dg = V dp - S dt (1.1) 2) There are two T ds equations that are useful in the analysis of thermodynamic systems. The first of these

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1 CHM 3410 Problem Set 5 Due date: Wednesday, October 7 th Do all of the following problems. Show your work. "Entropy never sleeps." - Anonymous 1) Starting with the relationship dg = V dp - S dt (1.1) derive the following. a) ( G/ p) T = V (1.2) b) ( G/ T) p = - S (1.3) c) ( V/ T) p = - ( S/ p) T (1.4) 2) There are two T ds equations that are useful in the analysis of thermodynamic systems. The first of these T ds = C p dt - TV dp (2.1) was derived in class. a) Starting with S S(V,T) (that is, writing entropy as a function of V and T) derive the second T ds equation T ds = T ( p/ T) V dv + C V dt (2.2) b) Use eq 2.2 to find q when the volume of one mole of a gas obeying the van der Waals equation of state is changed isothermally and reversibly from an initial value V i to a final value V f. 3) There are several general relationships for the difference between C p and C V. show that a) Starting with the relationship C p - C V = ( H/ T) p - ( U/ T) V (3.1) C p - C V = 2 VT (3.2) T where = (1/V) ( V/ T) p (3.3) and T = - (1/V) ( V/ p) T (3.4) (NOTE: This is a difficult problem, and so this part of the problem will not be part of your homework grade on this assignment.) b) Use eq 3.2 to find an expression for C p - C V for an ideal gas, and verify that the result is as expected. c) Use eq 3.2 and the data given below to find C p,m - C V,m for iron metal (Fe(s)) at T = 20. C. Give your answer in units of J/mol K. Substance M C p,m T (g/mol) (J/mol K) (g/cm 3 ) (K -1 ) (atm -1 ) Fe(s) x x 10-6

2 4) The vapor pressure for isopropanol (C 3H 8O(l)) for several different temperatures is given below. Based on this data, find the following: a) H vap, the enthalpy of vaporization for isopropanol. b) T vap, the normal boiling point for isopropanol. T (K) p vap (torr) T (K) p vap (torr) ) At room temperature and pressure the graphite form of carbon is thermodynamically more stable than the diamond form of carbon. However, at large values of applied pressure the diamond form becomes more stable. Using the information given below, estimate the pressure at which the diamond form of carbon becomes the thermodynamically more stable than the graphite form of carbon. Assume T = 298. K, and also assume that both the density and the coefficient of thermal expansion values for both forms of carbon are independent of temperature. Give your final pressure in units of atm. graphite diamond G f (kj/mol) (g/cm 3 ) T (atm -1 ) 3.04 x x ) A schematic phase diagram for the element sulfur is given below. Based on the information in this phase diagram, place the four phases of sulfur, in order, from most dense to least dense. Note that there are two solid forms of sulfur in the phase diagram, rhombic and monoclinic.

3 Solutions. 1) dg = V dp - S dt divide by dp dg/dp = V dp/dp - S dt/dp = V - S dt/dp ( G/ p) T = V - S ( T/ p) T = V Make T constant The last term drops out when T = constant dg = V dp - S dt dg/dt = V dp/dt - S dt/dt = V dp/dt - S ( G/ T) p = V ( p/ T) p - S = - S divide by dt Make p constant The last term drops out when T = constant Now, take / T) p of the first expression above and / p) T of the second expression above / T) p [ ( G/ p) T = V ] = ( 2 G/ T p) = ( V/ T) p / p) T [ ( G/ T) p = - S ] = ( 2 G/ p T) = - ( S/ p) T For "well behaved" functions the order of taking the derivatives doesn't matter, and so the left sides of the above expressions are equal. Therefore the right sides must also be equal, and so ( V/ T) p = - ( S/ p) T 2) If we write entropy as a function of volume and temperature, that is S S(V,T), then ds = ( S/ V) T dv + ( S/ T) V dt Multiply both sides by T T ds = T ( S/ V) T dv + T ( S/ T) V dt But (from the Chapter 3 handout) ( S/ V) T = ( p/ T) V Substitution of the above gives T = ( U/ S) V TdS = T ( p/ T) V dv + ( U/ S) V ( S/ T) V dt = T ( p/ T) V dv + ( U/ T) V dt = T ( p/ T) V dv + C V dt b) For an isothermal reversible expansion dt = 0, and so đq = T ds = T ( p/ T) V dv This is because đq rev = T ds For a van der Waals gas p = nrt - a n 2 (V - nb) V 2

4 So ( p/ T) V = nr (V - nb) Integrating both sides gives us f f i đq = i nrt/(v - nb) q = nrt ln[ (V f - nb)/(v i - nb) ] As a check, note that for b = 0 we get the result for an isothermal reversible expansion of an ideal gas. 3) a) We wish to prove the following C p - C V = 2 VT = - [ (1/V) ( V/ T) p ] 2 VT = - T ( p/ V) T ( V/ T) p ( V/ T) p T [ - (1/V) ( V/ p) T ] I have rewritten the right hand side in terms of partial derivatives so that I have an idea of what I am aiming for. So C p C V = ( H/ T) p ( U/ T) V (from the partial derivatives for C p and C V) Since Then U = H pv ( U/ T) V = ( H/ T) V ( (pv)/ T) V = ( H/ T) V V( p/ T) V (substitute for U, take V out of second term) So C p C V = ( H/ T) p ( H/ T) V + V( p/ T) V But ( H/ T) V = ( H/ T) p + ( H/ p) T( p/ T) V (partial derivative relationship, Handout 1) Substituting into the expression for C p C V gives C p C V = ( H/ T) p [( H/ T) p + ( H/ p) T( p/ T) V ]+ V( p/ T) V = - ( H/ p) T( p/ T) V + V( p/ T) V But ( H/ p) T = ( H/ p) S + ( H/ S) p( S/ p) T (partial derivative relationship, Handout 1) So C p C V = - [( H/ p) S + ( H/ S) p( S/ p) T]( p/ T) V + V( p/ T) V = - [ V + T( S/ p) T ]( p/ T) V + V( p/ T) V ( ( H/ p) S = V; ( H/ S) p = T, Handout 3) = - T( S/ p) T ( p/ T) V (first and last terms cancel) = T( V/ T) p ( p/ T) V ( ( S/ p) T = - ( V/ T) p, Handout 3) But ( p/ T) V( T/ V) p( V/ p) T = - 1 (partial derivative relationship, Handout 1) ( p/ T) V = - [( V/ T) p/( V/ p) T ] And so C p C V = - T( V/ T) p ( V/ T) p ( V/ p) T

5 If we multiply by 1 = [(1/V) 2 /(1/V) 2 ] we get our final answer C p C V = T(1/V) 2 ( V/ T) p ( V/ T) p = 2 VT (rearrange, use definitions for and T) (1/V) [ - (1/V) ( V/ p) T ] T This is not an easy derivation. However, the point of doing this is to show how one can start with a general relationship like C p C V and come up with an expression entirely in terms of things that can be measured in the laboratory. b) We can test the above result by applying it to an ideal gas. For such a gas, V = nrt/p, and so = (1/V) ( V/ T) p = (1/V) (nr/p) = (nr/pv) = (1/T) T = - (1/V) ( V/ p) T = - (1/V) (- nrt/p 2 ) = (nrt/p 2 V) = (1/p) Substituting into the expression obtained in part a (eq 3.2) we get C p - C V = (1/T) 2 VT = nr, as expected (1/p) c) To make things easier data above can be converted into MKS units, so that the final answer for C p - C V will also come out in MKS units (J/mol K). So V m = V/n = M/ = (55.85 g/mol)/(7.874 g/cm 3 ) = cm 3 /mol = x 10-6 m 3 /mol T = x 10-6 atm -1 (1 atm/ x 10 5 N/m 2 ) = x m 2 /N T = 20.0 C = K. The final result is then obtained by substituting into eq 3.2. C p C V = (0.354 x 10-4 K -1 ) 2 (7.093 x 10-6 m 3 /mol)(293.2 K) = J/mol K (5.813 x m 2 /N) 4) a) We organize the data below T (K) p (torr) 1/T (K -1 ) ln p The data are plotted on the next page. The slope and intercept of the best fitting line to the data are slope = K, intercept = Therefore H vap = - mr = - ( K)( J/mol K) = kj/mol

6 Plot of ln(p) vs (1/T) ln(p) (1/T) (K-1) b) To find the normal boiling point we solve our equation for the best fitting line for (1/T), and then insert p = 1.0 atm = 760. torr into the equation. ln(p) = m/t + b (1/T) = (1/m) [ ln(p) - b ] = - (1/ K -1 ) [ ln(760) ] = K -1 So T vap = 1/( K-1) = K = 85.7 C 5) There is an easier (and more approximate) and a more difficult (but more exact) way of doing this problem. Both are based on using the relationship ( G/ p) T = V For the phase transition C(g) C(d) it follows that ( ( G)/ p) T = V If we assume that we have one mole of substance, then for p = 1.00 bar V g = (12.01 g/mol) = cm 3 /mol = x 10-6 m 3 /mol (2.260 g/cm 3 ) V d = (12.01 g/mol) = cm 3 /mol = x 10-6 m 3 /mol (3.513 g/cm 3 ) V = (3.419 x x 10-6 ) cm 3 /mol = x 10-6 m 3 /mol

7 The simplest approximation to make is to assume V is constant. Then ( ( G)/ p) T = V dp = d( G pt)/( V) If we treat V as being independent of pressure, then p = ( G pt)/( V) Now, at p = 1.00 bar, G pt = kj/mol kj/mol = kj/mol The diamond and graphite phases will be in equilibrium when G pt = 0.00 Therefore ( G) = G f - G i = kj/mol kj/mol = kj/mol So p = ( J/mol) = x 10 9 Pa 1 atm = atm ( x 10-6 m 3 /mol) x 10 5 Pa Since p = p f - p i, and p f >> p i, p f p = atm What does this mean? Well, at p = atm, G pt for the graphite diamond phase transition goes from being positive (not spontaneous) to zero (equilibrium). Therefore, for p > atm, G pt becomes negative, and so diamond becomes thermodynamically more stable than graphite (the phase transition graphite diamond becomes spontaneous). We can redo the above in a slightly more precise manner by taking into account the isothermal compressibility of graphite and diamond. Since T = - (1/V)( V/ p) T dv/v = - T dp The left side of this equation is equal to d(ln V), which is inconvenient to work with. Since the volume changes are relatively small, we can say V V 0, the volume at p = 1.00 bar. Then dv = - V 0 T dp Integrating gives V - V 0 = - V 0 T p or V = V 0 - V 0 T p

8 So V = V d - V g = [ V 0,d - V 0,d T,d p ] - [ V 0,g - V 0,g T,g p ] = a + bp a = V 0,d - V 0,g = x 10-6 m 3 /mol b = [V 0,g T,g - V 0,d T,d ] = (5.314 x 10-6 m 3 /mol)(3.04 x 10-6 atm -1 ) - (3.419 x 10-6 m 3 /mol)(0.187 x 10-6 atm -1 ) = 1.55 x m 3 /atm mol (1 atm/ x 10 5 Pa) = 1.53 x m 3 /Pa mol Since d( G pt) = V dp = (a + bp) dp ( G pt) = ap + (b/2)p 2 or (b/2)p 2 + ap - ( G pt) = 0 where ( G pt) = J/mol (7.65 x )p x 10-6 p = 0 Good luck in solving this quadratic using the quadratic formula. We will use "trial and error" by guessing values for p and finding the value of the quadratic function. A correct guess for p will give a value of zero for the function. guess value of function 1.50 x x x x Interpolating gives p = x 10 9 Pa 1 atm = atm (about 7% larger than our previous answer) x 10 5 Pa A lot of work! 6) When dp/dt > 0 then the initial phase is more dense than the final phase, and when dp/dt < 0 the final phase is more dense than the initial phase. If we look at the phase boundaries, we see rhom > gas rhom > mono rhom > liq mono > gas mono > liq liq > gas So the phases, in order of density, are (rhom) > (mono) > (liq) > (gas)

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