WEBVTT chemistry/physical-chemistry/hovasapian
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Hello and welcome to www.educator.com and welcome back to Physical Chemistry.
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In the previous lesson, we discussed entropy as a function of temperature and volume but in this lesson we are going to do the same thing
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and we are going to discuss entropy as a function of temperature and pressure.
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Let us get started.
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Let us take entropy as a function of temperature and pressure and let us write our total differential expression
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which is going to be DS = DS DT at constant pressure + DS DP at constant temperature.
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This is going to be × DT and this is going to be × DP.
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Let us recall the fundamental equation.
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The fundamental equation is DS = 1/ T DU + P/ T DV.
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Let us recall the definition of enthalpy is H = U + PV.
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Therefore, I will go ahead and move this around and it becomes U = H – PV.
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Therefore, the differential DU = DH, we have this U = H – PV.
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I’m going to differentiate this expression so I get DU = DH or when I differentiate this is the product rule.
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This is going to be – P DV – V DP.
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I'm going to put this expression for DU into the fundamental equation.
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I’m going to put this for that and end up with the following.
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I'm going to end up with DS = 1/ T × DH – P DV – V DP + P/ T DV.
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I’m going to multiply it out DS = DH/ T - P/ T DV - V/ T DP + P / T DV – P/ T DV + P / T DV these terms cancel and I'm left with the DS = DH/ T - V/ T DP.
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This is just another version of the fundamental equation.
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I just manipulated with respect to enthalpy and energy.
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U was here and it also shows up in the definition of enthalpy so this is just a version of
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the fundamental equation of thermodynamics involving enthalpy instead of energy.
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This is a fundamental equation of thermodynamics.
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This is just the enthalpy version of it, that is all.
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There we go, in other words it expresses the change in entropy in terms of the changing of enthalpy and the change in pressure.
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From our discussion of energy we had the following, it was the constant pressure version of the DU.
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Let me go ahead and do this in red, we had DH is equal to CP DT + DH DP under constant T DP.
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Let me go back to black.
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I have this DH is this one so I'm going to end up putting this expression or I see DH in this expression and we get the following.
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We get DS = 1/ T × CP DT + DH DP constant T DP - V/ T DP.
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I get the DS = CP/ T DT + DH DP T DP - V/ T DP.
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DP and DP I'm going to combine some terms here so when I do that I end up with the following.
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I end up with DS = CP/ T DT + 1/ T × DH DP T - V DP.
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Let us go ahead and rewrite the differential.
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Rewrite the general total differential expression of S = S(T and P).
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I get DS = DS DT at constant P DT + DS DP at constant T DP.
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Notice both of these equations, I will call this one equation 1 and I will call this one equation 2.
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Both of these equations express entropy as a function of temperature and pressure DS temperature pressure,
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that means that this is identifiable with that and this is identifiable that.
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We are doing the same exact thing we did last time except now in terms of temperature and pressure.
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We have the following, we have DS DT under constant P = CP/ T and we have DS DP sub T = 1/ T × DH DP T – V.
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Once again, CP/ T is always positive.
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If I hold the pressure constant and raise the temperature, this is positive and this is positive that means if this is positive,
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if I hold the pressure constant and raise the temperature I raise the entropy.
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We already know it.
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At constant P, I will just say a rise in temperature implies a rise in the entropy.
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The temperature dependence on the entropy again is very simple, it is just the appropriate heat capacity
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in this case, constant pressure heat capacity divided by the temperature.
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Pressure dependence not so simple.
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Again, we have to form our mixed partials.
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Temperature dependence is simple, that is just this.
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Pressure dependence not so simple.
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For pressure dependence, we are going to do the same thing.
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We are going to form DDP of DS DT and we are going to form DDT of DS DP.
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These are mixed partials when I take the derivative of this, the derivative of this we set equal to each other we are going to find the relationship.
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Let us go ahead and go back to black for our purposes here.
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For DS DT under constant pressure equal to CP/ T we are going to differentiate with respect to P and we are going to hold T constant.
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When I do that, I get the following.
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Differentiate with respect to P so this is going to be DDP of DS DT of P and it is going to equal 1/ T × the derivative of the CP
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with respect to P which = the constant pressure heat capacity so DDP, by definition at constant pressure heat capacity was just DH DT.
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Therefore, D² S DP DT, DP DT = D² H DP DT.
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1/ T, sorry I forgot the 1/ T right there.
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1/ T DP DT.
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We have that one.
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For the other one, for our DS/ DP under constant T which we set = 1/ T × DH DP T - V
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this time we are going to differentiate with respect to T and we are going to hold P constant.
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This end up looking like also this equation, we are going to differentiate this with respect to T holding P constant.
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Here we differentiate with respect to P holding P constant or we are going to set it equal to each other to see what we get.
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We get D DT of DS DP T = this × the derivative of that + that × the derivative of this.
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We get 1/ T , D DT DH DP sub T - DV DT constant pressure -1 / T² × DH DP sub T – V.
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We get D² S/ DT DP = 1/ T × D² H DT DP.
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I will put this together - 1/ T DV DT P- 1 / T² × DH/ DP constant T – V.
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It is just mathematical manipulation.
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I will do this in red, D² S DT DP= D² S DP DT.
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I'm going to set this equal to this and when I do that I get the following.
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I get 1/ T D² H DP DT = 1/ T D² H DT DP -1/ T × DV DT sub P -1/ T² × DH DP sub T – V.
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D² H DP DT D² H DT DP, this cancel because mixed partial for enthalpy are the same.
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I'm going to move one of those over and I end up with the following.
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1/ T² DH DP sub T - V = - 1/ T × DV DT/ P.
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I'm going to cancel one of the T's so I going to end up with 1/ T × DH DP sub T - V = DV DT sub P.
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This thing right here, this thing happens to equal DS DP at constant T.
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We have DS DP at constant T, one of the differential coefficients that we wanted, we want to know the pressure dependence of entropy
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is actually equal to something that is easily measurable is equal to the relative is equal to
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the change in the derivative of volume with respect temperature DV DT at constant pressure.
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This is what we were looking for, the differential coefficient or expressing it in terms of something that is easily measurable.
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How much does the volume change when I change the temperature at constant pressure?
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Recall that A, the coefficient of thermal expansion = 1/ V DV DT P that is this thing right here.
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Therefore, V A = DV DT P.
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Therefore, wherever I see DV DT sub P I'm going to put that right there.
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Our final relation is pressure dependence of the entropy at constant temperature = - V/ A.
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This is the other differential coefficient, this is what we wanted.
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This equation right here is the pressure dependents of the entropy at constant temperature, in terms of very easily measurable quantities the volume
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and the A the coefficient of thermal expansion or final equation.
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The one that we want to memorize is the following.
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With respect to temperature and pressure, instead of temperature and volume, let me go back to black.
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I have DS = CP/ T DT – V A DP under conditions when I’m changing the temperature and the pressure
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and I want to know what the entropy change is, this is the equation that I use.
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If pressure is held constant it goes to 0.
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If temperature is held constant this goes to 0.
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Let us talk about the pressure dependence of the entropy of liquid and solids.
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The pressure dependents of entropy for liquids and solids.
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For liquids and solids, it is going to be the same thing as it was for volume.
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For liquids and solids, δ S with respect to pressure is very small.
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In effect, it is so small that you can pretty much ignore it.
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It is very small so we usually ignore it, in other words we are dealing with a system which happens to be liquid or solid
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and we change the pressure on it from let us say 1 atm to 10 atm, the entropy change
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that is contributed by the pressure change is just too small to be to be significant.
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Just ignore it.
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What that means is that this thing goes to 0.
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We just take the 0 and we just use this DS=CP/ T DT.
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In this particular case, let us see how small, let us actually get an example of this.
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Let us see how small.
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We have at constant temperature.
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I actually want to start with it again and do it the way we are supposed to do it.
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DS = CP/ T DT - V A DP.
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This time we want to see how much the entropy changes when I hold the temperature constant and when I change the pressure.
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I just want a sense of a how a thing is held constant.
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If I change the pressure how much does the entropy change?
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I want to see how small it is so we are going to be holding temperature constant so this is going to go to 0.
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This is the equation that I used, therefore, DS = -V A DP.
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I can integrate this so when I integrate this and that, this becomes δ S - V A that is the constant.
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It just becomes - V A × the integral of DP and I end up with δ S = - V A DP.
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If I want to know the change in entropy is when I change the pressure on a liquid or a solid, I take the volume
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and I multiply by its coefficient of thermal expansion then multiply by the change in pressure.
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This is the equation that I'm going to take a look at.
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For solids, A is approximately equal to 10⁻⁴.
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It is on the order of 10⁻⁴.
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For liquids, A is on the order of 10⁻³/°K.
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Let us go ahead and examine what happens when we look at a particular amount of liquid.
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We are going to take a look at 100 g of H2O when the pressure goes from 1 atm to 2 atm.
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If I change the pressure from 1 to 2 atm, I double the pressure what is the entropy change going to be in that 100 g of water?
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Let us see what we got, we have 100 g of water so let us go ahead and find out how many moles that is.
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That is going to equal 5.56 moles so our molar volume we have 100 g of water which is 100 ml/ 5.56 moles that is going to give me 18 ml/mol.
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This is the molar volume or it is going to be 0.018 L/ mol.
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Basically, 1 mol of liquid water occupies a volume of 0.018 L and we are going to need that.
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Δ S = our volume negative volume × A × the change in pressure where δ S = -0.018 L/ mol.
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We said that for a liquid, the A is an order of 10⁻³/°K.
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Our δ P is going to be final - initial so it was going to be 2 atm -1 atm.
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When I do this calculation, I get the δ S = -1.8 × 10⁻⁵ L/ atm /mol °K.
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Let us go ahead and convert this to Joules.
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This is going to be 8.314 J/ L/ atm.
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L/ atm and L/ atm cancels L/ atm and I'm left with an amount of -0.0018 J /mol-°K.
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Let me write that over here δ S= -0.0018 J /mol-°K.
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I double the pressure on something on 100 g of water and entropy only changes by 0.0018 J.
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This is very small.
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For liquid and solids, unless the change in pressure is massive and I do mean massive,
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you can pretty much ignore the second term on the equation DS = CP/ T DT – V A DP.
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The pressure dependence of the entropy for a liquid or a solid it does not even matter.
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It is the temperature that is going to have the deepest affect, the largest effect for a particular liquid or solid.
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I just want you to see how small it is.
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Let us go ahead with an example.
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1 mol of solid gas is raised from 25°C to 125°C at constant pressure.
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Its constant pressure molar heat capacity is 23.7 + 0.0052 T so this is temperature dependent.
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As the temperature changes that heat capacity changes so it is not constant anymore.
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What is the δ S for this transformation?
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Molar heat capacity is this.
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Let us start with our general equation which is this.
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We always want to start with our general and let the problem tell us what the constraints are so we can pair down the general equation if we need to.
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We are going to have CP/ T DT - V A DP.
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What is constant pressure?
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Constant pressure is if pressure does not change that means DP 0 so this term goes away.
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We are left with DS = CP/ T DT.
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The next thing we want to do is we want to integrate it because we want the finite change.
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The δ S is going to equal the integral from temperature 1 to temperature 2 of the CP/ T DT.
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Let us go ahead, this is molar heat capacity.
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Remember molar heat capacity is this, it is CP/ n so that is equal to 23.7 + 0.0052 T.
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We have to multiply by n so the actual the heat capacity itself is going to be 23.7 n + 0.0052 n × T.
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We are dealing with 1 mol this n is just 1.
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For all practical purposes I can either include it just put 1 in the computation or I can just know that I’m dealing with 1.
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I’m dealing with that heat capacity.
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It was very important in these problems.
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If they are talking about molar heat capacity, it is CP/ n.
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If they just say heat capacity it is already included in the n.
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Be very careful about that.
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This is going to equal, we are taking it from 25° to 125° that is going to be 298 to 398 and this is going to be 23.7 + 0.0052 T DT.
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The δ S is going to equal 23.7 × the integral of 298 to 398.
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This is over T because CP/ T so this is going to be DT/ T + the second integral which is this one.
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This was going to be 0.0052 T/ T, the T's are going to cancel so the 0.0052 was going to come out of the integral 298 to 398 of just DT.
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The δ S is going to equal this one was going to be 23.7 × the nat log of 398/ 298 + 0.0052 × 398 -298 just δ T.
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When I do this calculation, I end up with 6.86 + 0.52 I get δ S = 7.38 J/ mol-°K because we are dealing with 1 mol.
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It is that simple, you start off with your general equation, you take a look at what the constraint is, in this particular case pressure is constant so it goes to 0.
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We are left with this part of the equation, we integrate it, we put in our values,
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we check to see that we are dealing with molar heat capacity and see how many moles.
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As I take 1 mol of solid gold and I raise it from 25° to 125°, I'm changing the entropy, I'm raising the entropy 7.38 J /mol-°K.
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Thank you so much for joining us here at www.educator.com.
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We will see you next time, bye.