a) Use the following equation from the lecture notes: = ( J K 1 mol 1) ( ) 10 L


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1 hermodynamics: Examples for chapter One mole of nitrogen gas is allowed to expand from 0.5 to 10 L reversible and isothermal process at 300 K. Calculate the change in molar entropy using a the ideal gas law and b the van der Waals equation with a = atm L 2 mol 2, b = L mol 1. a Use the following equation from the lecture notes: S = nr ln V2 V 1 = J K 1 mol 1 10 L ln 0.5 L = J K 1 mol 1 b Start with the following equation see lecture notes: S he equation of state gives: S = P V P = nr V nb an2 P V 2 V = nr V 2 V nb S = nr V nb V 1 or S = R ln Plugging in the numerical values gives: V2 b V 1 b = nr V nb dv = nr ln V2 nb V 1 nb S = J K 1 mol 1 10 L 1 mol L mol 1 ln 0.5 mol 1 mol L mol 1 = J K 1 mol 1 1
2 2. Derive the relation for C P C V for a gas that follows the van der Waals equation. Based on the lecture notes we have: C P C V = V α 2 κ with α = 1 V V P and κ = 1 V V P And the van der Waals equation can be written in the following forms: P = R V b ā V or = 1 P 2 R V Pb + ā V ab V 2 Next α and κ must be calculated. Since the van der Waals equation of state cannot be written for the molar volume easily, one must use the reciprocal identity: V P = 1 / V P α 1 = V V P = V P R a R V + 2ab P V = 2 R V 3 R a R V + 2ab R V 2 It is convenient to eliminate pressure P from this equation by using the van der Waals equation: α 1 = Next we calculate κ: V V b 2a R V + 2ab R V 2 = α = V V b 1 V V V 2a b R V b 2 2 2a V b R V 2
3 κ 1 = V P V = V R V b 2 2a V 2 Plugging both α and κ into the expression for C P C V and simplifying, we get: C P C V = R 1 2a V b 2 R V 3 3. Estimate the change in molar entropy of liquid benzene at 25 C and atmospheric pressure when the pressure is raised to 1000 bar? he coefficient of thermal expansion α is K 1, the density is g cm 3, and the molar mass is g mol 1. Based on the lecture notes we have: S P = P Inserting the values into this equation we get: = V α S V α P S g mol m cm K Pa g cm 3 = J K 1 mol 1 4. Derive the expression for U/ the internal pressure for a gas following the virial equation with Z = 1 + B/V. Consider only PV  work and a reversible process. Note that B = B i.e. it depends on temperature. 3
4 Recall the first law of thermodynamics: du = dq + dw. For PV  work dw = PdV and by the second law: dq = ds. hus: du = ds PdV and dividing both sides by dv and imposing constant temperature: U = he equation of state gives: S P P = V P V R = 1 + B V P = R V + BR V 2 P Furthermore, differentiation of P with respect to gives: P V = R V + BR B R V + 2 V 2 By inserting these results into the expression derived above: U = R V + BR B V + 2 V R V 2 R V BR V 2 = B V R 2 V 2 5. Consider the process of freezing water at 10 C for which the enthalpy change is 5619 J mol 1 and entropy change is J K 1 mol 1. What is the Gibbs energy of freezing water at constant temperature of 10 C? he Gibbs energy can be written in terms of enthalpy: G = H S. At constant temperature we have: G = H S = 5619 J mol K J K 1 mol 1 = J mol 1. 4
5 6. a Integrate the GibbsHelmholtz to obtain an expression for G 2 at temperature 2 in terms of G 1 at 1, assuming that H is independent of temperature. b Obtain an expression for G 2 using a more accurate approximation that H = H C P, where 1 is an arbitrary reference temperature. Assume that C P is temperature independent. Hint: Integrate one side with respect to variable G/ and the other with respect to. a Integrate both sides of the GibbsHelmholtz equation see lecture notes: H 2 = G/ 2 H 2 d = G 2 / 2 d G/ 1 G 1 / 1 1 H 1 = G 2 / 2 G 1 / G 2 = H 2 + G 1 2 / b Assume temperature dependency for H = H C P : G 2 / 2 G 1 / 1 d G/ = 2 1 H 1 2 d C P G 2 G 1 1 = H 1 1 C P ln d 2 C P G 2 = G H C P 2 ln
6 7. An ideal gas is allowed to expand reversibly and isothermally 25 C from a pressure of 1 bar to a pressure of 0.1 bar. a What is the change in molar Gibbs energy? b What would be the change in molar Gibbs energy if the process occurred irreversibly? a Use the following equation see lecture notes: G 2 = R ln P2 P bar = J K 1 mol K ln 1 bar = kj mol 1 b Gibbs energy is a state function and depends only on the endpoints. hus the answer is the same as in a. 8. Helium is compressed isothermally and reversibly at 100 C from a pressure of 2 to 10 bar. Calculate a q per mole, b w per mole, c Ḡ, d Ā, e H, f Ū, g S, assuming that helium is an ideal gas. a and b Ideal gas: U = 0 and q = w. he integrated P V work is: w = R ln P2 P 1 10 bar = J K 1 mol K ln 2 bar = 4993 J mol 1 and hence q = 4993 J mol 1 c he following equation in the lecture notes yields: Ḡ = R ln P2 10 bar = J K 1 mol K ln P 1 2 bar = 4993 J mol 1 6
7 d A = U S and at constant : A = U S. U = 0 and A = S. Also by definition: G = U + PV S = U S = S. So Ḡ = Ā in this case and Ā = 4993 J mol 1. e H = U + PV = U = 0. f As already stated U = 0. g S = G/ = J K 1 mol oluene molecular weight g mol 1 is vaporized at constant external pressure and its boiling point, 111 C constant temperature. he heat of vaporization at this temperature is J g 1. For the vaporization of toluene, calculate a w per mole, b q per mole, c H, d Ū, e Ḡ, and f S. In part a assume that the volume of the liquid is negligible and that toluene vapor behaves according to the ideal gas law this assumption is not needed elsewhere. Note also that evaporation at the boiling point is a reversible process. a w = P V = PV fin V init = R 0 = J K 1 mol K = 3193 J mol 1. he final volume was obtained from the ideal gas law PV = nr. b and c For vaporization process we have q = H = J g g mol 1 = kj mol 1. d Ū = q+w = J mol J mol 1 = kj mol 1. Note that if ideal gas behavior was assumed here, U would have been zero as it would depend only on temperature. e and f G = U +PV S and hence G = U +V P +P V S S. At constant temperature and pressure P = = 0 and therefore G = U + P V S. From part a P V = 3193 J mol 1 and we need to calculate S. S = q rev / = kj mol 1 /384 K = 86.8 J K 1 mol 1. And finally G = 0 by combining all the required terms. 7
8 10. Calculate mix G and mix S for the formation of a quantity of air containing 1 mol of gas by mixing nitrogen and oxygen at K. Air may be taken to be 80% nitrogen and 20% oxygen in mol %. Assume ideal gas behavior. In the lecture notes, the following realtions were given: mix G = R n 1 lny 1 + n 2 lny 2 mix S = R n 1 lny 1 + n 2 lny 2 mix H = mix V = 0 n 1 = 0.80 mol of N 2 y 1 = 0.80 and n 2 = 0.20 mol of O 2 y 2 = By inserting the values in above equations, we get: mix G = 1239 J mol 1 and mix S = J K 1 mol 1. 8
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