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course ① Physicalchemistary Dr.tahseen altaei Chapter tow Fundamentals of thermodynamics 2.1 The zeroth law of thermodynamics: When two systems are in thermal equilibrium, i.e. no heat flows from one system to the other during their thermal contact, then both systems have the same temperature. If system A has the same temperature as system B and system B has the same temperature as system C, then system A also has the same temperature as system C. 2.2 internal energy, U: The total energy of a system the total kinetic and potential energy of the molecules in the system. The change in internal energy: ΔU = Uf – Ui (2.1) Uf – internal energy at a final state Ui – internal energy at an initial state Internal energy, heat, and work are all measured in the same units, the joule (J): 1 J = 1 kg m2 s-1 Changes in molar internal energy ΔUm are typically expressed in kilojoules per mole (kJ mol-1).An energy of 1 cal is enough to raise the temperature of 1 g of water by 1°C:1 cal = 4.184 J 2.3 First Law of thermodynamics: The internal energy of an isolated system is constant. ΔU = w + q (2,2) w – the work done on the system q – the energy transferred as heat to the system If we switch to infinitesimal changes, dU, dw, and dq, dU = dw + dq (2.3) Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 4 be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). course ① Physicalchemistary Dr.tahseen altaei Example3:A closed system performed work of 400 J, with 1000 J of energy in the form of heat supplied to it. How did the system’s internal energy change? Solution: If the system performed work, then, , the work W = −400 J. Hence the change in the internal energy is ΔU = Q +W = 1000 + (−400) = 600 J . 2.4 work Work = distance × opposing force The work needed to raise the mass through a height h on the surface of Earth: Work = h × mg = mgh (2.4) When energy leaves the system, i.e., the system does work in the surroundings, w < 0. When energy enters the system as work, w > 0. Energy leaves the system as heat, q < 0. Energy enters the system as heat, q >0 2.4.1 Irreversible volume work In this case the external pressure pex is different from the system pressure p. During expansion it is lower and during compression higher than the system pressure. We have dw = -F dz (2.5) The negative sign: when the system moves an object against an opposing force, the internal energy of the system doing the work will decrease. F = pex A dw = - pex dV dw = - pex A dz 𝑽𝒇 𝒘 − ∫𝑽𝒊 dV = A dz 𝑷𝒆𝒙 𝒅𝒗 (2.6) Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 5 be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). course ① Physicalchemistary Dr.tahseen altaei 2.4.2 Free expansion: Expansion against zero opposing force: pex = 0 w = 0 No work is done when a system expands freely. Expansion of this kind occurs when a system expands into a vacuum Expansion against constant pressure : 𝑽𝒇 𝒘 − ∫𝑽𝒊 𝑷𝒆𝒙 𝒅𝒗 pex (V f Vi ) ΔV = Vf – Vi w = -pex ΔV (2.7) Wvol = CV (T2 − T1) (2.7*) . 2.4.3 Reversible expansion The reversible volume work, Wvol is connected with a change of the system volume from the initial value V1 to the final value V2. dw = -pexdV 𝑽𝒇 𝒘 − ∫𝑽𝒊 dw = -pdV 𝒑𝒅𝒗 (2.8) 2.4.3.1 Isobaric process p = const Wvol = −p(V2 − V1) (2.9) 2.4.3.2 Isochoric process dV = 0 Wvol = 0 (2.10) 2.4.3.3 Isothermal process, the equation of state of an ideal gas p = nRT/V Wvol = −nRT ln(V2/V1) (2.11) Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). 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Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). - course ① Physicalchemistary Dr.tahseen altaei 2.4.3.4Isothermal process, the van der Waals equation of state 𝒏𝑹𝑻 𝑷 = 𝑽−𝒃 − 𝒂𝒏𝟐 𝑽𝟐 𝑾 == −𝒏𝑹𝑻 𝐥𝐧 𝑽𝟐 −𝒏𝒃 𝑽𝟏 −𝒏𝒃 + 𝒏𝟐 𝒂( 𝟏 𝑽𝟏 − 𝟏 𝒗𝟐 ) (2.12) 2.4.3.5 Reversible isothermal expansion work of a perfect gas at a temperature T For a gas to expand reversibly, the external pressure must be adjusted to match the internal pressure at each stage of the expansion 2.4.3.6 Adiabatic process for which Poisson’s equations: 𝑷 = 𝒄𝒐𝒏𝒔𝒕𝑽−𝒌 𝒘= . 𝒄𝒐𝒏𝒔 𝟏−𝒌 𝟏−𝒌 (𝑽 − 𝑽 𝟐 𝟏 ) = 𝐂𝐕 (𝐓𝟐 − 𝐓𝟏) 𝟏−𝒌 (2.13)) Note: Irreversible volume work is always higher than reversible volume work. Two cases may occur: • During irreversible compression, we have to supply more work to the system for compression from the initial volume V1 to the final volume V2 than in the case of reversible compression, • During irreversible expansion, the work done (Wdone = −Wvol) is smaller than during reversible expansion, and consequently the supplied work is greater. Example4: Calculate the work done when 50 g of iron reacts with hydrochloric acid to produce hydrogen gas in (a) a closed vessel of a fixed volume, (b) an open beaker at 25°C. a) the volume cannot change, no expansion work is done and w = 0. In (b) the gas drives back the atmosphere and therefore w = -pexΔV. We can neglect the initial volume because the final volume after the gas production is so much larger: ΔV = Vf – Vi ≈ Vf = nRT/pex n – the amount of H2 produced w = -pexΔV = -pex × (nRT/pex) = -nRT Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 7 be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). course ① Physicalchemistary Dr.tahseen altaei The reaction equation is Fe(s) + 2 HCl(aq) → FeCl2(aq) + H2(g) Therefore, 1 mol H2 is generated when 1 mol Fe is consumed and n can be taken as the amount of reacting Fe atoms. MFe = 55.85 g mol-1 nFe = 50 g / 55.85 g mol-1 w ≈ -(50 g / 55.85 g mol-1) × 8.31451 J K mol-1 × 298.15 K ≈ -2.2 kJ 2,4.3.7 Other kinds of work: • Surface work—work connected with a change of the system surface from the initial value A1 to the final value A2. 𝑨 𝑾𝑺𝑼𝑹𝑭 = ∫𝑨 𝟏 𝜸𝒅𝑨 where 𝜸 is the surface tension. 𝟐 (2.14) • Electrical work—work connected with the transfer of electric charge Q across the potential difference E. Q Wel = ∫0 EdQ • Shaft (2.15) work : Wsh is defined by the relation 𝑷 𝐖𝐬𝐡 = ∫𝑷 𝟐 𝑽𝒅𝑷 (2.16) 𝟏 2.5 Heat capacities: The heat capacity C for a given process is defined by the following relation 𝒄 ( 𝒅𝑸 𝒅𝑻 ) 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 = (2.17) The isochoric heat capacity, CV , and the isobaric heat capacity, Cp, are defined by the relations CV = 𝒅𝑸 ( 𝒅𝑻 ) izochoric Cp = ( = 𝒅𝑼 ( 𝒅𝑻 ) V 𝒅𝑸 𝒅𝑻 ) izobaric =( (218) 𝒅𝑯 𝒅𝑻 )P (𝟐. 𝟏𝟗) For CV and Cp we have Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 8 be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). course ① Physicalchemistary CV > 0 Cp > 0 lim CV = 0 𝑇→0 lim Cp = 0 𝑇→0 Dr.tahseen altaei (2.20) (2.21) 2.6 Enthalpy : H is a function of state defined by the relation H = U + Pv (2.22) Molar enthalpy, Hm = H/n Hm = Um + pVm For a perfect gas, Hm = Um + RT For a general infinitesimal change of state of the system,U changes to U + dU, p to p + dp, V to V + dV: H + dH = (U + dU) +( p + dp)(V + dV) = U + dU + pV + pdV + Vdp + dpdV dpdV – the product of two infinitesimally small numbers and can be ignored. U + pV = H H + dH = H + dU + pdV + Vdp dH = dU + pdV + Vdp dU = dq + dw dH = dq + dw + pdV + Vdp If the system is in mechanical equilibrium with its surroundings at a pressure p and does only expansion work, dw = -pdV dH = dq + Vdp The heating occurs at constant pressure: dp = 0 dH = dq When a system is subjected to a constant pressure, and only expansion work can occur, the change in enthalpy is equal to the energy supplied as heat Example5: A system containing 5 moles of an ideal gas was heated from temperature T1 = 300 K to temperature T2 = 400 K. The internal energy of the gas increased by ΔU = 800 J. How did the enthalpy of the system change? Solution: ΔH = ΔU + Δ(pV ) , Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 9 be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). Physicalchemistary course ① where Δ(pV ) = p2V2 − p1V1. For an ideal gas pV = nRT. Then Dr.tahseen altaei ΔH = ΔU + nR_T = 800 + 5 × 8.314 × (400 − 300) = 4957 J Example6: The internal energy change when 1.0 mol CaCO3 in the form of calcite converts to aragonite is +0.21 kJ. Calculate the difference between the enthalpy change and the change in internal energy when the pressure is 1.0 bar given that the densities of the solids are 2.71 and 2.93 g cm-3. We need to relate the density and the molar mass. For 1 mol, . ρ = M/Vm Vm = M/ρ The change in enthalpy when the transition occurs is ΔH = H(aragonite) – H(calcite) = {U(a) + pV(a)} – {U(a) + pV(c)} = ΔU + p{V(a) – V(c)} = ΔU + pΔV Vm(a) = 100 g / 2.93 g cm-3 = 34 cm3 Vm(c) = 37 cm3 pΔV = 1.0×105 × (34 – 37)×10-6 m3 = -0.3 J ΔH - ΔU = -0.3 J The difference is only 0.1% of the value of ΔU .Example7:Water is heated under p = 1 atm. When an electric current of 0.50 A from a 12 V supply is passed for 300 s, 0.798 g of water is vaporized. Calculate the molar internal energy and enthalpy changes at the boiling point (373.15 K). ΔH = qp = IVt ΔHm = 0.50 A × 12 V × 300 s / (0.798/18.02) mol ΔHm = +41 kJ mol-1 H2O(l) → H2O(g) Δng = +1 mol ΔUm = ΔHm – RT = +38 kJ mol2.6.1The variation of the enthalpy with temperature The enthalpy of a substance increases with temperature. The relation between the increase in enthalpy and the increase in temperature depends on the conditions (constant pressure or constant volume ). Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 10be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). course ① Physicalchemistary . Dr.tahseen altaei heat capacity at constant pressure, Cp :The heat capacity at constant - the analog of the heat capacity at constant volume, an extensive property. The molar heat capacity at constant pressure, Cp,m, - an intensive property. dH = CpdT (at constant pressure) (2.23) ΔH = CpΔT (at constant pressure) 𝑞𝑝 = CpΔT (2.24) The variation of heat capacity with temperature can sometimes be ignored if the temperature range is small - accurate for a monoatomic perfect gas. Otherwise, a convenient Approximate empirical expression is 𝐶𝑝, 𝑚 = 𝑎 + 𝑏𝑇 + 𝑐 (2.25) 𝑇2 The empirical parameters a, b, and c are independent of temperature. Example8: What is the change in molar enthalpy of N2 when it is heated from 25°C to 100°C? For N2(g) we find in Table 2.2 a = 28.58 J K-1 mol-1, b = 3.77×10-3 J K-2 mol-1, and c = -0.50×105 J K mol- Since dH = CpdT, 𝐶𝑝, 𝑚 = 𝑎 + 𝑏𝑇 + 𝑐 𝑇2 𝐻(𝑇2) 𝑇2 ΔH = H(T2 ) − H(T1) = ∫𝐻(𝑇1) 𝑑𝐻 = ∫𝑇1 ( T1=298K 𝑎 + 𝑏𝑇 + 𝑐2)𝑑𝑇 𝑇 T2=373K To carry out the integration, we use Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 11be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). ∫ dx = x + constant 1 𝑑𝑥 1 ∫ 𝑋𝑑𝑥 = 2 𝑥 2 + 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ∫ 𝑥 2 = 𝑥 + 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 course ① Physicalchemistary Dr.tahseen altaei HT2 HT1aT2 T1 𝑏(𝑇22 − 𝑇12 ) − 𝑐( − ) 2 𝑇2 𝑇1 1 1 1 H(373 K) - H(298 K) = 2.20 kJ mol-1 The relation between heat capacities : For a perfect gas, Hm - Um = RT ΔHm - ΔUm = RΔT ΔHm/ΔT - ΔUm /ΔT = R Cp,m – CV,m = R Cp,m – CV,m = large for gases, but negligible for most solids and liquids. •Heat capacity ratio and adiabats: We consider the change in pressure resulting from an adiabatic, reversible expansion of a perfect gas. We will show 𝑃𝑉 𝛾 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 γ - the heat capacity ratio γ = Cp,m/CV,m Because Cp,m > CV,m, γ > 1 𝛄 = 𝑪𝑽.𝒎 + 𝑹 𝑪𝒗.𝒎 𝑹 = 𝟏+ 𝑪𝑽.𝒎 •Adiabatic process—Poisson’s equations During an adiabatic process, the system does not exchange heat with its surroundings,i.e. Q = 0. Work during an adiabatic process (both reversible and irreversible) is equal tothe change of internal energy W = ΔU . Provided that the following conditions are fulfilled during an adiabatic process, i.e. that • the process is reversible, Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 12be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). • the system exchanges only volume work with its surroundings, • the system is an ideal gas, Physicalchemistary course ① • the heat capacities CV and Cp do not depend on temperature, Dr.tahseen altaei •the following relations, called Poisson’s equations, apply between T, p and V : 𝟏 𝜸 𝜸 𝑷𝑽 = 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝑇𝑉 𝛾−1 𝑻𝑷 𝟏−𝜸 𝜸 𝑷 𝑽 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝑽𝑻 𝟏 𝜸−𝟏 𝑷𝑻 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝜸 𝟏−𝜸 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 Example9: An ideal gas expanded adiabatically from temperature T1 = 300 K and pressure p1 = 1 MPa to pressure p2 = 100 kPa. Provided that Poisson’s equations hold and that Cpm = (5/2)R, find temperature T2 after the expansion Solution 1−γ P1 T2 = T1( P2 ) γ 𝛾= Cpm Cpm−𝑅 = 5 3 T2 = 119.3𝐾 Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 13be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). Physicalchemistary course ① Dr.tahseen altaei 2.7 Second law of thermodynamics:No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion to work Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 14be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). Physicalchemistary course ① Dr.tahseen altaei 2.7.1Entropy: A measure of the molecular disorder of a system 𝑑𝑞𝑟𝑒𝑣 𝑑𝑠 = 𝑇 𝐹 ∆𝑆 = ∫ 𝐼 𝑑𝑞𝑟𝑒𝑣 𝑇 (2.26) dS > 0 , [isolated system, irreversible process] dS = 0 , [isolated system, reversible process] units of entropy: J K-1; of molar entropy: J K-1 mol-1. Example10 : Calculating the entropy change for the isothermal expansion of a perfect gas .We need to find the heat absorbed for reversible path between the Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 15be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). initial and final states .The heat absorbed during a reversible isothermal expansion: ΔU = q + w ΔU = 0 q = -w qrev = -wrev 𝟏 𝟐 𝒒𝒓𝒆𝒗 ∆𝑺 = ∫ 𝒅𝒒𝒓𝒆𝒗 = 𝑻 𝒊 𝑻 Physicalchemistary 𝒒𝒓𝒆𝒗 course ① = −𝒘𝒓𝒆𝒗 = 𝐧𝐑𝐓 𝐥𝐧(𝐕𝟐/𝐕𝟏) Dr.tahseen altaei 2.7.2 Heat: That portion of internal energy which can be exchanged between a system and its surroundings only when there is a difference in temperature Q = ΔU –W (2.27) S2 T2 Q = ∫ 𝑇𝑑𝑆 = T2S2 − T1S1 − ∫ SdT S1 T1 Several typical examples of heat calculation are given below •Adiabatic process Q=0 •Isochoric process, no work done Q = ΔU = U(T2, V ) − U(T1, V ) •Isobaric process, only volume work done Q = ΔH = H(T2, p) − H(T1, p) •Isothermal reversible process, ideal gas 𝒒 = −𝒘 = 𝐧𝐑𝐓 𝐥𝐧(𝐕𝟐/𝐕𝟏) •Isothermal reversible process, the van der Waals equation of state 𝒒 = 𝒏𝑹𝑻 𝐥𝐧 𝑽𝟐 − 𝒏𝒃 𝑽𝟏 − 𝒏𝒃 Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 16be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). •General reversible isothermal process S2 T (S2 − S1) = TΔS𝑞 = 𝑇 ∫S1 𝑑𝑆 = Physicalchemistary course ① Dr.tahseen altaei 2.7.3 Entropy of phase transitions The entropy of a substance increases when it melts and when it boils because its molecules become more disordered as it changes from solid to liquid and from liquid to vapor Entropy of fusion at the melting temperature: ΔfusS = ΔfusH/T Entropy of vaporization at the boiling temperature: ΔvapS = ΔvapH/Tb 2.8 Helmholtz energy: The Helmholtz energy F is a function of state defined by the relation F = U − TS . (2.28) The change in the Helmholtz energy _F during a reversible isothermal process is equal to the work supplied to the system ΔF = W , [T, reversible process Example11 During a certain isothermal process, internal energy changed by ΔU and entropy by ΔS. Derive the relation for the change in the Helmholtz energy. Is it possible to calculate the change in the Helmholtz energy during a non-isothermal process if in addition to ΔU and ΔS we also know the initial and final temperatures? Solution For an isothermal process we obtain ΔF = ΔU − Δ(TS) = ΔU − TΔS, [T] . For a non-isothermal process we have ΔF = ΔU − Δ(TS) = ΔU − T2S2 + T1S1 . Since the values of entropy in the initial and final states, S1 , S2, are not specified and we only know that ΔS = S2 − S1, the change in the Helmholtz energy cannot be calculated. . 2.9 Gibbs energy Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 17be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). The Gibbs energy (or the Gibbs function) G is a function of state defined by the relation G = H − TS . (2.29) course ① Physicalchemistary Dr.tahseen altaei Example12: During a certain thermodynamic process, a system passed from its initial state defined by the values of volume V1 and pressure p1 to its final state defined by the values p2 and V2. The change in the Helmholtz energy was ΔF. Calculate the change in the Gibbs energy. Solution From the definitions for H, F, and G we obtain G = H − TS = U + pV − TS = F + pV , from which it follows: ΔG = ΔF + Δ(pV ) = ΔF + p2V2 − p1V1 . The change in the Gibbs energy _G during a reversible isothermal and isobaric process isequal to other than volume work, Wother, supplied to the system ΔG = Wother = W −Wvol = W + pΔV , [T, p, reversible process] Example13: The change in the Gibbs energy during the oxidation of one mole of glucose, according to the reaction C6H12O6 + 6O2 = 6CO2 + 6H2O is ΔG = −2870 kJ mol−1. How high can a person weighing 75 kg climb if he or she has eaten one mole (186 g) of glucose? The biological efficiency is 25%. Solution In our example, Wother is equal to mechanical work needed for lifting a load to a height h Wother = −mgh , where m is the mass of the load and g is the gravitational acceleration. If we project a human body as an isothermal and isobaric system in which reversible processes occur and which does not perform any volume work, then ΔG = −mgh ℎ= ∆𝐺 𝑚𝑔 = −2870000 75×9.8 = 3905𝑚 Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 18be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). 2.10 Fugacity Fugacity is defined by the relation 𝑮𝒎 (𝑻,𝑷)−𝑮°𝒎 (𝑻,𝑷𝒔𝒕 ) 𝒇 = 𝑷 𝐞𝐱𝐩( ) 𝑹𝑻 Physicalchemistary course ① 𝒔𝒕 (2.30) Dr.tahseen altaei where pst is the standard pressure (usually pst = 101 325 Pa), and Gm(T, p) is the molar Gibbs energy at temperature T and pressure p, and G_(T, pst) is the molar Gibbs energy of a system in an ideal gaseous state at temperature T and pressure pst. Fugacity is a thermodynamic quantity which is useful when solving phase and chemical equilibria. Note: For an ideal gas, fugacity equals pressure The fugacity/pressure ratio is called the fugacity coefficient ∅= 𝑓 _ 𝑝 Note: For an ideal gas = 1. Example14: Knowing the values of the molar Gibbs energy of carbon dioxide Gm(T, p) = −53 183 J/mol at T = 350 K and p = 10 MPa, and Gm(T, pst) = −65 675.14 J/mol, where pst = 101.325 kPa, calculate its fugacity and the fugacity coefficient at 350 K and 10 MPa. Solution: f = 0.101325 × exp( ∅= 𝟕. 𝟒𝟏𝟓 𝟏𝟎 −𝟓𝟑𝟏𝟖𝟑 − (−𝟔𝟓𝟔𝟕𝟓.𝟏𝟒) 𝟖.𝟑𝟏𝟒 × 𝟑𝟓𝟎 ) = 𝟕. 𝟒𝟏𝟓 𝐌𝐏𝐚 , = 𝟎. 𝟕𝟒𝟏𝟓. 2.11 Total differential: Let us consider functions M(x, y) and N(x, y) dz = M(x, y)dx + N(x, y)dy Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 19be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). to be the total differential of the function z = z(x, y) is the equality of the derivatives 𝜕𝑁 ( )x = ( )𝑦 ∂y 𝜕𝑥 ∂M Physicalchemistary at all points of the 𝑀=( 𝜕𝑍 )𝑦 𝜕𝑥 , (2.32) course ① region, where 𝑁=( Dr.tahseen altaei 𝜕𝑍 )𝑥 𝜕𝑦 (2.33) Hence for the total differential of the function z = z(x, y) it holds 𝜕𝑍 𝜕𝑍 ) 𝜕𝑥 + ( ) 𝜕𝑥 𝑦 𝜕𝑦 x ∂z = ( ∂𝑦 𝜕2z 𝜕2z = ∂x ∂y ∂y ∂x (2.34) Example15: Is the differential form dz = (10xy3 + 7)dx + 15x2y2dy the total differential of function z? Solution 𝜕𝑍 𝜕𝑍 𝑀 = (𝜕𝑦) 𝑦 = 10xy3 + 7 𝑁 = (𝜕𝑥) 𝑥 = 15x2y2 . It holds that 𝜕𝑀 𝜕2𝑧 ( )= = 30xy2 𝜕𝑦 𝜕𝑥𝜕𝑦 , ∂N 𝜕2𝑧 ( )= = 30xy2 . ∂x 𝜕𝑦𝜕𝑥 Sometimes we need to know the derivative of x with respect to y at a fixed z .0 𝜕𝑍 =( ) 𝜕𝑥 𝑦 𝜕𝑥 + (𝜕𝑍 ) 𝜕𝑦 ∂𝑦 x Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 20be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). 𝝏𝒙 (𝝏𝒚) = 𝜕𝑍 ) 𝜕𝑦 x 𝜕𝑍 ( ) 𝜕𝑥 𝑦 ( 2.11.1 Total differential and state functions: All state functions (p, V, T, n, U, H, S, F, G, Cp, CV , f, . . .) have total differentials. Heat and work are not state functions and they do not have any total differential. Physicalchemistary course ① Dr.tahseen altaei Example16: Prove that condition holds for the pressure of an ideal gas as a function of temperature and volume at a constant amount of substance. Proof 𝑛𝑅𝑇 𝑃= 𝑉 𝜕𝑃 𝜕𝑃 𝑛𝑅 𝑛𝑅𝑇 𝑑𝑃 = ( )𝑉 𝑑𝑇 + ( ) 𝑇 𝑑𝑉 = 𝑑𝑇 − 2 𝑑𝑉 𝜕𝑇 𝜕𝑉 𝑉 𝑉 1 𝜕( ) 𝑛𝑅 𝜕𝑇 𝑛𝑅( 𝑉 ) 𝑇 = − 2 ( )𝑉 𝜕𝑉 𝑉 𝜕𝑇 dp is thus a total differential and pressure is a function of state. A similar proof can be established for pressure calculated using any equation of state. Example17: By substituting the expression TdS for ¯dQ into equation (3.1) we obtain ¯dW = dU − TdS . Prove that work W is not a state function. Proof Comparison , shows that M = 1, N = −T, x = U, and y = S. Hence 𝝏𝑴 𝛛𝟏 𝛛𝐍 𝛛(−𝐓) ( ) 𝐱 = ( )𝐮 ≠ ( ) 𝐲 = ( ) 𝝏𝒚 𝛛𝐬 𝛛𝐱 𝛛𝐔 𝐬 𝛛𝐓 because ( )𝐬 is generally different from zero. This shows that ¯dW is not a 𝛛𝐔 total differential and therefore work is not a state function Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 21be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). Total differential of the product and ratio of two functions Similar rules as those governing differentiation hold for the total differential of the product and the ratio of functions x and y. d(xy) = xdy + ydx , 𝒙 𝟏 𝒙 𝒅(𝒚)= 𝒚dx− 𝒚𝟐 𝒅𝒚 For example, the total differential of the product of volume and pressure d(pV ) = pdV + V dp . course ① Physicalchemistary Dr.tahseen altaei 2.11.2:Integration of the total differential The integral of the total differential from point (x1,y1) to point (x2,y2) does not depend on the path between these points. We can, e.g., first integrate with respect to x at a fixed y1, and then with respect to y at a fixed x2 𝒙𝟐 𝐳(𝐱𝟐, 𝐲𝟐) = 𝐳(𝐱𝟏, 𝐲𝟏) + ∫ ( 𝒙𝟏 𝒚𝟐 𝝏𝒛 𝝏𝒛 )𝒚=𝒚 𝒅𝒙 + ∫ ( )𝒙=𝒙 𝒅𝒚 𝟐 𝟏 𝝏𝒙 𝝏𝒚 𝒚 (𝟐. 𝟑𝟓) 𝟏 𝒚𝟐 𝒙 𝝏𝒛 𝝏𝒛 𝐳(𝐱𝟐, 𝐲𝟐) = 𝐳(𝐱𝟏, 𝐲𝟏) + ∫ ( )𝒙=𝒙𝟏 𝒅𝒚 + ∫ ( )𝒚=𝒚𝟐 𝒅𝒙 𝒚𝟏 𝝏𝒙 𝒙 𝝏𝒙 2.12 Combined formulations of the first and second laws of thermodynamics: In this section we will assume (if not otherwise stated) that every system referred to is closed and homogeneous, that it exchanges only volume work with its surroundings, and that all the occurring processes are reversible. Gibbs equations: dU = TdS − pdV , (2.36) dH = TdS + V dp , (2.37) dF = −SdT − pdV , (2.38) dG = −SdT + V dp . (2.39) Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 22be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). internal energy is a function of variables S and V . Variables S and V will be called the natural variables of function U. The natural variables of enthalpy are S and p, those of the Helmholtz energy are T and V , and those of the Gibbs energy are T and p. 2.13Derivatives of U, H, F, and G with respect to natural variables If we consider the internal energy U as a function of S and V , its total differential 𝝏𝑼 𝝏𝑼 𝝏𝒔 𝝏𝑽 𝒅𝑼 = ( )𝒗 𝒅𝑺 + ( )𝒔 𝒅𝑽 Physicalchemistary course ① Dr.tahseen altaei By comparing () and () we obtain 𝝏𝑼 ( )𝑽 = 𝑻 𝝏𝑺 𝝏𝑼 ( )𝒔 = −𝑷 𝝏𝑽 , (𝟐. 𝟒𝟎) In a similar way we obtain for H = f(S, p), F = f(T, V ), G = f(T, p) 𝝏𝑯 ( )𝒑 = 𝑻 𝝏𝑺 𝝏𝑭 (𝝏𝑻)𝒗 = −𝑺 𝝏𝑮 (𝝏𝑻)𝒑 = −𝑺 𝝏𝑯 ( )𝑺 = 𝑽 𝝏𝑷 𝝏𝑭 (𝝏𝑽)𝑻 = −𝑷 𝝏𝑮 (𝝏𝑷)𝑻 = 𝑽 (𝟐. 𝟒𝟏) (𝟐. 𝟒𝟐) (𝟐. 𝟒𝟑) 2.13 Maxwell relations: By applying the equalities of mixed derivatives to the total differentials of the functions U, H, F, G, we obtain 𝝏𝑻 𝝏𝑷 ( )𝑺 = −( )𝑽 𝝏𝑽 𝝏𝑺 (𝟐.𝟒𝟒) Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 23be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). 𝝏𝑻 𝝏𝑽 ( )𝑺 = ( )𝑷 𝝏𝑷 𝝏𝑺 𝝏𝑷 𝛛𝐒 𝝏𝑻 𝛛𝐕 (𝟐.𝟒𝟓) ( )𝐕 = ( )𝐓 (𝟐. 𝟒𝟔) 𝝏𝑽 𝝏𝑺 ( )𝑷 = −( )𝑻 𝝏𝑻 𝝏𝑷 (𝟐.𝟒𝟕) course ① Physicalchemistary Dr.tahseen altaei 2.14 Total differential of entropy as a function of T, V and T, p: 𝝏𝑺 𝝏𝑺 𝒅𝑺 = ( )𝑽 𝒅𝑻 + ( )𝑻 𝒅𝑽 𝝏𝑻 𝝏𝑽 At a fixed volume 𝝏𝑺 𝟏 𝝏𝑸 ( )𝑽 = ( ) 𝝏𝑻 𝑻 𝝏𝑻 isochoric = 𝟏 𝝏𝑼 𝑪𝑽 ( )𝑽 = 𝑻 𝝏𝑻 𝑻 For the differentiation of entropy with respect to volume, Maxwell relation applies. 𝐶𝑉 𝜕𝑃 𝑑𝑆 = 𝑑𝑇 + ( )𝑉 𝑑𝑉 𝑇 𝜕𝑇 In the same way we obtain for entropy as a function of temperature and pressure 𝜕𝑆 1 𝜕𝑄 ( )𝑃 = ( ) 𝜕𝑇 𝑇 𝜕𝑇 isobaric 1 𝜕𝐻 𝐶𝑃 = ( )𝑃 = 𝑇 𝜕𝑇 𝑇 and using Maxwell relation 𝐶𝑃 𝜕𝑉 𝑑𝑆 = 𝑑𝑇 − ( )𝑃 𝑑𝑝 𝑇 𝜕𝑇 Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 24be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). 2.15 Conversion from natural variables to variables T, V or T, p Internal energy may be converted from the function of natural variables to the function of variables T, V 𝜕𝑃 dU = CV dT + [𝑇( )𝑉 ) − 𝑃] 𝑑𝑉 𝜕𝑇 In the same way, H = f(S, p) may be converted to H = f(T, p). Physicalchemistary course ① Dr.tahseen altaei 𝜕𝑉 dH = Cp dT + [𝑉 − 𝑇( )𝑃 ] 𝑑𝑝 𝜕𝑇 the total differential of the function U = f(T, V ) is equal to the expression 𝜕𝑈 𝜕𝑈 𝑑𝑈 = ( )𝑉 𝑑𝑇 + ( ) 𝑇 𝑑𝑉 𝜕𝑇 𝜕𝑉 𝜕𝑈 ( )𝑉 = 𝐶𝑉 𝜕𝑇 𝜕𝑈 𝜕𝑃 ( ) 𝑇 = 𝑇( )𝑉 ) − 𝑃 𝜕𝑉 𝜕𝑇 Similarly we obtain the respective partial derivatives of the dependence H = f(T, p) 𝜕𝐻 ( )𝑃 = 𝐶𝑃 𝜕𝑇 𝜕𝐻 𝜕𝑉 ( ) 𝑇 = −𝑇 + 𝑉( )𝑃 𝜕𝑃 𝜕𝑇 Example18: Prove that the internal energy of an ideal gas is only a function of temperature, i.e. that at a fixed temperature it depends neither on volume nor on pressure Proof For an ideal gas we have Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 25be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). 𝑇( 𝜕𝑃 𝑛𝑅 )𝑉 − 𝑃 = 𝑇 −𝑃 =0 𝜕𝑇 𝑉 𝜕𝑉 𝜕𝑉 𝑑𝑉 = ( )𝑃 𝑑𝑇 + ( ) 𝑇 𝑑𝑝 𝜕𝑇 𝜕𝑃 𝜕𝑃 𝜕𝑉 𝜕𝑃 𝜕𝑉 𝑑𝑈 = {𝐶𝑉 + [𝑇( )𝑉 − 𝑃] ( )𝑃 } 𝑑𝑇 + [𝑇( )𝑉 − 𝑃] ( ) 𝑇 𝑑𝑉 𝜕𝑇 𝜕𝑇 𝜕𝑇 𝜕𝑃 From this expression it follows that Physicalchemistary course ① Dr.tahseen altaei 𝜕𝑈 𝜕𝑃 𝜕𝑉 ( ) 𝑇 = [𝑇( )𝑉 − 𝑃] ( ) 𝑇 𝜕𝑃 𝜕𝑇 𝜕𝑃 As was shown above, the expression in brackets for an ideal gas is zero. Hence the internal energy of an ideal gas is independent of pressure. 2.16 Heat engines: A heat engine is a device that cyclically converts heat into work. According to the Second Law of thermodynamics, this conversion cannot be complete a heat engine means a device that cyclically receives heat Q2 (Q2 > 0) from the warmer reservoir of temperature T2. It converts part of it into work Wdone = −W, which it supplies to the surroundings. At the same time it supplies heat Qsupplied = −Q1 (Q1 < 0) to the cooler reservoir of temperature T1, and returns to the initial state. During this cyclic process, the change of internal energy is zero, and consequently we have Q1 +W + Q2 = 0 . (2.50) Example19: A heat engine received 600 J of heat from the warmer reservoir, supplied 500 J of heat to the cooler reservoir, and returned to the initial state. Ascertain the work performed by the engine. Solution It follows from the specification that Q1 = 600 J, Q2 = −500 J, Q = Q1 + Q2 = 100 J. Since the system returned to its original state, _U = 0. From equation (2.50)) we thus have W = −Q = −100 J . The work performed by the engine is Wdone = −W = −(−100 J) = +100 J Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 26be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). 2.16.1The Carnot heat engine: Let the temperature of the warmer reservoir be 𝑇ℎ and that of the cooler reservoir be𝑇𝑐 . One cycle is comprised of four partial reversible processes: 1. Reversible isothermal expansion from A to B at Th; the entropy change is qh/Th, qh – the heat supplied to the system from the hot source. 2. Reversible adiabatic expansion from B to C. No heat leaves the system, ΔS = 0, the temperature falls from Th to Tc, the temperature of the cold sink. 3. Reversible isothermal compression from C to D. Heat is released to theDr.tahseen altaei Physicalchemistary course ① cold sink; the change in entropy of the system is qc/Tc, qc is negative. 4. Reversible adiabatic compression from D to A. No heat enters the system, ΔS = 0, the temperature rises from Tc to Th. 𝐰𝐨𝐫𝐤 𝐝𝐨𝐧𝐞 −𝒘 ŋ= = 𝐡𝐞𝐚𝐭 𝐬𝐮𝐩𝐩𝐥𝐢𝐞𝐝 𝐟𝐫𝐨𝐦 𝐭𝐡𝐞 𝐰𝐚𝐫𝐦𝐞𝐫 𝐜𝐨𝐧𝐭𝐚𝐢𝐧𝐞r 𝑸ℎ Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 27be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). (2,51) 𝑇ℎ − 𝑇𝑐 ŋ= 𝑇ℎ (2.52) Note: A reversibly operating heat engine cannot exist in practice because, among other reasons, the processes in such an engine would progress at an infinitesimal rate. ŋreversible engine > ŋirreversible engine Physicalchemistary course ① Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 28be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). Dr.tahseen altaei Physicalchemistary course ① Dr.tahseen altaei Example20 The Carnot heat engine received heat Q2 = 100 J from a heat reservoir of temperature T2 = 600 K, performed work, delivered heat −Q1 to the cooler reservoir of temperature T1 = 300 K, and returned to the initial state. Calculate the engine’s efficiency, the performed work and the supplied heat. ŋ= 𝑇2 − 𝑇1 600 − 300 = = 0.5 𝑇2 600 −𝑊 = 𝑄2 ŋ = 100 × 0.5 = 50𝐽 Since the initial and final state are identical, the internal energy change is ΔU = 0 . 0 = Q +W = Q2 + Q1 +W Q1 = −Q2 −W = −100 − (−50) = −50 J . The Carnot engine supplied −Q1 = +50 J of heat to the cooler reservoir. 2.16.2 Heat engine with steady flow of substance: A heat engine with a steady flow of substance is a device that exchanges shaft work Wsh with its surroundings at the cost of the energy of the working medium Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This This text text only only appears appears in in the the demo demo version. version. This This text text can can be removed removed with with the the full full version. version. 29be Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com). This text only appears in the demo version. This text can be removed with the full version. Gewijzigd met de DEMO VERSIE VAN CAD KAS PDF editor (http://www.cadkas.com).