Thermodynamic Relation

aChapter 3 Thermodynamic RelationsClausis - Clapeyron's Equation ( I Latent Heat Equation)A substance can exist in three states i.e., solid, liquid, gas. Out of these three, only two states can co-exist in equilibrium. The state can change due to change in temperature. This temperature although depends upon pressure, is characteristic of each substance. When the change is from solid to liquid state, the characteristic temperature is called the melting point of the solid, but when the change takes place from liquid to vapor state, the temperature is called boiling point of the liquid. So, the value of melting point and boiling point is specific at particular pressure. DefinitionThe relation showing the variation of melting and boiling point with pressure by using second law of thermodynamics is called Clausius-Clapeyron equation. Theory: Let's consider two isothermal curves ABCD and EFGH at very close temperatures T and T+dT respectively as shown in figure.

Let AB and EF represent the liquid state of substance and GH and CD represents the vapour state of substance. At F and B, the substance is in pure liquid state. As we progress towards FG and BC, the liquid phase changes into vapour so that liquid and vapor co-exist along FG and BC. At C and G, the substance is in purely vapour state. Where P and P+dP represents the saturated vapour pressure of liquid for BC and FG at temperature T and T +dT respectively. Also, V1 and V2 be the volumes of substance at F and G respectively. Now, consider FGNMF be the reversible Carnot engine in which 1gm of working substance is used. Then along FG there is isothermal expansion, along GN there is adiabatic expansion, along CM there goes isothermal compression and finally along MF there is adiabatic compression.So, the amount of heat absorbed along FG is given by,Q1=L+dL [∵mL=lgm×(L+dL)=L+dL]The amount of heat rejected along NM is: Q2=L Then according to theory of Carnot cycle:Q1

T1=Q2

T2or, Q1

Q2= T1

T2or, Q1-Q2

Q2=T1-T2

T2or, (L+dL)-L

L=(T+dT)-T

T [∵T1=T+dT&T2=T]or, dL

L= dT

Tor, dL= L

TdT ..........................(i)Where, 'L: Latent heat of substance at T and L+dT: Latent heat of substance at T+dT. Now the work-done during the cycle FGNMF is given by: aW=Q1-Q2 =(L+dL)-L =dLAgain, the work-done during the Carnot's cycle ean also be calculated by calculating area of FGNMF, So, Workdone (W)=dL = Area of FGNMF =FG× Perpendicular distance between FG and BCor, dL=(V2-V1) dPor, L

TdT=(V2-V1)dP [From equation (i)]or, dP

dT=L

T(V2-V1)dP∴ dP

dT=L

T(V2-V1) … (ii)Where, V1 and V2 are specific volume of liquid and vapour respectively. Here, equation (ii) is the required Clausius-Clapeyron equation showing the variation of boiling point of liquid with pressure. This equation is also valid for the conversion of solid to liquid phase. In this case L represents the latent heat of fusion and T represents the melting point of the substance. V1 and V2 at that case represents the specific volume of solid and liquid phase of the substance, respectively. Importance application of Clausius Clapeyron's relation (I latent heat equation):• (i) Effect of pressure on boiling point of liquid: When a liquid boils, it changes into vapour. So that, the volurne of vapour is always greater than that of liquid. So, V2>V1 and hence (V2-| V1 ) is always positive quantity. This shows that dP

dT is positive This can be interpreted as when pressure increase, the boiling point of liquid increases and vice-versa. Thus, a liquid will boil at lower temperature under reduced pressure.• (ii) Effect of pressure on melting point of solid: When a solid melts, there may be increase in volume as in the case of certain substance like wax and sulphur but there may be decrease in volume as in the case of certain substances like ice, gallium and bismuth.• (a) In case of substances like wax and sulphur: (V2>Vi) is positive quantity, then dP

dT is also positive quantity This means that the melting point of such substantes rises with increase in pressure.• (b) In case of substances like ice, gallium, bismuth: (V2| - V1 ) is negative. So, dP

dT is also negative, which means that the melting point of such substances decrease with increase in pressure.Clausius II - Latent Heat Equation The equation showing the variation of latent heat with temperature in terms of specific heat at two different states is called II latent heat equation or Clausius equation. Let C1 : Specific heat of liquid in contact with its vapour. C2 : Specific heat of saturated vapour in contact with its liquid. Consider 1gm of substance is taken through a cycle BFGCB as shown in figure.

Now, in passing from B to F, amount of heat absorbed =C1dT In passing from F to G, amount of heat absorbed =L+ dL In passing from G to C, amount of heat lost =C2dT In passing from C to B, amount of heat lost =LSo, net amount of heat absorbed in a complete cycle =C1dT+(L+dL)-C2dT-L=(C1-C2)dT+dLThis amount of heat absorbed is equal to work done, which is in turns equal to area of FGNMF (approximation). i.e., dW=dP(V2-V1)=L

TdT...................(ii)From equation (i) & (ii), we get: (C1-C2) dT+dL=L

TdTa

	(C1-C2)+dL dT=L T
∴	C2-C1=dL dT-L T

This is the II-Latent heat equation. Triple Point

Definition: When three curves (like vaporization curve, fusion curve and sublimation (or Hoar frost) curve) of a certain substance are plotted on the same phasor diagram, then they are found to meet in a single point, called triple point. At the triple point, the pressure and temperature are such that solid, liquid and vapour states of the substance can exist simultaneously in equilibrium.Q. Show that trere exixts only one tripple point.

To show the existence of single triple point, consider if possible three curves i.e., (the steam line, ice line and hoar-frost line) do not meet single point but interest enclosing an area ACF as shown in P-T diagram. According to ice line CD, the substance must be entirely solid in the shaded area as it is to the left of CD. According to the steam line AB, the substance must be entirely liquid as it is above AB and according to Hoar frost line EF, the substance in the shaded area must be entirely vapour as it is below EF.But these three conclusions contradict one another and hence shaded triangle ACF cannot exist. That means, the three curves should meet in a single point called the triple point. Hence, there exists only one triple point.Thermodynamic Potentials The state of thermodynamical system can be represented by thermodynamic variables such as pressure P. volume V. temperature T and entropy S. Out of these four; two variables may vary independently and other considered as their functions. The relation between these thermodynamical variables can be provided by combination of Ist and 2nd law of thermodynamics as,a

Q=dU+PdV→ Ist law

and dQ=TdS→2nd law

∴ dU=TdS-PdV

For a complete knowledge of thermodynamic system, certain relations are required and for this purpose, some functions of thermodynamical variables P,V,T and S are introduced; are called thermodynamical potentials (or functions). There are four principal thermodynamical potentials, which are as below:i) Internal or intrinsic energy (U) The total energy of a system, when it is not subjected to the action of external forces; is called the internal energy (U).Expression:According to the Ist and 2 nd law of thermodynamics:dQ=dU+dW=dU+PdV→ Ist law and dQ=TdS→2 nd lawNow, from these equations, we get: dU=TdS-PdV... (i)Where, dW=PdV is the external work done. This equation gives the change in internal energy of the system in terms of four thermodynamical variables P,V,T, S and is called first thermodynamical potential.(ii) Helmholtz's function (F) or Helmholtz free energy (F) Helmholtz's free energy is defined as F=U-TS... (ii)Here, U,T and S are state variables, so F is also state variable and has dimensions of energy.Expression:According to the lst and 2 nd laws of thermodynamics, we get:dU= TdS-dWIf the system is maintained at a constant temperature by exchanging heat with surrounding, then TdS=d(TS).∴ dU=d(TS)-dWor, d(U-TS)=-dWor, dF=-dWWhere, F=U-TS is known as Helmholtz free energy or Helmholtz work function. iii) Enthalpy or total heat H Enthalpy is an extensive thermodynamical property and has particular significance, It is mathematically defined as,H=U+PV ........(iii)Here, U,P and V are state variables; so H is also a state variable. It has same dimension of energy. It is useful function for analysing the volume change during isobaric (i.e., P= Constant) Process.(iv) Gibb's potential (G):Gibb's function is defined as:G= H-TS .................(iv)Where, H=U+PVHere H,T and S are state variables. It has same dimension of energy and is useful to solve the problem involving change in state. If the system undergoes an reversible process, then aG=H-TS =(U+PV)-TSand dG=VdP-SdTSignificance of Thermodynamical Potentials

A mechanical system is in stable equilibrium when potential energy of the system is minimum i.e., mechanical process directed to acquire minimum potential energy. It is therefore, water flows from higher to lower level, heat flows from higher to lower temperature and electric current flows from higher to lower potential. The behaviour of internal energy U, Helmholtz free energy F, enthalpy H and Gibb's function G in thermodynamic is similar to potential energy in mechanics. e.g. The direction of isothermal-isochoric (TV) process is to make Helmholtz free energy minimum; in isothermal-isobaric process (TP), Gibb's energy tends to be minimum; in isobaric-adiabatic (PQ) process, the enthalpy H tends to be minimum. Thus, thermodynamic potentials has significant role to study equilibrium condition of a thermodynamic system. Maxwell's Thermodynamics RelationsMaxwell's four thermodynamic relations between the four thermodynamic variables P,V,T and S can be derived with the help of four thermodynamic potentials (or functions), which are:(i) internal energy (U), (ii) Helmholtz free energy (F), (iii) Enthalpy ( H ) and (iv) Gibb's function (G).(i) Internal energy (U):According to the lst and 2 nd law of thermodynamics; change in internal energy: dU=TdS-PdVHere, the variables S and V are independent. So, taking partial differential of intemal energy U with respect to S and V, we get(𝜕U

𝜕S)V=T and (𝜕U

𝜕V) S=-PThese are the relations which connect the internal energy U with the thermodynamic variables P,V, T and S. Since, U is the state function. So, dU is a perfect differential, then, we have:𝜕

𝜕V(𝜕U

𝜕S)V=𝜕

𝜕S(𝜕U

𝜕V) Sor, (𝜕T

𝜕V)S=-(𝜕P

𝜕S)V ........................(ii) This is Maxwell's first thermodynamical relation.(ii) Helmholtz free energy (F):From the first and second law of thermodynamics, we have:dU=TdS-PdV ..................(i)If we consider the process to be isothermal then:TdS=d(TS)Putting the above relation in equation (i), we get dU=d(TS)-PdVor. dU-d (TS)=-PdVor, d(U-TS)=-PdVor. dF=-PdVor. dF=-dW [a

where, F=U-TS, is called

Helmoltz free energy

]This shows that for reversible changes, the workdone by the system is equal to the decrease in this function F. Now taking Helmholtz function, F=U-TSor, dF=dU-d(TS)or, dF=dU-TdS-SdTWe have, dU=TdS- PdV∴ dF=TdS-PdV-TdS-SdTor, dF=-PdV-SdTHere T and V are independent variables. Taking partial differential of F we get, (𝜕F

𝜕V)T=-P and (𝜕F

𝜕T)V=-SWe know, F is the state function. So, dF is perfect differential. Then, we have: 𝜕2F

𝜕V𝜕T=𝜕2F

𝜕T𝜕Vor, 𝜕

𝜕V(𝜕F

𝜕T)V=𝜕

𝜕T(𝜕F

𝜕V)Tor, -(𝜕S

𝜕V)T=-(𝜕P

𝜕T)V∴ (𝜕S

𝜕V)T=(𝜕P

𝜕T)VThis is the second thermodynamical relation of Maxwell. (iii) Enthalpy (H) It is extensive thermodynamical property mathematically defined as H=U+PV.On differentiation, we getdH=dU+PdV+VdP dH=TdS+VdP (∵dU=TdS-PdV) and it gives the change in enthalpy.Since, S and P are independent variables. Then, taking partial differential of H with respect to S and P, we have:(𝜕H

𝜕S)P=T and (𝜕H

𝜕P)s=VSince, dH is a perfect differential, soa

𝜕 𝜕P(𝜕H 𝜕S)P=𝜕 𝜕S(𝜕H 𝜕S)S

or, (𝜕T 𝜕P)S=(𝜕V 𝜕S)P

… (iii)This is the Maxwell's third thermodynamical relation.(iv) Gibb's function (G) From the definition of enthalpy, we haveH=U+PVor, dH=dU+d(PV)=dU+PdV+VdPBut we have,dU=TdS-PdV ∴ dH=TdS-PdV+PdV+VdPor, dH=TdS+VdPIf the process is isothermal [i.e., TdS=d(TS)] and isobaric (i.e., dP=0 ) then dH=d(TS)or, dH-d(TS)=0or, d(H-TS)=0or, dG=0 [a

where, G=H-TS is called

Gibb's free energy or

Gibb's function

|i.e., G=H-TS= constantFurther, the funetion;G=H-TS=U+PV-TS[∵H=U+PV]=U-TS+PV∴G=U-TS+PV=F+PVThen, dG=dU-d(TS)+d(PV)or. dG=dU-TdS-SdT+PdV+VdPor, dG=TdS-PdV-TdS-SdT+PdV+VdP[∵dU=TdS-PdV]or, dG=-SdT+VdPor, dG=VdP-SdTHere, the independent variables are P and T. Taking partial differentials we get,(𝜕G

𝜕P)T=V, and (𝜕G

𝜕T)P=-SAnd since dG is a perfect differential,so:𝜕2G

𝜕T𝜕P=𝜕 2G

𝜕P𝜕Tor, 𝜕

𝜕T(𝜕G

𝜕P)T=𝜕

𝜕P(𝜕G

𝜕T)Por, (𝜕V

𝜕T) P=-(𝜕S

𝜕P)I∴ (𝜕V

𝜕T)P1=-(𝜕S

𝜕P)TThis is fourth thermodynamical relation of Maxwell. Derivation of Clausius-Clapeyron's equation from Maxwell's relation. We know the Second Maxwell equation as, (𝜕S

𝜕V)T=(𝜕P

𝜕T)VMultiplying on both sides by T, we getT(𝜕S

𝜕V)T=T(𝜕P

𝜕T )VBut from second law of thermodynamics: T\partialS=𝜕QThen we have,(𝜕Q

𝜕V)T=T(𝜕P

𝜕T)VHere, (𝜕Q

𝜕V)T represents the quantity of heat absorbed or librated per unit change in volume at constant temperature. This implies that heat energy must be associated with latent heat with respect to volume change at constant temperature. Let L be the latent heat of substance and V2-V1 be the change in volume. Also we suppose that one gram of substance is used, then we can write:dQ=mL=L and we have dV=V2-V1Thus equation (i) can be written as,a

(L V2-V1)T=T(𝜕P 𝜕T)V

or, (L V2-V1)T=T(𝜕P 𝜕T)V

or, L

V2-V1=TdP

dT[Since, V is constant so partial and full derivative are same]or, dP

dT=L

T (V2-V1)∴ dP

dT=L

T(V2-Vi)This is the Clausius-Clapeyron equation obtained from second Maxwell equation of thermodynamics.