Chapter 3 U(T), H(T), and Heat Capacities (C ) Thermodynamics is a collection of useful relations between quantities, every one of which is independently measurable. What do such relations tell one about one s system, or in other words what do we learn from thermodynamics about the microscopic explanations of macroscopic changes? Nothing whatever. What then is the use of thermodynamics? Thermodynamics is useful precisely because some quantities are easier to measure than others, and that is all. - M. L. McGlashan, J. Chem. Educ., 43, 226 232 (1966) 3.1 Mathematical properties of state fxns - Review State functions are functions of state, i.e. not how they got there! They have exact derivatives and Euler s reciprocity and cyclic relations hold. 3.1.1 Euler exercise with IGL Reciprocity [E ] = () ()= 1 [ ] = [ ] Applied to our function, [ ] =[ ] [ ( 1 )] =[ ( 2 )] 2 = 2. Cyclic [E ] = 1 ; = ; = 1 1 = =( 1 )()( 2 ) =( 1 )( 1 ) =( 1 )( ( ) 1 ) = 1 3.1.2 Some useful coefficients 1. The thermal expansion coefficient is the relative change in volume upon heating at fixed pressure. It is almost always 0. 1 (3.1) 2. The coefficient of isothermal compressibility is the relative change in volume with pressure at fixed temperature. It is always 0 for stable thermodynamic states. 1 (3.2) 3. The coefficient of adiabatic compressibility is the relative change in volume with pressure at fixed entropy. It is always 0 for stable thermodynamic states. 1 (3.3) 3.1.3 General relationships 1. Lets start with E 1 = 2. Flip the terms around to solve for = ) ) 3. Make the changes relative and identify the coefficients, = 1 ) 1 ) = (3.4) This is a general relationship, true simply due to the fact that we are dealing with state functions. With this definition we can write 26
= + = 1 = 1 ( 1 ) (3.5) Like all differential expressions, one can R to get to thechangeforafinite step. = R = R R 1 This general expression can be numerically evaluated if experimental values of ( )and () areknown (for the material under study.) However, if the finite step is not too large, the coefficients can be treated as constants. 1 ln 3.2 U(T,V) As U is a state function we can write, = + = + The constant-volume heat capacity is the increment function for the energy with T (at fixed volume.) The increment function for the energy with volume (at fixed temperature) is called the internal pressure (IP).Lets derive an expression for the IP. 1. To get an expression for the IP, lets start with the energy differential, =1 = 2. We now operate on this expression with 1. = ) thus = (3.6) 3. Now, as the Helmholtz free energy is a function of state, (and therefore has an exact differential) = thus = ( ) 4. Therefore a meterable expression for IP is IP = (3.7) A relation sometimes called the 1 thermo. EoS. 3.3 Example U(T,V) Consider the total differential of the internal energy = + 1. IG: = 1 (a) = = 3 2 (b) = = ( 1 ) =0 (c) Thus ( ), i.e. T ONLY! 2. vdw: = 2 (a) = ( ) (b) = = ( ) = 2 Thus the IP = a function of the attraction only. 3. hard-sphere gas ( =00) - well you guess. 4. Liquids and solids (a) = ( ) (b) can be large but dv is usually so small as to make this term negligible. Exercise 3.3.1 Recast IP in terms of and κ 3.4 H(T,P) Now lets do what we just did for U, to H. As H is a state function we can write, = + = + 1. To get the P dependence, start with the expression for the enthalpy differential, = + 2. We now operate on this expression with 1 ) = + (3.8) 3. Using the Gibbs free energy state function, = + thus = ( ) So 4. Therefore the meterable expression is = + = + = (1 ) (3.9) A relation sometimes called the 2 thermo. EoS. As was the expression for the internal energy, this is always true. You can deduce or for any analytic EoS. However, beware of analytics. Nature does not care about our analytic EOS s. 27
3.5 Heat Capacities - The T increment functions for U & H. 3.5.1 Overview = = The last relation is written for completeness. We shall run into this form when we get to the second law. const v const p = ) = = ) = = R = R From the firstrelationsweseethatthe and are the increment functions for U(T) and H(T) at constant and, respectively and and are the increment functions for and respectively. Exercise 3.5.1 Prove that = Hint: start with the compressibility ratio and use on both the numerator and demoninator, to get the heat capacity ratio. Rearrangebycollectingtermswiththesamequantity held constant. 3.5.2 vs I will now use molar quantities both for the energy functions and the heat capacities. 1. =[ (+) ] =[ ]+ 2. We can get an expression for the first differential by realizing that ( )withanexactdifferential of, = + and =[ 1+ ] 3. Using this in the expression for, =[ + ]+ = + { [ + ]} = { [ + ]} (3.10) 4. Cases (a) {} 0 The cases for = are when the volume does not change with (at fixed ). Most materials simply expand with T at fixed P. One might think (as the authors of the text do) that can be less than if the material contracts on heating. Water is such a case as it has a maximum density (minimum molar volume) at 4 However, with a little more analysis, we will see that the RHS is never (b) {} = Thus the extra cost of heating an IG at constant is (c) The last differential [ ]( IP) is zero for an ideal gas. The energy of an ideal gas ONLY depends on the temperature all partials at constant T equal zero (for an ideal/perfect gas.) 5. Using the result for the IP, = [ + ] = = ª= [ ( ] = 2 (3.11) 6. The RHS is a product of readily measurable quantities. Just as important, now we see that the thermal expansion coefficient is squared and thus its sign is irrelevant. (As we will prove later, the compressibilities, of either flavor, are strictly positive - for a stable thermodynamic state.) 7. Working through the enthalpy we can get another expression for =[ ( ) ] = ) Therefore, (a) = [ ] Using the chain rule, = [ ] = [ ] Now we can use Euler s cyclic relation, (b) = ) 1 [] = { [ ]} (3.12) (c) As before {} 0 (d) The last differential, [ ], is zero for an IG. (e) {} = The extra cost of heating an IG is 28
3.5.3 Heat Capacity Data 1. data for gases are smoothly varying (usually increasing) functions of and are often fit by polynomials. The reason to choose polynomials is two fold. First, they can be used to describe any smoothly varying function. Second, they are easily integrated. The difference is close to for real gases but the actual difference can be calculated with knowledge of the EoS. (a) Monatomic gases have: 3 2 (b) Diatomic gases have 5 2 at low and 7 2 at high. (c) Linear triatomics have 5 2 at low while non-linear triatomics have 6 2 at low. The heat capacities of both types of molecules increases with. (d) 1 5 3 complex(many) simple(few DoF) 2. (but )forliquids. Thetrendsare complex and no general prescription can be offered. However a few general comments can be made. The T dependence is usually rather weak and the values (of the molar heat capacity) of Hy-bonded liquids can be very large. (We will explain why when we discuss SM.) 3.6 u(t) and h(t) As and are the increment functions for and with T, they tell you how to increment or decrement values of these energy functions given in tables (at the standard T = 298 C) to get the values appropriate at any other temperature. Specifically, for, is the rate of increase of with temperature: = = ( ).Integrating, = R ( ) Where I have chosen to remind you that the heat capacity is a temperature dependent quantity. 3. for solids. Theformof( ) is universal if plotted using a reduced temperature scale (dividing by a material specific number.)einsteinanddebye explained this universal function. The following comments help to understand the dependence. (a) The high temperature limit for all solids is = 3, a universal result known by the middle of the 1800s and is called the Law of Dulong and Petit. (b) At low temperature the heat capacity scales as 3 This is a result of the turning on of crystal vibrations with increasing temperature. (All of them are turned on at high and the 3 result can be understood by just counting the number of vibrational modes. See SM.) (c) The scaling variable for q the (to make the plot universal) is Θ where is the mass of the atoms and is the atom-atom bond strength. Do you have an inkling about why there is a universal shape? How do you think you calculate how a reaction enthalpy changes with temperature? 3.7 Joule-Thomson effect 1. The Joule-Thomson coefficient is measured by an throttled expansion experiment. It can also be extracted from the EoS. The experimental apparatus is shown below. Gas moves from left to right changing both pressure and temperature. The gas moves slowly through a porous plug 29
and in an enclosure that prevents heat exchange with the environment. Why do we care about this experiment? As we will see =0but displays a rich functional dependence for real gases. This dependence reports on the interplay (as a fxn of and ) between attractive and repulsive terms in the intermolecular potential. On a more practical level, this is how many gases can be cooled and is in fact the method of choice for cooling N 2 and creating LN. 2. Why is = 0 in this experiment? Consider a slug of gas going from Left to Right. Thechangeofenergyofthisslugis, 2 1 = work on the gas slug = 1 + 2 = 1 1 2 2 = 1 1 2 2 (Recall that the volumes of the slug completely disappears on the left and appears on the right, so the = ) 2 1 =() 1 () 2. (Remember PV has the units of energy.) = 0 as the terms inside the [...] cancel. VDW: ( + )( ) = 2 Now differentiate wrt T holding constant. 2 3 ( )+( + ) 2 = We can plug in for ( + )andsolvefor, 2 = { 2( ) } 3 Therefore we find for a vdw gas = 1 [ ] { 2( ) 3 } Consider the limits: a=b 0 =0andjust a 0 = 0 One often speaks of the (upper) inversion line where, =0 After some algebra, 2 =2( ) 2 An interesting fairly general result is that the inversion temperature is about twice the Boyle temperature, 2 1 The cooling and ultimate liquefaction of gases is done by using the Joule-Thompson (isoenthalpic) expansion inside the loop, i.e. 0 Since one is cooling from a higher temperature, the critical issue is to start the expansion below the upper inversion temperature. 2 +() 2 = 1 +() 1 2 = 1. 3. Now we can go after the coefficient. = ( = 1 ) What does equal? = + = + Using the fact that has an exact differential,... well you make a substitution for the differential and seeifyoucanprovewhatyouwillsooncometo know and love as = + therefore, ANALYSIS: For decreasing pressure P& Inside: %%& & Outside:%&% % = 1 [ ] (3.13) 4. NowletsuseEOS stoget. IDEAL: 1 T is the temperature where the effects of attraction (vdw a ) and replusion (vdw b ) exactly cancel, yielding an ideal behavior for a gas that is not ideal at other temperatures. See chapter on gases. 30