Exam 3, Chemistry 481, 8 December 2017

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1 1 Exam 3, Chemistry 481, 8 December 2017 Show all work for full credit Useful constants: k B = J K 1 ; R (molar gas constant) = J K 1 mol 1 Helmholz free energy: A = U S, so that da Sd d Gibbs free energy: G = H S, so that dg Sd +d 1. One Maxwell relation is ( ) (1) a. Derive, in the case where and are the independent variables, the corresponding expression for ( S/). (12 pts) Since and are the independent variables, solution starts from the equation da Sd d. Comparing this with the general mathematical expression for the total differential da = ( A/ ) d + ( A/) d, we see S ( A/ ) and ( A/). Since the mixed 2nd derivatives of A with respect to and are identical, it follows that ( S/) = ( / ) Another, more circuitous, derivation is as follow: By the chain rule = ( ) (2) Now the cyclic rule is ( ) ( ) ( ) 1 which can be rearranged to give ( ) Substituting this into Eq. (2) gives ( ) ( ) ( ) ( ) We can now introduce the Maxwell relation given in the statement of this exam question

2 2 [( S/ ) (/ ) ] to show that ( ) ( ) ( ) = + he first and the second terms on the right hand side are reciprocals of each other, and hence cancel each other out, with the result that which is the desired answer. = ( ) b. We showed that ( ) U + ( ) Starting from the definition G = H S, obtain a similar expression for ( H/ ) in terms of the variables,, and. (13 pts) We know G = H S. Differentiation w/ respect to at constant gives ( G/ ) = ( H/ ) ( S/ ). Replace the 2nd term on the right hand side by the Maxwell relation given in problem 1, to get ( G/ ) = ( H/ ) (/ ) (1) Now wealso know thatdg Sd+d. Equatingthis with the mathematical relation dg = ( G/ ) d +( G/ ) d shows that ( G/ ) d =. Substitution of this result into Eq. (1) gives the desired answer ( H/ ) = +(/ ) c. In general, the energy U is a function of and. Show that the energy is a function of only the temperature for a gas that obeys the equation of state ( b) = R where b is a constant. (13 pts) o prove that the energy is a function of only the temperature, we need to show that ( U/) = 0. We know from part (b) that ( U/) + ( / ). For the given equation of state, = R/( b), so ( / ) = R/( b) and hence +( / ) R/( b)+r/( b) = 0. hus ( U/) =0.

3 3 d. In terms of the translational partition function, the definitions of the average energy and the pressure is E = k 2 ( lnq tr / ) and = k( lnq tr /), where ( ) 2πmk 3/2 q tr (,) = Starting from these expressions, derive an expression for ( E / ) in terms of the thermophysical variables,, and. (13 pts) Hint: should statistical mechanics, where the average energy is E, give the same result as thermodynamics? We have h 2 = k 2 ( ) lnqtr We can reverse the order of the differentiation in the mixed partial derivative to get Substituting in = k( lnq tr /) gives = k 2 = k 2 ( ) lnqtr ( ) ( ) + k his is the same result as given in part (b) but with U replaced by E 2. he Clapeyron equation states that the derivative of the coexistence curve of two phases α and β is ( ) = S β S α = S G=0 β α a. Derive the Clausius-Clapeyron equation, which relates ( / ) along a liquid-vapor coexistence curve, to the enthalpy of vaporization H vap. (13 pts) For a liquid vapor transition S = H/ vap. Also if we assume that the vapor is an ideal gas and that g l, then gas = R/ (for one mole). hus / = H/(Rvap 2 ). We can rearrange this equation to d/ = Hd/R 2. Integration gives ln/ o ( H/R)(1/ 1/ o ) 3. he phase diagram for water is: And a table of some thermodynamic and physical properties of water is a. What will be the melting point ( C) at the critical pressure? Assume the solid-liquid coexistence curve is linear, that fus and S fus are independent of temperature. (12

4 4 quantity ice water steam S (J/mol K) (m 3 ) R/ pts) Here we use the Clapeyron equation (roblem 2) ( / ) G=0 = S/. Here, for one mole, S (in the direction ice water) = =28.95 J/K. Similarly (also for one mole) = ( ) 10 5 m 3. So / = 28.95/ If we assume the solid-liquid coexistence curve is linear, then / ( crit norm )/( crit norm ) = ( ) /( crit ) a/k Solving this equation for crit gives or 1.22 C. b. he vapor pressures of solid and liquid water are given by ln( s /atm) = K ln( l /atm) = K Use this information to calculate the temperature ( C) and pressure (atm) at the triple point. (12 pts) hese two functions describe the solid-gas and solid-liquid interfaces in the figures. Where they intersect is the triple point. At that point s = l. So we can set these two equations equal at the triple point to get / = /. Solving gives = K = 0.01 C. Substituting this value into one of equations gives ln = /273.16=

5 5 Exponentiation gives = exp( )= atm. 4. In terms of thermodynamic properties, derive the value of the expressions: (Note, just stating the answer is not sufficient.) and ( ) (G/) [ ] (G/) (1/) (8pts) (4pts) By definition G = H S, so G/ = H/ S or ( ) G/ H ( ) H H 2 + C C H 2 Now let x = 1/ or = 1/x so by the chain rule, d/dx = (d/d)(d/x) (1/x 2 )d/d = 2 d/d. hus [ ] (G/) (1/) 2 ( H 2 ) = H

Homework Problem Set 8 Solutions

Homework Problem Set 8 Solutions Chemistry 360 Dr. Jean M. Standard Homework roblem Set 8 Solutions. Starting from G = H S, derive the fundamental equation for G. o begin, we take the differential of G, dg = dh d( S) = dh ds Sd. Next,