Statistical MolecularThermodynamics University of Minnesota

Homework Week 8

1. By comparing the formal derivative of G with the derivative obtained taking account of the first and second laws, use Maxwell relations to determine which of the following equations is valid. Start with the definition of the Gibbs energy,

G = H − TS. ∂T ∂P (a)

− ∂V S = ∂S V ∂V ∂T (b)

∂S P = ∂P S ∂S ∂P (c)

∂V T = ∂T V ∂S ∂V (d)

− ∂P T = ∂T P Answer:

Given that G = H − TS, using the first and second laws we can write,

dG =d(H − TS) =dH − T dS − SdT =d(U + PV ) − T dS − SdT =dU + (P dV + V dP ) − T dS − SdT =(T dS − P dV ) + (P dV + V dP ) − T dS − SdT = − SdT + V dP

and we can write the total derivative of G,

∂G ∂G dG = dP + dT ∂P T ∂T P in terms of its natural independent variables. We can compare terms in the two equa- tions and see that ∂G = V ∂P T and

∂G = −S ∂T P If we now take the cross derivatives of this result we have ∂V   ∂S  = − ∂T P ∂P T

¯ ◦ 2. Calculate ∆vapG for benzene at 77 C and 1 bar given that the molar enthalpy of vaporization of benzene at its boiling point of 80.09◦ C and 1 bar is 30.72 kJ·mol−1. ¯ ¯ Assume that the values of ∆vapH and ∆vapS are constant over the temperature range considered here. ¯ Hint: You will first need to calculate ∆vapS from the given information.

(a) ∆Gvap(350 K) = −0.6512 kJ/mol

(b) ∆Gvap(350 K) = 0.2687 kJ/mol

(c) ∆Gvap(350 K) = 0.3414 J/mol

(d) ∆Gvap(350 K) = 100.2 kJ/mol

(e) ∆Gvap(350 K) = 0.4414 kJ/mol 4 (f) ∆Gvap(350 K) = −3.04 × 10 kJ/mol 4 (g) ∆Gvap(350 K) = 3.04 × 10 kJ/mol Answer:

From lecture video 8.2, we know that ∆G = ∆H − T ∆S. For a given ∆H and ∆S (assuming they are independent of T ) we can compute ∆G at different temperatures. ¯ ¯ We are given only ∆vapH, so we first must determine ∆vapS. At the boiling point (and 1 bar) the liquid and gas phases are in equilibrium and thus ∆Gvap = 0. We can use ¯ this result to solve for ∆vapS as follows:

∆Gvap =∆Hvap − T ∆Svap

0 =30.72 kJ/mol − [(80.09 + 273.15)K]∆Svap 30.72 kJ/mol ∆S = vap (80.09 + 273.15) K −1 −1 −1 −1 ∆Svap =0.08697 kJ · mol · K = 86.97 J · mol · K

o Now we can compute ∆Gvap at 77 C (which is 350.15) K:

−1 −1 ∆Gvap(350.15 K) = 30.72 kJ/mol − 350.15 K · 0.08697 kJ · mol · K

∆Gvap(350.15 K) = 0.2687 kJ/mol (result carries all digits from the above calculation) 3. In lecture video 8.5, we derived for a thermoelastomer:

∂U  ∂S  = T + f. ∂L T ∂L T For a weight suspended on a tense rubber band and resting on a scale, assuming that the rubber band acts as an ideal elastomer and does not change volume when stretched, which of the following statements is TRUE if you increase the temperature of the rubber band?

(a) The force exerted by the rubber band, f, will not change. (b) The force exerted by the rubber band, f, will decrease. (c) The force exerted by the rubber band, f, will increase.

Answer:

In lecture video 8.5, we learned that the Helmholtz energy for the rubber band can be expressed as dA = −SdT + fdL. We can write the total derivative of A in terms of the natural, independent variables T and L as

∂A ∂A dA = dT + dL. ∂T L ∂L T And thus if we compare terms, ∂A = −S ∂T L and ∂A = f. ∂L T If we determine the cross derivatives, i.e.

∂ ∂A  ∂S  = − ∂L ∂T L T ∂L T and

∂ ∂A   ∂f  = . ∂T ∂L T L ∂T L These two mixed partial derivatives must be equal to one another. And, we know that ∂S
the entropy decreases with increasing length, so − ∂L T must be a positive value. ∂f
Thus, ∂T L must be positive, which means f must increase with increasing tempera- ture. (One can further manipulate the equations to show that the increase should be linear in temperature, but we will not do that here.) A final, intuitive way to see the answer to this question is to recognize that increasing temperature will favor lengths having greater entropy, i.e., shorter lengths (where the polymer chains can twist and turn and are not pulled straight). So, there will be increased resistance to losing entropy by stretching, and this is reflected in a higher force constant.

4. One of the Maxwell relations derived in the lecture videos is,

∂P   ∂S  = . ∂T V ∂V T Using this relationship, find the correct expression for the isothermal change in the entropy of a gas that obeys the van der Waals equation of state when you change the ¯ ¯ molar volume from V1 to V2.

 ¯  (a) ∆S¯ = R ln V2−b V¯1−b

 ¯  (b) ∆S¯ = R ln V2 V¯1  ¯    ¯ V2−b 1 1 (c) ∆S = R ln ¯ + 3aR ¯ 3 − ¯ 3 V1−b V2 V1  ¯  (d) ∆S¯ = R ln V2−b − aR V¯ 2 − V¯ 2
V¯1−b 2 1

Answer:

From lecture video 2.2, we know that the van der Waals equation of state is, RT a P = − V¯ − b V¯ 2 and therefore

∂P  R = ¯ . ∂T V¯ V − b We can substitute into the Maxwell relation above to obtain,

 ∂S  R ¯ = ¯ ∂V T V − b We can separate the variables and integrate this equation to yield

R dS = dV V¯ − b ¯ V2 − b ∆S = R ln ¯ . V1 − b 5. Write the total derivative of H(P,T ) as

∂H  ∂H  dH = dP + dT (1) ∂P T ∂T P and determine another expression, derived with the use of Maxwell relations, that expresses dH in terms of readily measured quantities. (Hint: equations involving the differential of the Gibbs free energy are likely to be the most useful to you in seeking partial derivative equalities.)

 ∂V
 (a) dH = V − T ∂T P dP + CP dT  ∂T
 (b) dH = T − V ∂V P dP + CP dT  ∂V
 (c) dH = V − T ∂T V dV + CV dT  ∂V
 (d) dH = V − T ∂T P dP + CV dT  ∂V
 (e) dH = V + T ∂T V dV + CP dT  ∂P
 (f) dH = P − T ∂T V dV + CP dT To solve this, we start with the definition of the Gibbs energy, G = H + TS and differ- entiate this with respect to pressure at constant temperature to obtain an expression ∂H
that contains the first term of the total derivative as written, ∂P T . We can then ∂V
re-arrange this term using Maxwell relations to include V and ∂T P . The second term of the original total derivative we can simply recognize as the definition of the constant pressure heat capacity. Taking the derivative of the expression for G,

∂G ∂H   ∂S  = − T (2) ∂P T ∂P T ∂P T

We know we can combine the first and second laws to write,

dG =d(H − TS) =dH − T dS − SdT =d(U + PV ) − T dS − SdT =dU + (P dV + V dP ) − T dS − SdT =(T dS − P dV ) + (P dV + V dP ) − T dS − SdT = − SdT + V dP.

From writing the total derivative of G,

∂G ∂G dG = dP + dT ∂P T ∂T P in terms of its natural independent variables, we see that ∂G = V ∂P T and

∂G = −S. ∂T P If we now take the cross derivatives of this result we have ∂V   ∂S  = − ∂T P ∂P T

Now we can substitute these results into Eq. 2 to obtain, ∂H  ∂V  V = + T ∂P T ∂T P

which we can solve to get

∂H  ∂V  = V − T . ∂P T ∂T P This result can now be substituted into the original total differential for H to yield,

 ∂V   dH = V − T dP + CP dT ∂T P

∂H
using the definition that CP = ∂T P . 6. Determine the Gibbs-Helmholtz equation for A, i.e determine, ∂(A/T )

∂T V

Hint: It was shown in lecture video 7.1 that

∂S  C = V ∂T V T

G (a) − T 2 − H H (b) − T 2 h ∂(S/V ) i (c) ∂T V U (d) − T 2 (e) U (f) G (g) none of the above

Answer:

We can begin with the definition A = U − TS and determine the derivative after division by T ,

∂ A ∂ U  = − S ∂T T ∂T T and therefore, ∂(A/T ) U 1 ∂U  ∂S  = − 2 + − ∂T V T T ∂T V ∂T V

From the definition of CV we know

1 ∂U  C = V T ∂T V T and using the hint given in the question we can write

  ∂(A/T ) U CV CV U = − 2 + − = − 2 ∂T V T T T T 7. For the following spontaneous endothermic reaction,

Ba(OH)2(s) · 8 H2O + 2 NH4NO3(s) → Ba(NO3)2(s) + 10 H2O(l) + 2 NH3(aq)

−1 at 300 K, ∆rH = 60 kJ and ∆rS = 400 J K . What is the change in the Gibbs free energy, ∆rG, for this reaction at 300 K? (a) 0 kJ (b) 30 kJ (c) -30 kJ (d) 60 kJ (e) -60 kJ (f) 120 kJ (g) -120 kJ (h) none of the above

Answer:

We can apply the formula ∆G = ∆H − T ∆S −1 ∆rG = 60 kJ − 300K · 0.4 kJ K = −60 kJ ¯ ◦ 8. Determine the molar free energy of vaporization, ∆vapG, for benzene at 75 C and 1 bar given that the molar enthalpy of vaporization of benzene at its normal boiling point of 80.09◦ C and 1 bar is 30.72 kJ·mol−1. Unlike a previous homework problem, do not assume that the molar enthalpies and molar entropies of vaporization are constant over all temperatures. Instead, make use of the molar heat capacities of liquid and gaseous benzene, 136.3 and 82.4 J · mol−1 K−1, respectively, to refine your answer.

(a) −40.2 kJ · mol−1 (b) 20.80 kJ · mol−1 (c) 0.102 kJ · mol−1 (d) −2.480 kJ · mol−1 (e) 0.282 kJ · mol−1 (f) 0.445 kJ · mol−1

Answer:

¯ ¯ We are given only ∆vapH, so we first must determine ∆vapS. At the boiling point (and 1 bar) the liquid and gas phases are in equilibrium and thus ∆Gvap = 0. We can use ¯ this result to solve for ∆vapS as follows:

∆Gvap =∆Hvap − T ∆Svap

0 =30.72 kJ/mol − [(80.09 + 273.15)K]∆Svap 30.72 kJ/mol ∆S = vap (80.09 + 273.15) K −1 −1 −1 −1 ∆Svap =0.08697 kJ · mol · K = 86.97 J · mol · K

Figure 1: Scheme used to compute ∆S as a function of T2: Starting from the left bottom corner upwards, there is a change in entropy in the liquid due to the change in temperature of the liquid. Then at this temperature the entropy changes due to the vaporization of the liquid to yield gas at the same temperature. Then the entropy changes again due to a change in the gas temperature. If we take ∆Svap as an illustration, we can think about this problem as is shown in figure 1. From figure 1, using the fact that S is a state function,

Z T1 Z T2 Cp, liquid dT Cp, gas dT ∆Svap(T2) = + ∆Svap(T1) + . T2 T T1 T The first and third terms are the changes in entropy (for the liquid and gas, respec- tively) due to simply a change in T , and have been obtained from the definition of entropy and the fact that dq = CpdT . We will take T2 to be the T for which we want to compute ∆Svap (and then ∆Gvap), and T1 to be that T for which ∆Svap is known, i.e. the boiling point of benzene in our case.

The same idea from figure 1 can be used for ∆Hvap. Now the change in enthalpy due to the change in T of a phase is given by ∆H = Cp∆T , which is obtained by using the fact that ∆H is the heat exchanged at constant pressure, and the definition of Cp. So, on the whole, for enthalpy we obtain:

∆Hvap(T2) = Cp, liquid∆T2→1 + ∆Hvap(T1) + Cp, gas∆T1→2

◦ So now we have all we need to compute ∆Gvap at 75 C. Let’s first compute ∆Svap,

Z 353.24 −1 −1 Z 348.15 −1 −1 136.3 J · mol · K −1 −1 82.4 J · mol · K ∆Svap(348.15 K) = dT + 86.97 J · mol · K + dT 348.15 T 353.24 T 353.24 348.15 = 136.3 J · mol−1 · K−1 ln + 86.97 J · mol−1 · K−1 + 82.4 J · mol−1 · K−1 ln 348.15 353.24 = 87.75 J · mol−1 · K−1

Next we’ll compute ∆Hvap at 348.15 K:

−1 −1 −1 ∆Hvap(348.15) = 136.3 J · mol · K · [(353.24 − 348.15) K] + 30.72 kJ · mol + +82.4 J · mol−1 · K−1 · [(348.15 − 353.24) K] = 30.99 kJ · mol−1

And finally we can compute ∆Gvap,

∆Gvap(348.15 K) = ∆Hvap(348.15 K) − 348.15 K · ∆Svap(348.15 K) = 30.99 kJ · mol−1 − 348.15 K · 87.75 J · mol−1 · K−1 = 0.445 kJ · mol−1 (result carries all digits from the above calculations) 9. In a Joule-Thomson device, a constant pressure, P1, is applied to a gas in a cylindrical chamber, with volume V1, so that the gas flows through a porous plug into another chamber where a lower constant pressure, P2, is maintained. In the initial state V1 > 0 and V2 = 0. In the final state V1 = 0 and V2 > 0, i.e., all of the gas is transferred from the first chamber to the second in the course of a compression cycle. The apparatus is constructed such that the entire process is adiabatic. The so-called Joule-Thompson coefficient is defined as ∂T  1 ∂H  = − ∂P H CP ∂P T and is a measure of the change in temperature of a gas with respect to the change in the pressure. Which of the following Joule-Thompson coefficients applies to the case of an ideal gas being used in the device.

(a) ∂T
= − 1 ∂G
∂P H CP ∂P T (b) ∂T
= − 1 ∂H
∂P H CV ∂P V ∂T
nR (c) ∂P H = P (d) ∂T
= 1 ∂P H CP ∂T
(e) ∂P H = 0 ∂T
∂A
(f) ∂P H = ∂S V Answer:

There is an easy way to do this problem, and a hard way. The hard way is completely general, so let’s do it first, and then we can note the easy way. Start with the definition of the Gibbs energy, G = H − TS and differentiate this with respect to pressure at constant temperature to obtain an expression that contains the ∂H
first term of the total derivative as written, ∂P T . We can then try to solve for this term using a Maxwell relation. Taking the derivative of G with respect to pressure at constant temperature,

∂G ∂H   ∂S  = − T (3) ∂P T ∂P T ∂P T Now combining the first and second laws,

dG =d(H − TS) =dH − T dS − SdT =d(U + PV ) − T dS − SdT =dU + (P dV + V dP ) − T dS − SdT =(T dS − P dV ) + (P dV + V dP ) − T dS − SdT = − SdT + V dP. The total derivative of G,

∂G ∂G dG = dP + dT ∂P T ∂T P is written in terms of its natural independent variables, such that

∂G = V ∂P T and

∂G = −S. ∂T P Taking cross derivatives of this result yields ∂V   ∂S  = − . ∂T P ∂P T

Now we can substitute these various results into Eq. 3 to obtain, ∂H  ∂V  V = + T ∂P T ∂T P

which we can solve to get

∂H  ∂V  = V − T . ∂P T ∂T P If we substitute this into the definition for the Joule-Thompson coefficient, we obtain

∂T  1  ∂V   = − V − T . ∂P H CP ∂T P T This is completely general. But, for an ideal gas,

∂V  nR = , ∂T P P so ∂T  1  nRT  = − V − = 0. ∂P H CP P And the easy way? Remembering that the enthalpy of an ideal gas depends only on ∂H
temperature, in which case ∂P T must be zero, and so the Joule-Thompson coefficient is zero. 10. A hand warmer produces heat when a concentrated sodium acetate solution crystal- lizes spontaneously at constant pressure. The observed spontaneity of the reaction implies ... (a) exothermic release of heat reduces the energy of the universe. (b) exothermic release of heat increases the energy of the universe. (c) exothermic release of heat reduces the enthalpy of the universe. (d) exothermic release of heat increases the enthalpy of the universe. (e) exothermic release of heat reduces the entropy of the universe. (f) exothermic release of heat increases the entropy of the universe. (g) crystallization reduces the entropy of the universe (h) crystallization reduces the entropy of the system (i) none of the above

Answer:

As the sodium acetate crystallizes, the entropy of the system (the sodium acetate) does decrease, i.e, ∆S for the system is negative, but spontaneous processes do not occur simply to reduce the entropy of an isolated system (to the contrary, they oc- cur to increase the entropy of an isolated system, and, in any case, this system is not isolated). As this is a constant pressure process for a non-isolated system, the relevant state function is the Gibbs free energy. Given that ∆G = ∆H − T ∆S, and that ∆S is negative, it must be the negative change in enthalpy, which is associated with the liberation of heat, that causes the free energy to be negative and makes this process spontaneous. This exothermic release of heat, released spontaneously to the surrounding universe, contributes to the increase in entropy of the universe. Indeed, all spontaneous processes increase the entropy of the universe, because the universe is the ultimate isolated system, and the Second Law says spontaneous processes occur to increase the entropy of isolated systems. (Note that it is not enough simply to be exothermic to be spontaneous; if the entropy of a system decreases by more than the entropy increase of the rest of the universe provided by an exotherm, that process will not be spontaneous.)