THERMODYNAMICS Thermodynamics: First Law of Thermodynamics: energy can neither be created nor destroyed but can be converted from one form to another. Mathematical Form of First Law of Thermodynamics: Consider a system having energy E1 .Let q be the amount of heat added to the system and w is the work done by the system. Total energy E2=E1+q+w E2-E1=q+w ΔE= q+w

THERMODYNAMICS If work done is the work of expansion. w= -PΔV ΔE= q- PΔV q= ΔE+ PΔV For small change, Enthalpy Or Heat Content: Under condition of constant pressure qp= ΔE+ PΔV……………i δq=dE +PdV

THERMODYNAMICS Suppose when system absorbs q calorie of heat its internal changes fromE1 to E2 and volume increases from V1 toV2 ΔE= E2-E1………………ii ΔV= V2-V1…………………iii Then qp=( E2-E1)+ P(V2-V1) qp= (E2+PV2)-(E1+PV1) …………..iv But H=E+PV qp=H2-H1

THERMODYNAMICS qp= ΔH…………….v Enthalpy change of a system is equal to the heat absorbed by the system at constant pressure. From equation (I) and (v) we get The Enthalpy change during a process may also be defind as the sum of the increase in internal energy of the system and the pressure volume work done. ΔH= ΔE+P ΔV

THERMODYNAMICS Heat Capacity: The amount of heat required to raise the temperature of the system through 1 ˚C. C=δq/dT ………….1 heat capacity are of two type: 1 Heat capacity at Constant Volume CV According to first law of themodynamics ………………….2 …………………….3 δq=dE +PdV

THERMODYNAMICS dV=O…………………………4 then above equation becomes When the volume is kept constant, dV=O…………………………4 then above equation becomes Thus heat capacity at constant volume is the rate change of internal energy with temperature at constant temperature 2 Heat capacity at Constant Pressure C p When the pressure is kept constant during absorption of heat, then equation (3) becomes CP = (E/T)P + P(V/T)P………………..6 Cv= (E/T)v

THERMODYNAMICS H=E+PV differentiate w.r.t T at constant P (dH/dT)P = (E/T)P + P(V/T)P. …………….7 Combining equation (6) and (7 ) Thus, CP = (H/T)p Simply equation can be written as CP=dH/dT

THERMODYNAMICS Thus, heat capacity at constant pressure may be defined as the rate of change enthalpy with temperature at constant pressure Relationship between CP and CV. We know that μ > CV CP-CV= dH/dT-dE/dT…………8 But H=E+PV and PV=RT

THERMODYNAMICS H=E+RT Differentiate this equation w.r.t T, we get dH/dT=dE/dT+R dH/dT-dE/dT=R……………..9 combining equation (8) and (9) we get CP-CV=R

THERMODYNAMICS Joule’s Law: It Deals with the study of variation of internal energy of a gas with volume at constant temperature. The change in internal energy of an ideal gas with volume at constant temperature is equal to zero. Mathematically, (∂E/∂V)T =o Proof: Taking E as a function of volume and temperature i.e. E=f(V,T) dE= (∂E/∂T)V dT + (∂E/∂V)T dV But dE=0, dT=0 Hence,

THERMODYNAMICS (∂E/∂V)T dV=0 As dV≠ 0, It means (∂E/∂V)T =o Which is Joule’s Law. Joule-Thomson Effect(Adiabatic Expansion of Real Gas): “ When a real gas is made to expand adiabatically from region of high pressure to region of low pressure, a change of temperature (generally cooling except for hydrogen and helium) is produced”

THERMODYNAMICS Experimental Set –Up : According to first law of thermodynamics ΔE=q+w Since the process is carried out adiabatically, q=0 ΔE=w………….1

THERMODYNAMICS Work done on the system on the left side=P1V1 Work done by the system on the right side=-P2V2 Net work done by the system is w= P1V1-P2V2………….2 Putting this value in equation (1). We get ΔE = P1V1-P2V2 E2-E1= P1V1-P2V2 E2+P2V2 =E1+P1V1 H2=H1 H2-H1=0 ΔH=0 When the expansion of a gas takes place adiabatically through a porous plug or a fine hole, the enthalpy of the system remain constant.

THERMODYNAMICS Joule Thomson Coefficient: μ= (∂T/∂P) Temperature change in degrees produced by a drop of one atmospheric pressure when the gas expands under conditions of constant enthalpy. 1.For cooling , μ will be positive 2. For heating, μ will be negative 3. If μ=0, the gas gets neither heated up nor cooled on adiabatically expansion Inversion Temperature: The temperature at a particular pressure at which μ=0 i.e. the gas gets neither cooled nor heated up on a adiabatic expansion and below which μ is positive and above which μ is negative.

THERMODYNAMICS thermodynamic quantity: Relationship between μ and other thermodynamic quantity: H= f(T,P)……………1 dH= (∂H/∂T)P dT + (∂H/∂P)T dP…………2 In Joule Thomson Effect, H remain Constant i.e dH=0 Putting this value in above equation, we get (∂H/∂T)P dT + (∂H/∂P)T dP=0………3 Dividing above equation throughout by dP and keeping enthalpy constant. (∂H/∂T)P (dT/dP)H + (∂H/∂P)T=0……………..4

THERMODYNAMICS But (∂H/∂T)P = CP, heat capacity at constant pressure, and (∂T/∂P)P= μ, Joule Thomson coefficient Putting these value in equation (4) (∂H/∂P)T + μ CP=0 μ= -1/CP. (∂H/∂P)T Joule Thomson Coefficient for an ideal gas: μ=-1/CP. (∂H/∂P)T ……………..1 Putting H=E+PV in above equation μ=-1/CP. (∂E+PV/∂P)T…………..2 =- 1/CP.[ (∂E/ ∂ P)T+ ∂ (PV)/ ∂ P)T]

THERMODYNAMICS =- 1/CP.[ (∂E/ ∂V. ∂V/ ∂P )T+ ∂ (PV)/ ∂ P)T]………….3 But for an ideal gas ,(∂E/ ∂ V)T=0 Therefore [ (∂E/ ∂V. ∂V/ ∂P )T=0………….4 Also for an ideal gas , PV=constant , at constant temperature. ∂(PV)/ ∂ P)T=0…………….5 Substituting the values of equation (4) and (5) in (3) We get μ=0

THERMODYNAMICS Isothermal Reversible Expansion of Ideal Gas: (Calculation of w, q, ΔE and ΔH) An isothermal process is conducted in such a manner so that temperature remains constant during the entire operation. Consider an ideal gas contained in cylinder fitted with frictional piston, and placed in a thermostat. Let the pressure of gas which is undergoing isothermal expansion by reversible process be P. Let the external pressure be reduced by a very small quantity dP. The gas expand in the cylinder until the internal pressure and external pressure become equal. Suppose the volume dV changes when gas expands. The work done δw by the gas in reversible process is given by

THERMODYNAMICS δw=-(P-dP) dV………………i = -PdV+dP dV…………..ii Product dP.dV being small can be neglected. Equation i become δw= -P dV…………iii Total work done w When volume changes from V1 toV2 is given by v2 w= …………..iv v1 For n moles PV= n RT P=n RT/V, Substitute this value in equation (iv)

THERMODYNAMICS V2 V2 w= =n RT V1 V1 On integrating, w= -n RT ln V2/V1 ……………v w= -2.303 n RT log V2/V1………..vi Since P1V1=P2V2i.e, V2/V1=P1/P2 Equation( vi) become w=-n RT ln P1/P2 w=-2.303 n RT log P1/P2……………..vii

THERMODYNAMICS Expression for ΔE: For ideal gas, energy depends upon temperature only. Since Δ T=o for isothermal process. Hence Expression for q: Δ E=q +w For isothermal Process, Δ E=o Therefore q=-w It shows that work is done at the expense of heat absorbed q=2.303 n RT log V2/V1…………..viii q= 2.303 n RT log P1/P2……………..ix Δ E=o

THERMODYNAMICS Expression for ΔH: ΔH= Δ(E+PV)= ΔE+ Δ(PV) = ΔE + Δ(n RT)= ΔE +n R(ΔT) = ΔE+n R((o) ΔH =o

THERMODYNAMICS Adiabatic Reversible Expansion of Ideal Gas: In adiabatic process δq=0 According to first law of thermodynamics dE=δq+w…………..i We get w=δq If work involved is the work of expansion only, then if dV is small increase in volume and P is pressure of gas. Then w= -P dV……………..ii Putting this value in equation i,we get dE= -P dV…………iii For an ideal gas Cv=dE /dT or dE=Cv dT…………….iv

THERMODYNAMICS Calculation of ΔH: We know that For finite Changes, ΔE =Cv ΔT…………….v Calculation of ΔH: We know that H=E+PV ΔH = ΔE + Δ(PV) = ΔE + Δ(RT) =ΔE+R ΔT = Cv ΔT+ R ΔT =(Cv+R) ΔT ΔH =Cp ΔT……………….vi

THERMODYNAMICS PV=RT or P=RT/V Comparing equation iii and iv we get -P dV=Cv dT……….vii Relationship between temperature and volume: PV=RT or P=RT/V Putting this value in equation vii , we get -Cv dT= RT dV/V Cv DT = -RT dV/V or Cv dT/T = -RdV/V……………..viii Let volume of gas change from V1 to V2 and temperature change from T1 to T2. Assuming Cv to be independent of temperature and integrating equation viii between limits V1,V2 and T1and T2, we get

THERMODYNAMICS T2 V2 Cv = - T1 V1 or Cv ln T2/T1 = R ln V1/V2……….ix We know that Cp-Cv =R Putting this value in equation ix, we get Cv ln T2/T1 = (Cp-Cv) ln V1/V2 Dividing throught by Cv, ln T2/T1 = (Cp/Cv- 1) ln V1/V2 Putting Cp/Cv= γ , ratio of two heat capacities

THERMODYNAMICS Relation between temperature and pressure: ln T2/T1= (γ-1) ln V1/V2 Taking natural log on both side, T2/T1= (V1/V2) (γ-1) ………………x T2V2 (γ-1) = T1V1 (γ-1) Relation between temperature and pressure: P1V1=RT1 and P2V2=RT2 From above equation, we get V1/V2 = P2T1/P1T2 Putting this value in equation x, we get T2/T1 = (P2T1/P1T2) (γ-1) After solving this we get (T2/T1) = (P1/P2) 1- γ …………………….xi Taking γ th root on both side T V γ-1 = Constant TP 1- γ/ γ = constant

THERMODYNAMICS From Equation ix and x , we get (V1/V2) (γ-1) = (P1/P2) 1- γ/ γ …………..xi (V2/V1)1- γ = (P1/P2) 1- γ/ γ V2/V1 =( P1/P2 )1/ γ (V2/V1) γ = P1/P2 P1V1 γ = P2V2 γ PV γ = constant