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Increasing Temperature Increases Disorder, Because the Entropy Dominates the Free Energy at High Temperatures, Whereas Enthalpy Dominates at Low Temperatures

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Increasing Temperature Increases Disorder, Because the Entropy Dominates the Free Energy at High Temperatures, Whereas Enthalpy Dominates at Low Temperatures Increasing temperature increases disorder, because the entropy dominates the free energy at high temperatures, whereas enthalpy dominates at low temperatures. That is, in the expression ΔG = ΔH TΔS, − the entropy term (TΔS) becomes more important (either for ΔS positive or ΔS negative) as T increases; hence at high temperatures the system will become more disordered to increase its entropy, even at the cost of increasing enthalpy. 3. The difference between the energy and enthalpy changes in expanding an ideal gas. How much heat is required to cause the quasi-static isothermal expansion of one mole of an ideal gas at T = 500 K from PA =0.42 atm, VA = 100 liters to PB =0.15 atm? (a) What is VB? (b) What is ΔU for this process? (c) What is ΔH for this process? (a) For an ideal gas, in any isothermal process ΔU = 0. Therefore ΔU = q + w =0 = q = w. ⇒ − For such a process, when it is also quasi-static, V2 nRT V w = pdV = dV = nRT ln 2 , − − V1 V − V1 so V q = nRT ln 2 V1 1 1 V2 = (1 mol)(2 cal K− mol− )(500 K) ln 100 L Using the ideal gas law, V2 p1 p1V1 = nRT = p2V2 = = . ⇒ V1 p2 79 Therefore we can write p q = nRT ln 1 p2 0.42 atm = (1000 cal) ln 0.15 atm = 1030 cal 1000 cal. ≈ (b) From the ideal gas law, p1 0.42 atm V2 = V1 = (100 L) p2 0.15 atm = 280 L. (c) For any isothermal process of an ideal gas, ΔU =0 [ΔU = C (T T )]. v 2 − 1 (d) For any isothermal process of an ideal gas, ΔH =0. This can be shown as follows. ΔH = ΔU + Δ(pV ). But ΔU = 0 and Δ(pV )=p V p V . Using the ideal gas law, 2 2 − 1 1 P2V2 = nRT, P1V1 = nRT. Therefore ΔH = 0+(nRT nRT ) − =0. 80 4. The work and the heat of boiling water. For the reversible boiling of five moles of liquid water to steam at 100◦C and 1 atm pressure, calculate q.Isw positive or negative? From tables that give the heat of vaporization of water, q = q = ΔH boiling − condensation − condensation 1 1 = (540 cal g− )(18 g mol− )(5 mol) =48, 600 cal, qboiling =48.6 kcal. At constant pressure, V2 w = pdV − V1 = p(V V )= p(V V ), − 2 − 1 − gas − liquid so V V = V V gas liquid ⇒ gas − liquid is positive, and so w<0. 81 5. The entropy and free energy of gas expansion. Two moles of an ideal gas undergo an irreversible isothermal expansion from VA = 100 liters to VB = 300 liters at T = 300 K. (a) What is the entropy change for this process? (b) What is the Gibbs free energy change? dq (a) ΔS = rev . T Since S and U are state functions, we needn’t be concerned that the actual process is irreversible (read: complicated). We can choose any path to calculate state functions, so we invent an isothermal, quasi-static process to get qrev. V2 ΔU =0 = qrev = wrev = nRT ln . ⇒ − V1 For an isothermal process, qrev V2 1 1 300 1 ΔS = = nR ln = (2 mol)(2 cal K− mol− )ln =4.39 cal K− . T V1 100 (b) ΔG = ΔH TΔS.For an isothermal process on an ideal gas, ΔH =0. − (Note that ΔH = ΔU + Δ(PV), and for an isothermal process, ΔU =0 and Δ(PV)=Δ(nRT ) = 0; therefore ΔH =0.) 1 ΔG = TΔS= (300 K)(4.39 cal K− )= 1318.33 cal. − − − 82 Divide the numerator and denominator by CB to get CA TA + TB Cb CA T = TA if 1. CA −→ C +1 B CB 12. ΔS for an adiabatic expansion of a gas. In an adiabatic quasi-static expansion of an ideal gas, how do you reconcile the following two facts: (1) the increase in volume should lead to an increase in entropy, but (2) in an adiabatic process, δq = 0, so there should be no change in entropy (since dS = δq/T = 0)? Adiabatic expansion of a gas involves both an increase in volume, which increases the entropy, and a decrease in temperature, which decreases the entropy. To show this more quantitatively, consider the total entropy change as a volume increase at constant T and then a temperature change at constant V : δq ∂S ∂S S3 S1 = dS = =0= dV + dT − T ∂V T ∂T V 2 Change V S Change T 1 3 V The two terms in the integral are 2 ∂S Nk V2 V3 dV = dV = Nkln = Nkln , since V3 = V2, 1 ∂V T V V1 V1 3 ∂S Cv dT T3 T3 dT = = CV ln = CV ln , since T1 = T2. 2 ∂T V T T2 T1 But, according to Equation (7.20) of the text, these terms must exactly cancel in a quasi-static adiabatic expansion. 88 14. Free energy of an ideal gas. (a) For an ideal gas, calculate F (V ), the free energy versus volume, at constant temper- ature. (b) Compute G(V ). t (a) dF = SdT pdV + μ dN = pdV − − i i − i=1 for constant T and Ni’s. So NkT F (V )= pdV = dV = NkT ln V. − − V − (b) G = H TS =(U + pV ) TS =(U TS)+pV = F + pV. − − − So G(V )=F (V )+pV = NkT ln V + NkT = NkT(ln V 1). − − − 15. Heat capacity of an ideal gas. The energy of an ideal gas does not depend on volume, ∂U =0. ∂V Use this fact to prove that the heat capacities Cp =(∂H/∂T )p and CV =(∂U/∂T )V for an ideal gas are both independent of volume. (a) Use ∂2U ∂2U = ∂V ∂T ∂T∂V 90.
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