ln( P vap(s) / torr) = T / K ln( P vap(l) / torr) = T / K

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Lionel Quinn
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1 Chem 4501 Introduction to Thermodynamics, 3 Credits Kinetics, and Statistical Mechanics Fall Semester 2017 Homework Problem Set Number 9 Solutions 1. McQuarrie and Simon, 9-4. Paraphrase: Given expressions for the liquid and solid vapor pressures of Cl 2, what is the temperature and what is the pressure for Cl 2 at the triple point? We are provided expressions for the temperature-dependent vapor pressures of solid and liquid Cl2 of ln( P vap(s) / torr) = T / K ln( P vap(l) / torr) = T / K At the triple point, solid and liquid and vapor are all in equilibrium with one another, implying that the vapor pressure above the solid must be identical to that above the liquid (it s all one vapor ) so that we may equate the two expressions above and solve for T. Doing so provides a value of K. Using that value in either vapor pressure expression provides Pvap(tp)=11.1 torr. 2. McQuarrie and Simon, 9-6. Paraphrase: Given a dependence of methane melting pressure on temperature and (T,P) values of K and bar at the triple point, what is the melting pressure of methane at 300 K. Along the (T,P) vapor/liquid coexistence curve (melting curve) for methane, we are told that the slope of the curve varies as

2 dp dt = ( bar)t 0.85 / K 1.85 Given an initial value pressure value of bar at K, the value at 300 K may be computed by integrating over the given temperature range, i.e., P f & 300 % dp( bar = T 0.85 & % dt ( bar K 1.85 ' ' 300 K & P f % 1.85 T1.85 ( bar K K' ( ) bar = [ ( )] bar P f = = bar Notice that it takes vastly more pressure to liquefy methane at 300 K compared to K. 3. McQuarrie and Simon, Paraphrase: Consider an isolated set of two rigid boxes that permit material, but not heat, flow between the two. How will the chemical potentials in the two compartments dictate the direction of material flow? The implications of the problem set up are that dvj = 0 for j = 1, 2 (rigid boxes) and duj = 0 for j = 1, 2 (dv = 0 implies δw = 0 and no heat flow implies δq = 0, hence du = 0). Simple mass balance requires that n1 + n2 must be a constant. The next part of the question asks about chemical potentials, which are related to the Gibbs free energy as µdn = ( dg) P,T

3 Expanding dg neglecting terms in dp and dt (based on the definition above holding P and T constant, we have µdn = du TdS + PdV but, the nature of the system has du and dv also equal to zero, so that the net result is µdn = TdS In which case, considering the net entropy change over both boxes we have ds = ds 1 + ds 2 = µ 1 T dn 1 µ 2 T dn 2 = 1 ( T µ 2 µ 1)dn 1 So, since the system is isolated, ds must be non-negative, implying that if µ2 > µ1, then dn1 must be positive. That is, if the chemical potential in box 1 is lower than in box 2, material will flow into box 1. Similarly, if µ2 < µ1, then dn1 must be negative, implying that if the chemical potential in box 1 is higher than in box 2, material will flow out of box 1. Lastly, when the two chemical potentials are equal to one another (inter-box equilibrium) there will be no net flow. 4. McQuarrie and Simon, Paraphrase: Determine the boiling point of water at 2 atm given the densities of liquid and gaseous water and the molar enthalpy of vaporization at K. We are asked first to determine dt/dp for water at K given a molar enthalpy of vaporization of kj mol 1 and

4 liquid and vapor densities of g ml 1 and g L 1 at 1 atm (notice the difference in per ml and per L the gas is clearly much, much less dense, as expected). Given a molecular weight for water of g mol 1, that makes the molar volumes for liquid and gaseous water equal to and L mol 1, or ΔvapV = = L mol 1. This value can be used in the Clapeyron equation (in inverse form) to determine dt dp = TΔ vapv Δ vap H ( ) kj mol 1 ( K) L mol 1 = = 27.9 Katm kj Latm Thus, if we want to know the boiling point at 2 atm, we would add 27.9 K to that at 1 atm, resulting in an estimate of K. 5. McQuarrie and Simon, Paraphrase: Given the molar enthalpy of vaporization of water at its normal boiling point, what is the vapor pressure predicted for water at 110 C from the Clausius-Clapeyron equation? The Clausius-Clapeyron equation provides ln P % 2 ' = Δ vap H & R P 1 T 2 T 1 T 1 T 2 % ' & plugging in the appropriate values for all knowns gives

5 P ln 2 % kj mol 1 ) 10 K ' = 1atm& J mol 1 K 1 + * K = ( )( K) P 2 / atm = e =1.408 atm =1070 torr This result is in close agreement with the experimental value of 1075 torr. 6. McQuarrie and Simon, Paraphrase: Determine the molar enthalpy of vaporization for lead (gulp) given some (T,P) data. The (T/K, P/torr) data points provided are (1500,19.72), (1600,48.48), (1700,107.2), (1800,217.7), (1900,408.2). If one determines the slope from a linear regression of lnp on 1/T, one obtains a value of and an intercept of That is, based on the form of the Clausius-Clapeyron equation given in the book as ln P = Δ vaph ' & % R ( ) 1 ' & % T ) + C ( = ' & ) % T ( with P in torr and T in K. Noting that the P we are interested in is 1 atm or 760 torr, and solving for T gives a T value of 2010 K. The enthalpy of vaporization is trivially computed by multiplying the slope by R in units of kj mol 1 K 1 ( ) to derive a value of kj mol 1., McQuarrie and Simon, Paraphrase: Determine how the vapor pressure of a spherical droplet differs from that of a flat surface as a function of droplet radius and surface tension.

6 The transfer of an ideal gas from a pressure P1 to another P2 involves a change in Gibbs free energy according to ΔG = nrt ln P & 2 % ( ' So, if the vapor pressure above the droplet is P and that above the surface is P0, then the transfer of an infinitesimal amount from surface to droplet would be P 1 dg = dnrt ln P % ' P 0 & It s worth thinking about signs to be sure that this is right. It says that the process should be spontaneous (dg < 0) if P0 > P. That is, if the vapor pressure over the surface is larger than over the droplet, then water should want to move from the surface to the droplet (easier to vaporize over the surface and easier to condense over the droplet). OK, that seems to check. Over an infinite surface, the change in energy from loss of dn moles is essentially zero, but over the droplet, the change is related to the surface tension (units of energy per area) according to dg = γd( Area) ( ) = γd 4πr 2 = 8πγrdr In order to set this equal to dg from the earlier expression, we need a relationship between dn and r. The number of moles n

7 times the molar volume gives the total volume, which for a sphere is equal to (4/3)πr 3. Thus dnv = d 4 & % 3 πr3 ( ' = 4πr 2 dr Based on these results we can write 8πγrdr = 4πr2 dr V RT ln P ' & ) % P 0 ( or, simplifying ln P % ' = 2γV P 0 & rrt So, what does this mean? Well, every quantity on the righthand-side is a positive constant, so the argument of the logarithm must be greater than 1, implying P, the vapor above a droplet, must be greater than P0, the vapor above a flat surface. This holds true unless r approaches infinity, at which point the two vapor pressures must be equal to one another so that the logarithm can equal 0. This is consistent a droplet of infinite radius loses its curvature and becomes, for all practical purposes, flat. So, droplets spontaneously evaporate to increase the size of bulk liquids.

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