Thermodynamics Lecture Notes PDF

11/28/2024 dq rev  T 0 2 dq dq rev rev S  S2  S1 dS  1 T T mathematical statements of the second law of the...

11/28/2024 dq rev  T 0 2 dq dq rev rev S  S2  S1 dS  1 T T mathematical statements of the second law of thermodynamics. T2 V V  Isothermal S=nCVln  nRln 2 ΔS  nRln  2  T1 V1  V1  S=nCPln T2 P  nRln 1 P  T1 P2 ΔS  nR ln  1   T2  reversible adiabatic Reversible Adiabatic Changes  S  0 processes are called ΔH trans isentropic processes. Phase changes. ΔS trans  Ttrans Entropy Change for Isolated System (dS)rev = 0 and (dS)irr > 0 Or (S)rev = 0 and (S)irr > 0 1  11/28/2024 Entropy Change on Heating or Cooling of a Substance T  T  ΔS  nC V ln  2  ΔS  nC P ln  2   T1   T1  ΔSuniverse  ΔSsystem  ΔSsurroundings Example 1 mole of an ideal gas is compressed isothermally at 298oK. If the gas is compressed from 1 atm to 10 atm reversibly and irreversibly against an external pressure of 10 atm, calculate the entropy change for both cases. ΔS For the reversible isothermal process P 1 S system  nR ln  1   1*8.314 ln    19.14 EU  P2   10  ΔS (surroundings) = +19.14 EU ΔS (universe) = ΔS (system) + ΔS (surroundings) = -19.14 + 19.14 = 0 2  11/28/2024 ΔS For the irreversible isothermal process ΔS (system) in the irreversible = reversible ΔS (system) Because the Entropy is a state function ΔSsystem,irreversible= -19.14 EU  P   10  q  nRT 1  2   1* 8.314 * 2981    1* 8.314 * 298 * 9 irreversib le  P   1  1 q  22298.15 J irreversib le This is the heat lost by the system and gained by the surrounding qsurroundings,irreversible = + 22298.15 J ΔSsurroundings,irreversible = qsurr/T= +22298.15/298= +74.9 EU ΔS (universe) = ΔS (system) + ΔS (surroundings) =-19.14 +74.9 = +55.83 EU Example What is the change in entropy when argon at 25oC and 1 atm pressure is expanded isothermally from 500 cm3 to 1000 cm3 and simultaneously heated to a temperature of 100oC, CV=3/2 * R PV 1* 0.5 n   0.020 moles RT 0.082 * 298 V  T  S  nR ln  2   nCV ln  2   V1   T1   1000  3  373  S  0.020 *8.314* ln    0.020* *8.314 ln    500  2  298  S  0.1153 .05599  0.172 J / K 3  11/28/2024 Combined Form of the First and Second Laws of thermo-dynamics The mathematical form of the first law of thermodynamic is, dq = dE + dw = dE + PdV The mathematical expression for second law is dq rev or dqrev = TdS dS  T substitution of equation in the first law equation give, TdS = dE + PdV dE = TdS – PdV combined form of the first and the second laws of thermodynamics. According to equation dE = T dS - PdV  E   E    T (1)    P (2)  S  V  V S Differentiate (1) with respect to V at differentiate equation (2) with respect to S at constant S constant V 2E  T     2E  P VS  V S    SV  SV  2E 2E  T   P         Maxwell equation(1) VS SV  VS  SV 4  11/28/2024 Enthalpy and Entropy: Enthalpy is expressed as H = E + PV dH = dE + PdV + VdP using equation , dE = TdS – PdV dH = TdS – PdV + PdV + VdP  dH = TdS + VdP dH = TdS + VdP H H    T (1)    V (2)  S P  PS Differentiate (1) with respect to P differentiate equation (2) with respect to S at at constant S constant P 2H  T   2H  V        PS  P S SP  S  P  T V Maxwell equation (2)     PS  SP 5  11/28/2024 Energy as a Function of S and V dE = T dS - PdV E = f (S,V) The total differential of E is given as  E   E  dE    dS    dV  S  V  V S Comparing the coefficients with this equation and dE = T dS - PdV E  E    T    P  SV  V S  E  Equation   T can be written as  S  dE = TdS For a given temperature this equation can be integrated to give a new expression + Integration constant  dE  T  dS E = TS + A Or A = E – TS A = Integration constant and called Helmholtz free energy function. 6  11/28/2024 Enthalpy as a Function of S and P From equation dH = TdS + VdP H = f(S, P)  H   H  H  H dH    dS    dP  S  p  P  S   T   V  Sp  P S  dH   TdS Constant (G) H = TS + G The function G is the same as the G = H – TS Gibbs free energy function Chapter V 7  11/28/2024 We have studied three thermodynamic function for any system; these are the energy content (E), the heat content (H) and the entropy (S). Although the entropy concept is the fundamental consequence of the second law of thermodynamics, there are two other function, which utilize the entropy in their derivation. The free energy (G) will be employed extensively in connection with the study of equilibria, both physical and chemical, and the direction of chemical change. The second is called the work function (A)or Helmholtz free energy function. 8  11/28/2024 Free Energy Functions A and G Entropy is a measure of unavailable energy. entropy is multiplied by absolute temperature the product (TS) is equal to the amount of heat (energy) which is not free to be used for useful work. Then out of total heat absorbed by a system the amount less by TS must be the available amount of heat which can be put for useful work. Thus the free energy (X) can be expressed as (X) = q – TS (1) Since heat is absorbed either at constant V or at constant P the term q may be replaced either by E or H. Therefore, the free energy would be equal to E – TS or H-TS. Two new symbols are given to these free energy terms. 9  11/28/2024 1-Helmholtz free energy A A = E - TS 2- Gibbs free energy G G = H - TS Properties of A 1- A = E – TS dA = dE – TdS – SdT (Differential change) But dE = dq – dw dE = TdS – dw dA = TdS – dw – TdS – SdT = -dw – SdT If T is constant dT = 0, and –(dA)T = dw The decrease in Helmboltz free energy is or -(A)T = wrev equal to the maximum reversible work which a system can be do isothermally. 10  11/28/2024 2- A = E – TS dA = dE – TdS – SdT = TdS-dw – TdS -SdT dA = -PdV – SdT At constant volume, dV = 0 At constant temperature, dT = 0  A   A     S   P  T V  V T Differentiating with respect to V at constant T Differentiating with respect to T at constant V 2 2A  P  A  S       TV  TV V T  V T 2A 2A P  S applying Euler's theorem      Maxwell`s equation (3) TV VT Tv VT P  S     Maxwell`s equation (3) Tv VT This equation is very useful to study the variation of entropy with volume at constant temperature. Experimentally it is impossible to measure the change of entropy with change of volume as we do not have any apparatus which can measure entropy directly. But it is convenient to study the variation of pressure with temperature at constant volume. Hence, using Maxwell's equation, the experimentally difficult quantities can be calculated easily 11  11/28/2024 3- A = E – TS dA = TdS – dw – TdS – SdT (dA)T = - dW (dA)T = -PdV For ideal gas, nRT P V dV dA nRT V V  P A  nRT ln  2   nRT ln 1 Integration gives  V1  P2 4 - Variation of A with T at constant V A = E – TS   A  E     2 T  T  v T 12  11/28/2024 4 - Variation of A with T at constant V A = E – TS dA = dE – TdS – SdT dA = -SdT – PdV A but  S  T  A   A  E  T  (13)  T v  A  or A  T   E  T  v 1 Multiplying both sides by  T2 we get  A   A  T   T  v E   T2 T2 13  11/28/2024  A 1 A E  ( ) V  (14) T2 T T T2   A L.H.S. is the differential of equation 14   and hence T  T   A E     (15) T  T  v T2 Consider a system in state 1. Let it is changed isothermally to state 2. Thus equation 13 changed to, A  A1 E1 T 1   T V  A  and A2  E2  T  2   T V The change in A is given as 14  11/28/2024  A2 A1 A(E2 E1)T  A2 A1V T   A    E  T     T V   A  Or A  T    E  T  V 1 On multiplying by  and replacing L.H.S. by differential T2   A    form, we get T  T    A  E (16)     T  T  v T2 Properties of G 1.G = H – TS dG = dH –SdT – TdS H = E + PV dH = dE + PdV + VdP  dG = dE + PdV + VdP – SdT - TdS but dE = dq – dw = TdS - dwrev  dG = TdS – dw + PdV + VdP – SdT – TdS  dG = – dw + PdV + VdP – SdT At constant temperature and pressure dT = 0 = dp, (dG)T,P = -(dw-PdV) 15  11/28/2024 (dG)T,P = -(dw-PdV) where dw includes all types of work. PdV = mechanical work and therefore, dw-PdV = dw' = net work = useful work Thus, -(dG)T,P = dwnet - (G)T,P = wnet The decrease in free energy is equal to the reversible maximum work which a system can be isothermally and isobarically over and above the mechanical work. 2. G = H-TS dG = dH-TdS-SdT But H=E+PV dH=dE+PdV+VdP dG = TdS – dw + PdV + VdP – SdT – TdS dG = TdS – PdV + PdV + VdP – SdT – TdS  dG = -SdT+VdP 16  11/28/2024 dG = -SdT+VdP At constant P At constant T  G  G    S   V  P T  T P Applying Euler's theorem for dG  2G  S   2G   V              PT T  P T  TP P  T P  S   V       Maxwell's (4) P  T  T P 1.Variation of free energy with pressure at constant T: dG = -SdT+VdP At constant temperature, dT = 0, the equation will reduced to (dG)T =VdP If the state of the system is changed from 1 to 2, then 2 P2 nRT P2 dP  dG   VdP V G2 G1 G  nRT  1 P1 P P1 P P2 V1 G  nRT ln G  nRTln P1 V2 This equation can be used to calculate the free energy for expansion or compression. 17  11/28/2024 4. Calculation of Free Energy Changes with temperature at constant Pressure Derivation of Gibbs-Helmholtz Equation Gibbs-Helmholtz equation relates the free energy change to the enthalpy change and the rate of change of free energy with temperature. G=H-TS   (G / T)   H  T   P T2 There are two forms of Gibbs-Helmholtz equation. These are (1) The relation between E and A is given by equations A = E -TS   A E    T  T v T2 (2) The relation between G and H, will be derived here. Consider the relation, G=H-TS ,   ( G / T)   H    2  T P T 18