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 example problems on free energy.
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Let us dive right on in.
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The first problem says, demonstrate that for a real gas the constant pressure heat capacity ×
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the Joule-Thompson coefficient = RT²/ P DZ DT constant P,
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where the entity is the Joule-Thompson coefficient and Z = PV/ RT, the compressibility factor for the gas.
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Z which is pressure × volume/ RT is the standard variable that you run across in thermodynamics.
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It is called the compressibility factor so we want to demonstrate that this is the case given
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the mathematics and all of the equations that we have at our disposal.
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Let us see what we can do.
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We know from back when we did on energy and we first talked about the Joule-Thompson coefficient that we have the following relation.
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Let us go ahead and actually do this in blue.
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We have the following.
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We have that CP C sub P × the Joule-Thompson coefficient = - DH DT at constant P.
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This was the definition that we have for this particular relationship earlier on.
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Let me go ahead and put Z = PV/ RT over here.
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We have by the thermodynamic equation of state which we just a few lessons ago, we have the following.
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The DH DT sub P = that molar V with a line over any variable that means per mole.
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- T × DV DT sub P so we have that Z sub P μ of JT = negative of this which is the negative of this which is the negative of this.
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I’m just substituting this expression in for that.
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I'm going to flip these to change the negative so I end up with T DV DT sub P – V.
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This is by the thermodynamic equation of state.
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Let us go ahead and we see this thing here.
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Let us go and find what DZ DT is when we hold P constant.
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Here is my expression.
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I’m going to differentiate this with respect to T holding the P constant.
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When I do that, I get the following.
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We do with over here separate, I get DZ DT holding P constant.
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When I run the full derivative of this and I will go through it entirely step by step, I will just write up what the final result is.
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It will be T × DV DT sub P - V/ T².
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We get this thing for the derivative.
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Now what I'm going to do is, I’m going to take this side and I’m going to move some things around.
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I’m going to multiply by R, I’m going to multiply by T².
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I’m going to divide by P so I’m going to end up with the following.
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I end up with RT²/ P × DZ DT sub P = T × DV DT sub P – V.
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This right here is that right there.
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If this is equal to that and that is equal to this, that means this equal this.
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I'm done, I’m just going to substitute this expression in for here and I end up with what it is that I wanted.
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CP μ Joule-Thompson coefficient = RT²/ P DZ DT sub P which is absolutely what I wanted.
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I use the original definition of the heat capacity × the Joule-Thompson coefficient
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which was the rate of change of enthalpy with respect to temperature at constant pressure.
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I use the thermodynamic equation of state.
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I set these two equal to each other.
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I have differentiated Z, manipulated that and then substituted this expression back here to get what it is that I wanted.
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Nice and straightforward.
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Relatively speaking, there are a lot of symbols floating around here.
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Let us see what is next, calculate δ A, the change in the Helmholtz energy for the isothermal expansion.
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We have something that is going to happen isothermally of 1 mol of an ideal gas.
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This was nice and easy.
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25° C from a volume of 20 L to a volume of 80 L.
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We are going to expand this gas, this ideal gas.
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We are going to keep the temperature constant and we want to know what the change in Helmholtz energy is.
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We go back to our basic fundamental thermodynamic equations.
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For DA, it is the following.
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It is DA = - S DT – P DV.
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It is an isothermal process so DT = 0.
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This is 0 because it is isothermal that means that DA is just equal to - P DV.
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We are talking about an ideal gas so an ideal gas is PV = nRT which means that P = nRT/ V.
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I can go ahead and put this expression in for that and I get DA = -nRT/ V DV.
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Then I go ahead and I integrate my expression and I end up with δ A = -nRT.
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I will go ahead and do this here V1 to V2 of DV/ V.
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We have seen this 1000 times already, nRT LN.
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This is not pressure, we are not doing δ G here.
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This is V2/ V1.
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We just put our numbers in δ A = - 1 mol and we have 8.314 J/ mol °K.
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I’m going to go ahead and leave the units off, I hope you will forgive me.
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25° C so this is going to be 298 °K and the log of 80 L.
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As long as the units match we are okay.
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20 L we do not have to make any conversions.
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And if I have done my arithmetic correctly, which I hope you will confirm -3435 J/ mol for this ideal gas.
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That is the change in Helmholtz energy δ A = this.
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This much energy is available to do a certain specified amount of work under the conditions of isothermal change in volume.
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Calculate δ G for the isothermal expansion of 1 mol of an ideal gas at 400°K from a pressure of 7000 kPa to a pressure of 100 kPa.
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It was going from a really high pressure situation to a low pressure, it is expanding.
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Again, the process is going to be isothermal and we are going to do this at 400°K.
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It is the same exact thing as before, except now, instead of Helmholtz energy we are calculating the change in free energy.
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This is going to be a question of temperature and pressure.
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You remember Helmholtz energy is under conditions of constant temperature and volume.
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The Gibbs free energy is under conditions of constant temperature and pressure.
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Since, most of the things that we do in chemistry happen under constant temperature and pressure,
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when we really do not have to walk anything, we just run the experiment.
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Temperature is constant, pressure is constant, this is why the Gibbs energy is important in chemistry.
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We go back to our fundamental equation of thermodynamics in the form of DG = - S DT + V DP.
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Again, this is 0 because we are talking about isothermal process.
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Therefore, we have the DG = V DP.
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We will take our ideal gas law PV = nRT and we solve for V.
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This is nRT/ P and we substitute this into here and we integrate.
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We get DG = nRT/ P DP.
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When you integrate this expression you end up with the following.
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The same thing as before, you get DG = nRT the log this time it is going to be the final pressure - the initial pressure.
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P2/ P1, I'm sorry final pressure/ initial pressure.
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We get 1 × 8.314 × 400°K × log.
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The final pressure now was 100 kPa and this is 7000 kPa.
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The units are the same so you do not have to make any conversion, it is not a problem.
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If I did my arithmetic correctly, I should get - 14,129 J and this is because it is 1 mol we can go ahead and put the per mole.
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I actually put the 1 mol here so technically this is -14,129 J.
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Since, it is 1 mol let us go ahead and put the per mol.
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We can do that, not a problem.
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There you go, that is δ G.
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For this particular process, there are 14,129 J of free energy available to do work.
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That is the maximum amount of work that this particular process, this expansion can accomplish.
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Will it accomplish this much work?
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No, the maximum has never reached.
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It is an ideal, it is the maximum that you can get.
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If everything were perfect and there is nothing lost as heat, then yes you can recover this much work.
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But again, this is the maximum amount of work that is derivable from this process.
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That is what δ G is, it tells you how much work you can extract from a process that is why we call it free energy.
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It is free energy, it is energy that you can use freely to do work because the other energy of the system
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that is not available for work is tied up in the entropy of the system.
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Things seem to be moving along pretty nicely here.
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Use the following form of the Van Der Waals equation to derive an expression for the δ G of 1 mol
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of a gas as it is compressed isothermally from 1 atm to Pa atm.
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We are going from 1 atm to some random Pa atm.
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We do not want actual number, we want an expression.
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In the previous problems, we work with the ideal gas, the PV = nRT.
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We are going to do something with the Van Der Waals gas.
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We are going to see what happens when we deviate slightly from ideal behavior.
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Z = PV/ RT this is the compressibility factor and this is the version of the Van Der Waals equation expressed in this form.
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Let us write out, we have got this expression.
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I’m going to go ahead and express volume.
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I’m going to go ahead and move the RT over here and I'm going to divide by P.
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I’m going to rewrite this expression as V = RT/ P + B - A/ RT.
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This is just simple algebra, moving things around so I can leave V alone on that side.
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We are looking for δ G so again we use the same relation that is universally available.
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It is DG the fundamental equation of thermodynamics for free energy,
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the relationship is DG in other words the total differential for this is - SDT + V DP.
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This is isothermal so we can go ahead and dismiss this term because DT is 0.
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We have DG = V DP.
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V is right here so now I take this expression and I put in here like I did for the ideal gas.
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The ideal gas is just nRT/ P.
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Here it is a little bit more complicated because we are dealing with a Van Der Waals gas.
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What should I do here?
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That is fine, I’m going to rewrite it.
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We have DG = the integral from P1 to P2 of V DP = the integral from P1 to P2 of this expression RT/ P + B - A/ RT.
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Do not let the symbolism intimidate you, it is just stuff.
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And notice, you are integrating with respect to P.
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There is no P in this one, the P only shows up here.
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This is a simple integration, this is just a constant.
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Do not let the symbols intimidate you.
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When we do this, we are just going to integrate this expression because the integral of the sum is
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the sum of the integrals we end up with the following.
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We end up with RT LN P/ 1 because we are 1 atm + B - A/ RT × δ P -1.
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This is a very simple integration.
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It is the integral of this with respect to P and the integral of this with respect to B.
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This is just a constant so you end up with just the integral of DP which end up being the DP which is P-1 final – initial.
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You have your expression.
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For an ideal gas, for a Van Der Waals gas this is the expression.
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For the ideal gas, we have this part P2/ P1.
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The Van Der Waals gas, there is a little bit of an adjustment to it.
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Notice, you are actually adding some so the amount of free energy that you have for an ideal gas
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is going to be a certain amount like our previous problem.
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The amount of energy that you have for a Van Der Waals gas is actually going to be a little bit less.
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You have a negative number and you are going to add some to it.
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A real gas behaves in a non ideal fashion, that is all this is saying.
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For a Van Der Waals gas this is the expression for δ G.
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The Van Der Waals constants for oxygen are as follows.
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A= this, B = this, find δ G for the isothermal expansion O2 as a Van Der Waals gas at 400 °K from 7000 kPa to 100 kPa.
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What it actually do, example problem 3 except now we are going to treat it as a Van Der Waals gas
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instead of an ideal gas to see what the difference is.
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We have our expression for the Van Der Waals gas DG = RT LN P2/ P1 + B - A/ RT × P2 - P1.
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When I put my numbers in, δ G is going to equal 8.314 and we are doing this at 400°K.
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We are taking the logarithm, we are doing 100 kPa/ 7000 kPa + then we have the B which is 3.18 ×
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10⁻⁵ - 0.138 which is A/ R which is 8.314 and T which is 400.
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This is just numerical stuff × final is 100 -7000.
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I’m going to multiply it by 10³ because we have to express it in Pascal.
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These are in kPa and I need the actual value to be in Pascal, that is why I multiply by 10⁻³.
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When I do this, I get δ G = -14,129 which is the same value that we got for the ideal gas.
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This part is the ideal gas portion of it and then I have + 66.9, this is the adjustment that we make.
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Our final δ G for oxygen as a Van Der Waals gas is -14,062 J/ mol.
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We have less energy available in this particular expansion once it expands as a Van Der Waals gas.
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You can imagine a real gas is going to have even less free energy available.
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The maximum amount available is the -14,129 that comes from ideal behavior.
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Treating it as a Van Der Waals gas, we have to make a little bit of adjustment.
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If we use other equations of state, whatever they happen to be, we are going to end up with less and less and less.
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The real gas is going to be the least of all, that is all that is happening.
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Let us see what we have got.
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1 mol of the following substances at 298°K are subjected to isothermal pressure increases from 1 atm to 100 atm,
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calculate δ G for each substance.
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An ideal gas and liquid water whose molar volume is 18 cm³ / mol and iron metal with density of 7.9 g/ cm³.
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Let us just jump right in and do these.
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Let me go back to blue here, no worries.
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DG = -S DT, you always want to start with your fundamental equations.
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Those are the ones that you want to know, to memorize.
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From those, based on what the problem is asking whether it is isothermal or isobaric, whatever, you knock off terms, you adjust terms.
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You do not have to learn 50,000 equations.
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There is a handful of equations that you should know, that is your starting point.
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Not to mention the fact that you are always consistent when you do your problems.
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You are always starting at the same point and the problem will go in the direction that
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it needs to go based on the constraints that you are putting off of the problem is putting on it.
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The same equation.
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It is a nice way of learning the equations because you are only starting with them.
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DG = - S DT + V DP this is isothermal so that goes away.
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We already know what this is, this is for an ideal gas.
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Let us just go ahead and integrate this so we can just write δ G = nRT LN P2/ P1.
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Nice and simple.
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When I put these particular values in here, I get δ G = 1 mol 8.314 temperature at 298°K, nat log of 100 atm / 1 atm.
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100 atm is actually a lot of pressure.
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When we do this, we end up with δ G = 11,409 J/ mol.
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The δ G is positive, this is not a spontaneous process.
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Clearly, it is not a spontaneous process.
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I have to do work on the system to compress it from 1 atm to 100 atm.
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If the other way around, 100 atm and I just let it expand that would be a spontaneous process.
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It is going to go from high pressure to low pressure.
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I do not have to do anything, it will expand by itself.
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I have to put the pressure on to actually compress it.
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That takes care of the ideal gas.
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Let us go ahead and see what we have for liquid water.
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I have got some space available that is not a problem.
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Part B well, I know already that δ G = the integral from P1 to P2 of V DP, that just come from the equation from part A.
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For a liquid, I have to put a lot of pressure on a liquid to change its volume.
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When we are talking about liquid and solids, for all practical purposes,
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unless you are talking about a huge pressure difference and even then, the volume changes very little.
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For all practical purposes, the volume of liquid or solid is constant.
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It is not like that for a gas, you know that from experience.
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For a liquid and a solid, we can treat the volumes constant so we can actually pull this out of the integral sign.
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δ G is just equal to whatever the volume happens to be × the change in pressure P2 - P1, this is our equation.
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We just need to find the volume of the 1 mol of water and I multiply by the pressure difference.
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This is a really easy problem.
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We just have to make conversions because the molar volume given to us was in cm³/ mol.
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We need to either deal in cubic meters if we are dealing in Pascal’s or in this case we are dealing with atmospheres,
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you want to work in liters which is the same as a dm³.
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We just need to make some basic conversions.
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Let us go ahead and do that first, let us find out what our volume is.
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18 cm³/ mol × I’m just going to do the conversion so you see it.
00:25:33.700 --> 00:25:39.100
1 dm is 10 cm, I have cubic centimeter which is cm cm cm.
00:25:39.100 --> 00:25:45.400
You remember, when you are dealing with squares and cubes and power to the 4th, you have to cancel each unit.
00:25:45.400 --> 00:25:52.900
I have 3 cm so I have to multiply by 3 times, it cancel cm 3 times.
00:25:52.900 --> 00:26:01.100
I apologize for being so elementary but I think it is always nice to remember these things because it is easy to forget.
00:26:01.100 --> 00:26:09.200
1 dm/ 10 cm, when I do this it is cm cm cm cancels that.
00:26:09.200 --> 00:26:15.900
When I do that, I get 18 × 10⁻³ dm³.
00:26:15.900 --> 00:26:18.600
Dm³ is a liter.
00:26:18.600 --> 00:26:24.000
The volume = 18 × 10⁻³ L.
00:26:24.000 --> 00:26:44.300
Well δ G = V × P2 - P1, volume is 18 × 10⁻³ L and my difference is 100 atm and 1 atm.
00:26:44.300 --> 00:26:51.100
When I do this, I’m going to get δ G = 1.782, but this is liter atmosphere.
00:26:51.100 --> 00:26:54.600
Liter atmosphere is not a Joule, I need to convert it to Joule.
00:26:54.600 --> 00:27:04.300
The conversion factor is 8.314 J to 0.08206 L atm.
00:27:04.300 --> 00:27:10.300
You just use the two values for the gas constant, J/ mol °K, L atm/ mol °K.
00:27:10.300 --> 00:27:22.100
The mol °K cancels, you are left with a conversion factor, liter atmosphere cancels liter atmosphere, I get the δ G =180 J/ mol.
00:27:22.100 --> 00:27:29.200
That is not a lot compared with the previous number which was in the thousands.
00:27:29.200 --> 00:27:31.300
What is the number specifically?
00:27:31.300 --> 00:27:37.000
It is 11,409, so 11,500, 180.
00:27:37.000 --> 00:27:43.300
Clearly, pressure does not have the same effect on liquid and solids that it has on a gas.
00:27:43.300 --> 00:27:45.700
This is a very small difference.
00:27:45.700 --> 00:27:47.600
100 atm is huge.
00:27:47.600 --> 00:27:51.400
We are putting a lot of pressure on this water, very little change.
00:27:51.400 --> 00:27:56.300
The δ G that is pretty insignificant.
00:27:56.300 --> 00:27:59.800
Let us go ahead and do part C here.
00:27:59.800 --> 00:28:03.300
We are going to treat it the same way.
00:28:03.300 --> 00:28:10.600
We have δ G = V × δ P which is the 99.
00:28:10.600 --> 00:28:17.100
This time they gave us the density, we are going to have to do a couple of extra conversions.
00:28:17.100 --> 00:28:27.900
They said that the density was 7.9 g/ cm³, we need to find the molar volume.
00:28:27.900 --> 00:28:38.700
We are going to multiply this by iron is 55.85 g/ mol so now we have cm³/ mol.
00:28:38.700 --> 00:28:47.700
We multiply by 10⁻³ in order to, that is dm³/ 1 cm³.
00:28:47.700 --> 00:29:03.700
We can cancel that so we end up with 0.00707 dm³/ mol,
00:29:03.700 --> 00:29:08.000
which is the same as 0.00707 L.
00:29:08.000 --> 00:29:30.400
δ G is going to be 0.00707 L × 99 atm which is 100 - 1 and we end up with δ G of 0.700 L atm.
00:29:30.400 --> 00:29:36.400
We do the multiplication, multiply by a 8.314 ÷ 0.08206.
00:29:36.400 --> 00:29:44.800
We get a δ G is equal to 70.9 J/ mol, even less.
00:29:44.800 --> 00:29:56.900
A solid response to pressure increases even less.
00:29:56.900 --> 00:29:59.900
Free energy changes with volume.
00:29:59.900 --> 00:30:05.300
When you change the volume of something, the free energy available is going to change.
00:30:05.300 --> 00:30:14.000
A gas is very sensitive to volume changes, the pressure changes, volume changes which is why you have 11,500.
00:30:14.000 --> 00:30:22.600
Liquids and solids are virtually non responsive to volume changes upon when you change the pressure.
00:30:22.600 --> 00:30:25.600
You can squeeze them as much as you want, the volume is not the change that much.
00:30:25.600 --> 00:30:29.900
Because the volume does not change that much, the free energy does not change that much.
00:30:29.900 --> 00:30:33.300
It is 71 J/ mol is completely insignificant.
00:30:33.300 --> 00:30:35.200
A 100 atm is a lot.
00:30:35.200 --> 00:30:46.400
70.9 for a solid and 100 that we got for the liquid, they are actually kind of high relative to the fact that they are liquid and solids.
00:30:46.400 --> 00:30:57.200
When you are dealing with liquid and solids, most of the time you are talking about changes of like 1, 2, 3 J/mol in J not in kJ, J/ mol.
00:30:57.200 --> 00:31:04.900
Clearly, liquid and solids demonstrates that they just do not respond to pressure increases.
00:31:04.900 --> 00:31:10.200
Their volume does not change all that much, but we knew that anyway from experience.
00:31:10.200 --> 00:31:12.700
Thank you so much for joining us here at www.educator.com.
00:31:12.700 --> 00:31:16.500
We will see you next time for a continuation of some example problems in free energy.
00:31:16.500 --> 00:31:17.000
Take care, bye.