas with the constant volume heat capacity of 3 Rn/ 2 is initially in the following state 350°K. 1.5 atm.
00:29:07.300 --> 00:29:20.500
The gases expanded and this time isothermally and reversibly until the pressure of the gas drops 2.75 atm.
00:29:20.500 --> 00:29:28.400
Isothermally the temperature stays the same but the pressure does not 1.5 to 0.75 atm, the gas itself.
00:29:28.400 --> 00:29:33.000
Calculate these variables and compare δ S with Q/T.
00:29:33.000 --> 00:29:34.400
Let us see what is going on here.
00:29:34.400 --> 00:29:40.200
The important thing is isothermal and reversible so this is going to affect the mathematics.
00:29:40.200 --> 00:29:47.800
Let us go ahead and draw this one real quickly so we know what we are dealing with.
00:29:47.800 --> 00:29:54.800
This is pressure and this is volume so this is an isotherm.
00:29:54.800 --> 00:30:00.500
This is state 1 and this is state 2, whatever that happens to be.
00:30:00.500 --> 00:30:09.200
State 2, we know this is 350°K and it is going to be 1.58 atm.
00:30:09.200 --> 00:30:17.900
This is going to be 350°K because it is isothermal and it is going to be 0.75 atm.
00:30:17.900 --> 00:30:26.200
This is your P1, this is your 1.5 and this is your 0.75.
00:30:26.200 --> 00:30:29.600
I know it is not the scale and this is not exactly half of it but that is fine.
00:30:29.600 --> 00:30:34.400
Here we have volume 1 and volume 2, it is expanding.
00:30:34.400 --> 00:30:42.800
The gas is going from one volume and it is expanding to a larger volume that is what this is.
00:30:42.800 --> 00:30:48.500
What we are doing is we are keeping the temperature constant so the expansion just not happened randomly.
00:30:48.500 --> 00:30:53.600
This path or that path it actually happens isothermally, it happens along this path.
00:30:53.600 --> 00:30:56.300
The temperature stays along this path.
00:30:56.300 --> 00:30:59.600
Even more than that, it happens reversibly.
00:30:59.600 --> 00:31:03.600
Reversibly means we absolutely follow this path.
00:31:03.600 --> 00:31:10.900
In other words, we do not go this way, we do not go that way.
00:31:10.900 --> 00:31:15.200
We can have an isothermal transformation that is not reversible.
00:31:15.200 --> 00:31:20.400
That is not a problem, we can go ahead and expand the gas and keep the temperature constant.
00:31:20.400 --> 00:31:28.700
It follows an isotherm but the actual path that it follows, the work that is done is going to be different depending on the path we follow,
00:31:28.700 --> 00:31:31.100
this work is a path function.
00:31:31.100 --> 00:31:34.500
In this case, it is reversible so we are absolutely following this path.
00:31:34.500 --> 00:31:40.200
What we are doing is we are making a bunch of little of micro changes all the way around.
00:31:40.200 --> 00:31:41.600
Again this a reversible.
00:31:41.600 --> 00:31:50.500
Remember what reversible means in terms of mathematics, it means that the external pressure = to the pressure of the system,
00:31:50.500 --> 00:31:52.700
the internal pressure if you will.
00:31:52.700 --> 00:32:00.000
The pressure on the outside = the pressure on the inside so the system is always at equilibrium.
00:32:00.000 --> 00:32:05.400
Any new small change that you make, all you have to do is go back that little small differential change.
00:32:05.400 --> 00:32:07.500
Essentially, the system is in equilibrium.
00:32:07.500 --> 00:32:10.200
Mathematically, it means that P external = P.
00:32:10.200 --> 00:32:13.200
That is what reversible means.
00:32:13.200 --> 00:32:17.000
Let us go ahead and see what is going on here.
00:32:17.000 --> 00:32:20.000
I will just this label this so this is an isotherm.
00:32:20.000 --> 00:32:30.000
It is very important so you can do something isothermally or you can do something isothermally and reversibly.
00:32:30.000 --> 00:32:39.200
DW = P external × DV that is the basic, that is the definition of work.
00:32:39.200 --> 00:32:50.300
Because it is reversible, the P external = P so we have DW =P DV.
00:32:50.300 --> 00:33:13.300
This is an ideal gas so we have PV =nRT volume =nRT/ P.
00:33:13.300 --> 00:33:18.500
My apologies, I’m going to do this, this is going to be P =nRT/ V.
00:33:18.500 --> 00:33:35.800
I’m going to have put this into here so I get DW =nRT / V DV then when I integrate this, I end up with the following.
00:33:35.800 --> 00:33:40.400
I get the integral of DW is W not δ W.
00:33:40.400 --> 00:34:03.100
I get that the work = nRT × the integral volume 1 to volume 2 of DV/ V= work.
00:34:03.100 --> 00:34:18.500
When I do this integral, I get nRT × the log of volume 2 / volume 1 so this is my fundamental equation.
00:34:18.500 --> 00:34:19.700
Let us deal with this.
00:34:19.700 --> 00:34:23.800
I do not have V2 and V1, what I have is P2 and P1.
00:34:23.800 --> 00:34:27.600
Again, this is an ideal gas, let me express it this way.
00:34:27.600 --> 00:34:35.200
I have PV =nRT, I have V =nRT/ P.
00:34:35.200 --> 00:34:45.500
Therefore, V2 =nRT / P2, V1 =nRT/ P1.
00:34:45.500 --> 00:34:52.900
This is isothermal so T stays the same, the pressure that is going to be different.
00:34:52.900 --> 00:35:01.800
V2/ V1, I will go ahead and do on the next page.
00:35:01.800 --> 00:35:16.900
V2/ V1 = nRT/ PQ / nRT/ P1 you end up with P1/ P2.
00:35:16.900 --> 00:35:37.800
Therefore, our equation work = nRT Ln of V2/ V1 ends up becoming work = nRT × the log of P1/ P2.
00:35:37.800 --> 00:35:43.200
We are just fiddling things around, just basic mathematics that is all it is.
00:35:43.200 --> 00:35:45.300
Now we go ahead and solve it.
00:35:45.300 --> 00:36:09.400
We have 1 mol, we have 8.314 J/mol-°K, we have a temperature of 350°K, and we have a log of 1.5 atm/ 0.758 atm.
00:36:09.400 --> 00:36:21.600
This cancels giving us the log of a pure number and we get that the work = 2017 J.
00:36:21.600 --> 00:36:26.700
That much work is done as the gas expands.
00:36:26.700 --> 00:36:37.500
The system is doing work on the surroundings, work is positive.
00:36:37.500 --> 00:36:41.100
It says that it is isothermal.
00:36:41.100 --> 00:36:47.800
Isothermal implies that δ U =0.
00:36:47.800 --> 00:36:54.100
However, be careful with this so δ U = 0 and let me tell you why.
00:36:54.100 --> 00:37:09.700
In this particular case, this is the basic equation DU = CV DT + DU DV at constant temperature DV.
00:37:09.700 --> 00:37:14.100
The temperature is constant so this term goes to 0.
00:37:14.100 --> 00:37:25.300
DU DV we are dealing with an ideal gas so this term goes to 0 that is why DU = 0 or δ U =0.
00:37:25.300 --> 00:37:27.900
You are going to run across a lot of problems in thermodynamics.
00:37:27.900 --> 00:37:31.100
Most of the problems that you do are going to involve an ideal gas.
00:37:31.100 --> 00:37:37.300
When you see the word isothermal, you can just automatically set δ U = 0, it is not a problem.
00:37:37.300 --> 00:37:43.500
However, if you understand that δ U = 0 under isothermal conditions only for an ideal gas.
00:37:43.500 --> 00:37:55.200
If you are not dealing with an ideal gas this term is not 0, if it is small or whatever happens to be but it is not 0.
00:37:55.200 --> 00:37:59.200
Isothermal does not automatically mean that δ U is 0.
00:37:59.200 --> 00:38:04.100
Isothermal for an ideal gas implies that δ U is 0.
00:38:04.100 --> 00:38:12.200
There are times when we can set something to 0, isothermal δ U= 0, adiabatic Q = 0.
00:38:12.200 --> 00:38:21.600
It is for an ideal gas, if this is not an ideal gas, if it is a Van Der Waals gas or another kind of gas,
00:38:21.600 --> 00:38:27.300
or the solid or liquid because this equation works for any system at all this is not to be 0.
00:38:27.300 --> 00:38:34.100
We cannot just automatically set isothermal means δ U = 0 please remember that.
00:38:34.100 --> 00:38:36.300
Let us go ahead and finish up here.
00:38:36.300 --> 00:38:49.200
δ U=Q - W well δ U was 0 and Q the work was 2017 J.
00:38:49.200 --> 00:38:56.800
Therefore, Q = 2017 J.
00:38:56.800 --> 00:39:02.300
Let us go ahead and deal with δ H and δ S.
00:39:02.300 --> 00:39:16.400
Let us do it over here, δ H we said was δ U + δ PV.
00:39:16.400 --> 00:39:41.000
This is 0, therefore we have δ PV which is nothing more than P2 V2 - P1 V1 which is nothing more than nR T2 – nR T1, which is equal to nR δ T.
00:39:41.000 --> 00:39:46.200
It is isothermal, the temperature stays the same so this is 0.
00:39:46.200 --> 00:39:52.500
The δ H for this process is 0.
00:39:52.500 --> 00:40:07.900
Another way of doing this, this δ H business, we could use the δ H equation DH = CP DT + DH DP at constant T DP,
00:40:07.900 --> 00:40:13.300
change in temperature δ T, this is an isothermal process so this term goes to 0.
00:40:13.300 --> 00:40:16.000
For an ideal gas, this term goes to 0.
00:40:16.000 --> 00:40:21.900
Again, we have DH = 0 which means that δ H =0.
00:40:21.900 --> 00:40:24.500
Either one of these processes are absolutely fine.
00:40:24.500 --> 00:40:28.000
You are going to end up getting the same answer.
00:40:28.000 --> 00:40:31.000
Let us go ahead and take care of the δ S.
00:40:31.000 --> 00:40:43.800
We have DS = CP/ T DT- nR/ P DP.
00:40:43.800 --> 00:40:48.500
This is isothermal so that takes this term to 0.
00:40:48.500 --> 00:40:59.000
We are left with DS = - nR / P DP.
00:40:59.000 --> 00:41:08.200
We are going to integrate this like we always do and we end up with δ S =.
00:41:08.200 --> 00:41:19.300
The nR comes out of the integral you going to be left with the integral from DP/ P so what you get is the following.
00:41:19.300 --> 00:41:28.500
You get - nR × the nat log of pressure 2 / pressure 1.
00:41:28.500 --> 00:41:31.400
We can go ahead and put the numbers in.
00:41:31.400 --> 00:41:38.200
This 8.314 × the nat log of pressure 2/pressure 1.
00:41:38.200 --> 00:42:01.200
Pressure 2 is 0.75 atm, pressure 1 is 1.5 atm, δ S is going to end up equaling -5.76.
00:42:01.200 --> 00:42:08.500
The log 0.75/ 1.5 is 1/2 , the log of the number less than 1 is negative.
00:42:08.500 --> 00:42:14.400
This number is negative × the negative gives you a positive.
00:42:14.400 --> 00:42:24.200
δ S =5.76 J/°K.
00:42:24.200 --> 00:42:27.100
Let us compare δ S with Q/ T.
00:42:27.100 --> 00:42:37.400
δ S which is calculated δ S 5.76 J/°K.
00:42:37.400 --> 00:42:45.100
Let us calculate Q/ T, in this case we can do T at Q/ T because T stays the same, it is just 350°K.
00:42:45.100 --> 00:42:53.200
In this particular case, it is actually Q reversible because we are doing this process along the reversible paths so Q/ T = Q reversible / T.
00:42:53.200 --> 00:43:01.100
That is the definition of entropy.
00:43:01.100 --> 00:43:15.100
The Q reversible we have 2017 J, we have 350°K ends up being 5.76 J/°K.
00:43:15.100 --> 00:43:28.100
What about 5.76? it is exactly the same because we are dealing with a reversible path.
00:43:28.100 --> 00:43:31.000
A quick review of what is happening here physically.
00:43:31.000 --> 00:43:34.300
Notice, how the work and the heat are equal.
00:43:34.300 --> 00:43:42.400
We ended up with a work and the heat being equal to 2017 J, they ended up being equal.
00:43:42.400 --> 00:43:47.300
Here is what is going on.
00:43:47.300 --> 00:43:58.600
As the system expands, it does work on the surroundings.
00:43:58.600 --> 00:44:10.800
It is W on the surroundings and the amount was I think 2017 J.
00:44:10.800 --> 00:44:31.500
As it expands, as a gas expands it is going to cool but the problem is we are doing this process isothermally so we do not want it to cool.
00:44:31.500 --> 00:44:35.800
We are going to keep the temperature the same as it expands.
00:44:35.800 --> 00:45:10.200
Because the process is to happen isothermally, the system requires energy in the form of heat
00:45:10.200 --> 00:45:14.800
in order to keep the temperature up so the gas is expanding, the gas wants to cool.
00:45:14.800 --> 00:45:18.900
It is happening isothermally so something, somehow, energy has to come into the system to keep the temperature from dropping,
00:45:18.900 --> 00:45:24.000
to keep it from cooling because we want to keep the temperature the same.
00:45:24.000 --> 00:45:38.900
Because the process happens isothermally, the system requires energy in the form of heat to keep the temperature constant, to keep the T constant.
00:45:38.900 --> 00:45:41.900
The question is where is this heat going to come from?
00:45:41.900 --> 00:45:53.000
There is only one source, it has to come from the surroundings.
00:45:53.000 --> 00:46:14.300
This heat as energy has to come from somewhere.
00:46:14.300 --> 00:46:35.700
It comes from, it is pulled from the surroundings.
00:46:35.700 --> 00:46:54.100
Since Q is positive, when it enters the system that is what we get Q =2017 J.
00:46:54.100 --> 00:46:58.100
What is happening here is the following under isothermal conditions.
00:46:58.100 --> 00:47:12.200
Work is done on the surroundings, the surroundings get colder.
00:47:12.200 --> 00:47:15.200
This is our experience with the first law.
00:47:15.200 --> 00:47:20.900
The surroundings get colder.
00:47:20.900 --> 00:47:26.300
The gas expands does work on the surroundings 2017 J of work.
00:47:26.300 --> 00:47:31.400
In the process of expanding the gas inside wants to cool, the system wants to cool.
00:47:31.400 --> 00:47:33.100
We are doing this process isothermally,
00:47:33.100 --> 00:47:37.700
In order for it to maintain its temperature, energy has to come into the form of heat.
00:47:37.700 --> 00:47:42.000
Heat has to come from somewhere so it pulls that heat from the surroundings.
00:47:42.000 --> 00:47:48.000
As the gas is doing work on the surroundings it is actually pulling heat from the surroundings.
00:47:48.000 --> 00:47:59.600
When you pull heat from the surrounding, the surroundings are going to get colder.
00:47:59.600 --> 00:48:03.700
Energy is transferred to the surroundings as work.
00:48:03.700 --> 00:48:07.500
Energy is taken from the surroundings as heat.
00:48:07.500 --> 00:48:19.100
That is all that is happening here.
00:48:19.100 --> 00:48:27.000
Let us see what is going on, let us go to our last example here.
00:48:27.000 --> 00:48:33.200
Example 5, the same starting conditions, different set of circumstances.
00:48:33.200 --> 00:48:40.400
1 mol of an ideal gas, constant volume heat capacity 350°K, 1.5 atm, the gas expands isothermally.
00:48:40.400 --> 00:48:51.000
We are isothermal, against a constant external pressure of 0.75 atm until the pressure of the gas achieves 0.75 atm.
00:48:51.000 --> 00:48:53.200
The same exact initial conditions.
00:48:53.200 --> 00:48:57.500
The same exact same exact initial state, same exact final state.
00:48:57.500 --> 00:49:02.900
The final state is going to be 0.75 atm but now it is not going to expand reversibly.
00:49:02.900 --> 00:49:06.400
It is going to do it isothermally but that is not going to be reversible.
00:49:06.400 --> 00:49:13.100
Let us see what this one looks like.
00:49:13.100 --> 00:49:22.100
We have our pressure, we have our volume, we have our isotherm, we have our state 1, we have our state 2.
00:49:22.100 --> 00:49:34.800
This is state 1, this is state 2, this is pressure 1 that is the 1.5 atm and this is going to be P2, this is going to be 0.75 atm.
00:49:34.800 --> 00:49:37.200
1.5 and 0.75 atm.
00:49:37.200 --> 00:49:39.900
This is volume 1 and this is volume 2.
00:49:39.900 --> 00:49:47.500
It is going to expand so isothermally the temperature is going to stay the same so we are still going to expand.
00:49:47.500 --> 00:49:55.600
This is the isotherm but now the particular path that we are going to take is the following.
00:49:55.600 --> 00:50:02.000
The external pressure is 0.75 so we are going to expand this way.
00:50:02.000 --> 00:50:04.400
This is the path that we are going to take.
00:50:04.400 --> 00:50:10.400
It is at constant external pressure so this is the amount of work that is done.
00:50:10.400 --> 00:50:17.900
The last problem we did isothermal reversible work.
00:50:17.900 --> 00:50:24.200
We follow this path, we kept the temperature the same but we did along the isotherm itself.
00:50:24.200 --> 00:50:25.900
There was all this extra.
00:50:25.900 --> 00:50:34.600
We can only expect that a work in this particular example is going to be less than the work that we had in the previous example.
00:50:34.600 --> 00:50:38.100
Let us hope it actually turns out that way.
00:50:38.100 --> 00:50:41.100
It is a different path.
00:50:41.100 --> 00:50:45.700
The temperature is the same so we can do something isothermally but now we are following a different path so
00:50:45.700 --> 00:50:56.800
the work is going to be different, the heat is going to be different.
00:50:56.800 --> 00:50:58.300
Where shall we start?
00:50:58.300 --> 00:51:01.800
Let us start the same way, the definition of work.
00:51:01.800 --> 00:51:09.100
The definition of work is external pressure × the change in volume.
00:51:09.100 --> 00:51:19.800
That means that work = the external pressure because the external pressure is constant, constant external pressure × δ V.
00:51:19.800 --> 00:51:32.600
We just integrated this differential equation so we get work = P external.
00:51:32.600 --> 00:51:43.600
This is going to be P external volume 2 - volume 1, we are dealing with an ideal gas therefore V = nRT / P.
00:51:43.600 --> 00:52:03.100
Volume 2 = nRT/ P2 - nRT/ P1.
00:52:03.100 --> 00:52:11.000
We get that the work = the external pressure × the nR T.
00:52:11.000 --> 00:52:13.400
I’m going to pull out the nRT because it is going to be constant.
00:52:13.400 --> 00:52:22.000
It is going to be 1/ P2 -1/ P1 and now we can go ahead and put our numbers.
00:52:22.000 --> 00:52:29.300
work = the constant external pressure is 0.75 atm.
00:52:29.300 --> 00:52:35.300
n is 1 mol, R is 8.314 J/°K.
00:52:35.300 --> 00:52:40.800
Our temperature is isothermal, it is happening at 350 K.
00:52:40.800 --> 00:52:50.700
1/ the external pressure P2 is 0.75 -1/1.5.
00:52:50.700 --> 00:52:58.000
If the arithmetic comes out correct you should have 1455 J.
00:52:58.000 --> 00:53:01.800
This work is definitely less than the 2017 J.
00:53:01.800 --> 00:53:07.500
This 1455 accounts for this work was done against a constant external pressure.
00:53:07.500 --> 00:53:12.300
In the previous problem we did it along the isotherm itself, we did this process reversibly.
00:53:12.300 --> 00:53:21.500
Therefore, remember reversible processes are important because they represent maxima and minimal, things like that.
00:53:21.500 --> 00:53:32.300
In the case of an expansion, it is the maximum amount of work that you are going to get from a particular expansion.
00:53:32.300 --> 00:53:45.800
We are dealing with an ideal gas so isothermal again in the case of an ideal gas, isothermal definitely implies that the change in energy of the system is 0.
00:53:45.800 --> 00:53:56.900
δ U =Q - W so 0 = Q – W.
00:53:56.900 --> 00:54:10.500
Therefore, Q = W = 1455 J and it is still happening isothermally.
00:54:10.500 --> 00:54:13.000
The heat and the work are still the same.
00:54:13.000 --> 00:54:15.100
The same thing is happening, it is still expanding.
00:54:15.100 --> 00:54:22.700
It is expanding and wants to get cooler, we are not letting it get cooler, it is doing or forcing it to do isothermally and it has to pull energy from somewhere.
00:54:22.700 --> 00:54:24.100
It is going to pull it from the surroundings.
00:54:24.100 --> 00:54:32.800
Work is done on the surroundings in expansion, the surroundings end up getting colder.
00:54:32.800 --> 00:54:39.500
Let us go ahead and do DH.
00:54:39.500 --> 00:54:56.500
DH =CP DT + DH DP DP constant temperature that goes to 0, ideal gas that goes to 0.
00:54:56.500 --> 00:55:03.600
Therefore, our enthalpy change δ H for this process is 0.
00:55:03.600 --> 00:55:13.300
δ S I’m going to calculate because δ S is a state property.
00:55:13.300 --> 00:55:17.700
State 1 and state 2, the first path we took was this way.
00:55:17.700 --> 00:55:21.400
This path now we are going this way.
00:55:21.400 --> 00:55:23.600
This way and this way.
00:55:23.600 --> 00:55:27.100
A state property does not depend on the path.
00:55:27.100 --> 00:55:33.400
Therefore, the change in entropy is still to be 5.76.
00:55:33.400 --> 00:55:37.900
It is the same number that we got from the previous example J/ K.
00:55:37.900 --> 00:55:42.500
It is going to be 5.76 J/°K.
00:55:42.500 --> 00:55:49.500
Heat and work are different because they are not state functions, they are state properties, they are path properties.
00:55:49.500 --> 00:55:53.500
But because δ S is a state function the path that we taken together does not matter.
00:55:53.500 --> 00:55:57.900
We can go ahead and use it from the previous example.
00:55:57.900 --> 00:56:00.600
Let us go ahead and calculate Q and T.
00:56:00.600 --> 00:56:13.000
Q in this case is 1455 J, temperature is 350°K, in this particular case we get 4.16 J/°K.
00:56:13.000 --> 00:56:19.000
In this particular case because of a different path the entropy is actually less.
00:56:19.000 --> 00:56:20.600
It is not the 5.76.
00:56:20.600 --> 00:56:35.300
You are going to get the same entropy Q/ T and δ S when you run a reversible process because the definition of entropy is DQ reversible / T.
00:56:35.300 --> 00:56:39.000
This reversible part is very important.
00:56:39.000 --> 00:56:41.100
Thank you so much for joining us here at www.educator.com.
00:56:41.100 --> 00:56:44.000
We will see you next time for more example problems, bye.