Behaviors of gasesBehaviors of gasesSupervised by Supervised by

Dr fawzyDr fawzy

7.17.1 IntroductionIntroduction The application of thermodynamics is the determination of The application of thermodynamics is the determination of the equilibrium state of a chemical reaction system the equilibrium state of a chemical reaction system requires a knowledge of the thermodynamic properties of requires a knowledge of the thermodynamic properties of the reactants in, and the products of the reactionthe reactants in, and the products of the reaction

The thermodynamic properties of these individual The thermodynamic properties of these individual reactants and products are most expressed by means of reactants and products are most expressed by means of their equations of state which relate the thermodynamic their equations of state which relate the thermodynamic properties of interest (e.g. free energy, enthalpy, entropy, properties of interest (e.g. free energy, enthalpy, entropy, etc.) are the operational independent variables (pressure, etc.) are the operational independent variables (pressure, temperature, and composition)temperature, and composition)..

This chapter is concerned with the thermodynamic This chapter is concerned with the thermodynamic behavior of the simplest states of existence namely gasesbehavior of the simplest states of existence namely gases

1.21.2 The P-V-T relationships of The P-V-T relationships of gasesgases

For all gases it is experimentally found thatFor all gases it is experimentally found that: : V PV/RT=1 (limP→0)V PV/RT=1 (limP→0)This isotherms plotted on a p-v diagram become This isotherms plotted on a p-v diagram become

hyperbolic as P→0 being given by equation (1.)hyperbolic as P→0 being given by equation (1.)

PV=RTPV=RTA gas which obeys equation 1 over a range of A gas which obeys equation 1 over a range of

states is said to behave ideally in this range of states is said to behave ideally in this range of states and the gas is called a perfect gasstates and the gas is called a perfect gas

The variation of V with P at several temperatures for a The variation of V with P at several temperatures for a typical real gas is shown in (figuretypical real gas is shown in (figure

1.8.11.8.1 ( (This figure shows that as the temperature is decreased , This figure shows that as the temperature is decreased , the character of the P-V isotherms changes and eventually of the character of the P-V isotherms changes and eventually of T=T(crit) is reached at which at same fixed values of T=T(crit) is reached at which at same fixed values of P=P(critical) and V=V(critical) an inflexion occurs in the isotherm P=P(critical) and V=V(critical) an inflexion occurs in the isotherm

; (δP/δV)=0; (δP/δV)=0

At temperature below T(cr) two faces can existAt temperature below T(cr) two faces can exist..Consider for example one mole of vapor initially A at T Consider for example one mole of vapor initially A at T

(8); if the vapor is compressed isothermally at t(8),the (8); if the vapor is compressed isothermally at t(8),the state of vapor move along the isothermal(ABCD)………; state of vapor move along the isothermal(ABCD)………; at state B the saturated vapor pressure of this liquid at T at state B the saturated vapor pressure of this liquid at T

(8) is reached(8) is reached . .Further decrease in the volume of the system causes Further decrease in the volume of the system causes

condensation of the vapor and consequent appearance condensation of the vapor and consequent appearance of the liquid phaseof the liquid phase . .

At state C the system occurs at 100% liquid At state C the system occurs at 100% liquid phase lphase l

Vphase, at the range B-C. Liquid phase Vphase, at the range B-C. Liquid phase at state C is in equilibrium with vapor phase at state C is in equilibrium with vapor phase at state B, i.e. at this range we have two at state B, i.e. at this range we have two phases equilibriumphases equilibrium..

V(c) is the molar volume of the liquid at T V(c) is the molar volume of the liquid at T (8) and P (8)(8) and P (8)

P (8) is the saturated vapor pressure at T P (8) is the saturated vapor pressure at T (8)(8)

The compressibility factor of the gas, − (δP/δV) t is higher than the compressibility factor of the liquid phase, − (δP/δV) t

)Figure 1 (show also that as the temperature increase, the molar volume of the liquid increase, the molar volume of gas decrease,

and the saturated vapor pressure of the gas increaseAt T(cr), the molar volumes of the coexistent phases coincide

Above T (cr) distinct two-phase equilibrium does not occur and the gases state cannot be liquefied by isothermal compression alone.

)Figure 2.8.2 :(shows the phase regions ; liquification of a gas require that the gas be cooled as shown in the process (1→2) shows in figure(2)at which the

temperature of the gas should be cooled to T which is less than T(cr)

**The deviation from ideality and equation state of Real The deviation from ideality and equation state of Real gasesgases..

Let Z to be a factor that measured the deviation of a real gas Let Z to be a factor that measured the deviation of a real gas from ideality ; the Z factor is defined as : Z=(PV / RT) from ideality ; the Z factor is defined as : Z=(PV / RT)

(2) (2)Figure (3). shows the variation of Z values with pressure for Figure (3). shows the variation of Z values with pressure for

same gases at 0’c; these experimental results shows that at same gases at 0’c; these experimental results shows that at pressures less than 10 atm , the value of Z in linearly related pressures less than 10 atm , the value of Z in linearly related

to pressure ; i.e. Z=mP+1to pressure ; i.e. Z=mP+1 PV/RT=mP+1PV/RT=mP+1 **Which can be written asWhich can be written as?? P (v-mRt) =RTP (v-mRt) =RTOROR P (v-b') = RT Where b'=mRTP (v-b') = RT Where b'=mRT

it should be noted that Z=Z(P,T,type of gaseous special) and that b' it should be noted that Z=Z(P,T,type of gaseous special) and that b' is a positive value through the slope of the z-p relation in the range is a positive value through the slope of the z-p relation in the range

p< 10 atm is negative, this means that the constant of purely p< 10 atm is negative, this means that the constant of purely empirical equationempirical equation

the fact that the particles of a real gas occupy a finite volume and the fact that the particles of a real gas occupy a finite volume and the fact that interactions occur among the particles of a real gas the fact that interactions occur among the particles of a real gas

imply the need for a correction to the volume term in the ideal gas imply the need for a correction to the volume term in the ideal gas equation of state,equation of state,pv=RT, since an ideal gas comprises a system of pv=RT, since an ideal gas comprises a system of

non interacting ,volume less particlesnon interacting ,volume less particles..the z value become independent of the type of species of z is the z value become independent of the type of species of z is

plotted versus the reduced pressure where Pr=P/Pcm ,for fixed plotted versus the reduced pressure where Pr=P/Pcm ,for fixed value of reduced temperature ,Tr=T/Tcf figure(4.8.4)value of reduced temperature ,Tr=T/Tcf figure(4.8.4)

Thus if two gases have identical values of two reduced variables, Thus if two gases have identical values of two reduced variables, then they have approximately equal values of the third, and the two then they have approximately equal values of the third, and the two

gases are then paid to be in corresponding statgases are then paid to be in corresponding stat

****1.41.4 The van der Waals gasThe van der Waals gasthe most celebrated equation of state of non ideal gases, which was the most celebrated equation of state of non ideal gases, which was

derived on the basis of the two facts that the particles of a real gas derived on the basis of the two facts that the particles of a real gas occupy a finite volume and interactions occur among the particles of a occupy a finite volume and interactions occur among the particles of a

real gas, is the real gas, is the Vander waal equation which for 1 mole of the gas is Vander waal equation which for 1 mole of the gas is written aswritten as : :

) ) p+a/v2p+a/v2)()(V-BV-B=(=(RTRT Where a and b are empirical constant for each type of gasWhere a and b are empirical constant for each type of gasThe wan der waal equation can be rewritten as pv3-(bp+Rt) v2 +v-The wan der waal equation can be rewritten as pv3-(bp+Rt) v2 +v-

ab=0ab=0At tcr we can writeAt tcr we can write Pcr=rPcr=rThus Tcr 8a/27br,Thus Tcr 8a/27br,VCR=3b, Pcr=a/27b2VCR=3b, Pcr=a/27b2the critical states,van der waals constans,and the values of z at the the critical states,van der waals constans,and the values of z at the

critical points for several gases are listed in critical points for several gases are listed in table (1)table (1) the van der waals equation is cubic in v, thus it has three rootsthe van der waals equation is cubic in v, thus it has three rootsPlotting v against p for different values of t gives the series of Plotting v against p for different values of t gives the series of

isotherms shown in (figure 5.8.6)isotherms shown in (figure 5.8.6) . .

Consider the isothermal p-v line shown in figure 6, when the pressure Consider the isothermal p-v line shown in figure 6, when the pressure exerted on the system is increased, the volume of the system exerted on the system is increased, the volume of the system decreases, that id (dp/dv) <this is a condition of intrinsic stability of the decreases, that id (dp/dv) <this is a condition of intrinsic stability of the systemsystem..

It can be seen also that in figure (6.8.7), the condition of (dp/dv) <0 is It can be seen also that in figure (6.8.7), the condition of (dp/dv) <0 is violated over the portion JHF; this portion of the curve thus has no violated over the portion JHF; this portion of the curve thus has no physical significancephysical significance..

by considering(figure 7a) as it can be shown that the correction term by considering(figure 7a) as it can be shown that the correction term for the finite volume of the particles, is four times the volume of all for the finite volume of the particles, is four times the volume of all particles present,ie b=4/3*4 r3 *n where r is the radius of the gas particles present,ie b=4/3*4 r3 *n where r is the radius of the gas particles and n is the number of particles present in the systemparticles and n is the number of particles present in the system..

since the decrease in the exerted pressure by the gas on the wall of since the decrease in the exerted pressure by the gas on the wall of containing vessel in proportional to the number of the particles it the containing vessel in proportional to the number of the particles it the surface layer and the number of particles the 'next-to-the surface surface layer and the number of particles the 'next-to-the surface layer' due to the interaction between the two layers as shown in (figure layer' due to the interaction between the two layers as shown in (figure 7b) And both of these quantities are proportional to the density of the 7b) And both of these quantities are proportional to the density of the gas (n/v). thus the net inward pull is proportional to the sequence of gas (n/v). thus the net inward pull is proportional to the sequence of the gas density, or is equal to a/v2 where a is constantthe gas density, or is equal to a/v2 where a is constant,,

1.51.5 other equations of state for non other equations of state for non ideal gasesideal gases

Other example of empirical equation of state are the dieterici Other example of empirical equation of state are the dieterici equation: equation:

p(v-b')e =RT p(v-b')e =RT The Berthelot equation: (p+a/tv2) (v-b) =RTThe Berthelot equation: (p+a/tv2) (v-b) =RT The ones or virual, equation: PV/RT=1+BP+CP2The ones or virual, equation: PV/RT=1+BP+CP2.………+.………+ OR PV/RT=1+B'/V+C'/V2OR PV/RT=1+B'/V+C'/V2++

..……..……Where B or B' is termed the first visual coefficient, c or c' is termed Where B or B' is termed the first visual coefficient, c or c' is termed

the second visual coefficient, etc...and the these coefficient are the second visual coefficient, etc...and the these coefficient are function if temperature ,it worth noting that as p— 0 and hence v— function if temperature ,it worth noting that as p— 0 and hence v—

∞ then pv/RT—1∞ then pv/RT—1The virial equation converges in the gas phase ,thus the visual The virial equation converges in the gas phase ,thus the visual

equation of state represents the p-v-t relations for real gases over equation of state represents the p-v-t relations for real gases over the entire range of preserves and densitiesthe entire range of preserves and densities

It worth noting that these equations of state have no real It worth noting that these equations of state have no real fundamentalfundamental

1.6 The Thermodynamic 1.6 The Thermodynamic Properties of Ideal Gases Properties of Ideal Gases &Mixtures of Ideal Gases&Mixtures of Ideal Gases

1.6.1The Isothermal Free Energy-1.6.1The Isothermal Free Energy-Pressure Relationship of an Pressure Relationship of an

Ideal GasIdeal GasThe Free Energy of an Ideal gas at Temperature (T) and The Free Energy of an Ideal gas at Temperature (T) and

Pressure (p) is given byPressure (p) is given by:: G = G˚ + R T ln(p) G = G˚ + R T ln(p)

(5)(5) Where G˚ is the standard free energy of ideal gas at Where G˚ is the standard free energy of ideal gas at

temperature T and pressure equals 1atm of Ptemperature T and pressure equals 1atm of P

Equation 5 is derived from the relation: Equation 5 is derived from the relation: ∂∂G =V ∂PG =V ∂P ∂∂: At constant T: At constant T

THUS: G (P2‚T) =G (P1‚T) + R T ln THUS: G (P2‚T) =G (P1‚T) + R T ln p2/p1p2/p1

1.6.21.6.2 Mixtures of Perfect GasesMixtures of Perfect Gases

Mole FractionMole Fraction : :

) ) The mole fraction of constituent I in the The mole fraction of constituent I in the mixture is defined as the number of mole mixture is defined as the number of mole

of constituent I divided by the total number of constituent I divided by the total number of moles of all constituents in the systemof moles of all constituents in the system .( .(

Xi = Ni /Σ NiXi = Ni /Σ Ni

Based on this definition: Σ Xi = 1Based on this definition: Σ Xi = 1

Dalton’s Law of partial Dalton’s Law of partial pressurespressures

It states that the pressure p excited by a mixture of perfect gases is It states that the pressure p excited by a mixture of perfect gases is equal to the sum of the pressure excited by each of the individual equal to the sum of the pressure excited by each of the individual

component gases; the contribution made to the total pressure p by component gases; the contribution made to the total pressure p by each individual gas is called the partial pressure that gas pieach individual gas is called the partial pressure that gas pi

P=Σ piP=Σ pi

Since the component I of the gas mixture behaves ideally and Since the component I of the gas mixture behaves ideally and occupies the total volume of the systemoccupies the total volume of the system

Thus: Pi = Ni RT/VThus: Pi = Ni RT/V

Thus: P=Σ Pi=RT/VΣ NiThus: P=Σ Pi=RT/VΣ Ni

Hence: Pi/P=Ni/Σ Ni=XiHence: Pi/P=Ni/Σ Ni=Xi

**Therefore: Pi=Xi P Therefore: Pi=Xi P ) ) 66 ( (

The Partial Moles QuantitiesThe Partial Moles Quantities : :The partial moles quantity is given by the relationThe partial moles quantity is given by the relation::QQ‾‾i= (∂Qi= (∂Q‾‾/∂Ni) T, P, Ni ≠ i/∂Ni) T, P, Ni ≠ i..Where Q represents the extensive property of the system Thus, Where Q represents the extensive property of the system Thus,

the partial moles free energy of the constituent I of the system is the partial moles free energy of the constituent I of the system is given bygiven by : :

GG‾‾i= (∂Gi= (∂G‾‾/∂Ni) T, P, Ni≠i = μi/∂Ni) T, P, Ni≠i = μi

Where μi is the chemical potential of constituent I in the solutionWhere μi is the chemical potential of constituent I in the solution

Since ∂ GSince ∂ G‾‾i= RT ∂ ln Pi , and by integrating this equation, we i= RT ∂ ln Pi , and by integrating this equation, we

havehave GG‾‾i-Gi˚=RT ln Pii-Gi˚=RT ln Pi ==RT ln Xi PRT ln Xi P

The Heat of Mixing of perfect gasesThe Heat of Mixing of perfect gases Since GSince G‾‾i-Gi˚=RT ln Xi +RT ln Pi-Gi˚=RT ln Xi +RT ln P

Then (∂ (GThen (∂ (G‾‾i/T)/ ∂ (I/T))P, Xi = (∂ (G˚/T)/ ∂ (1/T))p, xi +R(∂ ln Xi/ ∂(1/T))p ,xi+ i/T)/ ∂ (I/T))P, Xi = (∂ (G˚/T)/ ∂ (1/T))p, xi +R(∂ ln Xi/ ∂(1/T))p ,xi+ R(∂ ln P/∂(1/T))p ,xiR(∂ ln P/∂(1/T))p ,xi

Thus: -HThus: -H‾‾i/T² =-H˚/T²+zero+zeroi/T² =-H˚/T²+zero+zero Or: HOr: H‾‾i=Hi=H ˚ ˚ ThereforeTherefore: Δ H: Δ H‾‾μi=Hi-Hi˚=zeroμi=Hi-Hi˚=zero

Since Gi˚ is a function only of temperature, by definition, thus Hi is Since Gi˚ is a function only of temperature, by definition, thus Hi is

independent of pressure and compositionindependent of pressure and composition **The zero value of heat of mixing of an ideal gas is a consequence of the The zero value of heat of mixing of an ideal gas is a consequence of the

fact that ideal gases are assemblies of no interacting particlesfact that ideal gases are assemblies of no interacting particles . .

Based on the previous analysis we haveBased on the previous analysis we have : :

ΔHμ=Σ Xi .ΔH-μ=zeroΔHμ=Σ Xi .ΔH-μ=zero

The free energy of mixing of perfect The free energy of mixing of perfect gasesgases

Since ΔGμ =X1(G-μ1-G1 (int)) +X2(G2-Gμ2 Since ΔGμ =X1(G-μ1-G1 (int)) +X2(G2-Gμ2 (int))(int)) .……+ .……+

= = RT [X1ln (P mix/Pint) 1 +X2ln (P RT [X1ln (P mix/Pint) 1 +X2ln (P mix/Pint) 2+……]mix/Pint) 2+……]

If: (Pint) 1= (Pint) 2=……=P mix=PIf: (Pint) 1= (Pint) 2=……=P mix=P

And: V1+V2+…..=V mix=VAnd: V1+V2+…..=V mix=V

Then: ΔGμˉ=RT Σ Ni lnXiThen: ΔGμˉ=RT Σ Ni lnXi

Or: ΔGμ=RT Σ Xi lnXiOr: ΔGμ=RT Σ Xi lnXi

Since the values of Xi are less than unity, the Since the values of Xi are less than unity, the value of ΔGˉ is negative which means that the value of ΔGˉ is negative which means that the

mixing is spontaneous processmixing is spontaneous process . .

The entropy of mixing of perfect gasesThe entropy of mixing of perfect gases

Since Δ Hμ ˉ =0 and ΔGμˉ= Δ Hμ ˉ-T ΔSμˉ, Since Δ Hμ ˉ =0 and ΔGμˉ= Δ Hμ ˉ-T ΔSμˉ, ThusThus : :

ΔSμˉ=-R Σ Ni ln Xi OrΔSμˉ=-R Σ Ni ln Xi Or

For the case of mixing gases such thatFor the case of mixing gases such that : :

V mix = V1+V2+…… = VV mix = V1+V2+…… = V

))PintPint ( (1= (Pint) 2 =……=P mix= P1= (Pint) 2 =……=P mix= P

1.7The Thermodynamic 1.7The Thermodynamic Treatment of Imperfect Treatment of Imperfect

GasesGases

1.7.1Fugacity of an Imperfect 1.7.1Fugacity of an Imperfect GasGas

For ideal gases, the relation between the free energy of For ideal gases, the relation between the free energy of the gas and the logarithm of its pressure is G=G˚+RT ln the gas and the logarithm of its pressure is G=G˚+RT ln

(P) (P) Linearity, this is a direct result Linearity, this is a direct result

of the ideal gas lawof the ideal gas law . .Since the equations of states of real gases deviate from Since the equations of states of real gases deviate from

the ideal gas law, the linearity between the free energy the ideal gas law, the linearity between the free energy and the logarithm of the gas pressure is not valid and the logarithm of the gas pressure is not valid

anymore, so it is important to define a function of the anymore, so it is important to define a function of the state of the real gas which when used in the free energy-state of the real gas which when used in the free energy-

pressure equation of the pressure ensures linearity pressure equation of the pressure ensures linearity between G and the logarithm of this function in any state between G and the logarithm of this function in any state

for any gasfor any gas . .

Between G and the logarithm of this function in any state for any gas

This function is called fugacity, f, and is partially defined by the equation :

dG = RT d ln (T) and

F/p — 1 as P— 0

Therefore G=G+RT ln (T)

Where G is the partial free energy of the gas in its attendant state

which now defined as that state at which f=1 and the temperature of concern .

Consider now that gas obeys the equation of state, V=RT/P –α and since Consider now that gas obeys the equation of state, V=RT/P –α and since dg=VdP at constant temperature, we can prove thatdg=VdP at constant temperature, we can prove that::

F/p=e – α P/RTF/p=e – α P/RT= = 11 - -αP/RT =P/Pi (OR) P=√f PrdαP/RT =P/Pi (OR) P=√f Prd * *Which means that the actual pressure of the gas is the geometric mean of Which means that the actual pressure of the gas is the geometric mean of

the fugacity and the pressurethe fugacity and the pressure?? It can also be shown thatIt can also be shown that : : d ln f/p= -α/RT dp= (V/RT -1/P) dPd ln f/p= -α/RT dp= (V/RT -1/P) dP) = ) = Z-1 / PZ-1 / P ( (dPdP Thus (ln f/p) → p = p = ∫0Z-1 /P dP (8)Thus (ln f/p) → p = p = ∫0Z-1 /P dP (8)The variation of the ratio (f/p) which P for N2 at 0С˚ is shown in( figure.8.12)The variation of the ratio (f/p) which P for N2 at 0С˚ is shown in( figure.8.12)The free energy of anon ideal gas resulting from an Isothermal Pressure The free energy of anon ideal gas resulting from an Isothermal Pressure

changechange This can be calculated from either of the following equationsThis can be calculated from either of the following equations : : dg=VdPdg=VdPOr dg=RT d ln f=RT dln f/p +RT d lnPOr dg=RT d ln f=RT dln f/p +RT d lnP The integration of these equation can be carried The integration of these equation can be carried

eithereither

1.7.31.7.3 The effect of pressure in the The effect of pressure in the equilibrium state of van der Waals equilibrium state of van der Waals

gasesgases..The isothermal P-V variation of a van der Waals gas at The isothermal P-V variation of a van der Waals gas at

temperature below the critical temperaturetemperature below the critical temperature..By considering the states A, B, C.D.F.E.G.H.I.J.K, L, M, By considering the states A, B, C.D.F.E.G.H.I.J.K, L, M,

N.O and taking state A as the reference point, the free N.O and taking state A as the reference point, the free energy of the system at the other states can be calculated energy of the system at the other states can be calculated

using the following equationusing the following equation:: G (i) =G (A) +∫VdPG (i) =G (A) +∫VdP

= = G (A) + (VI+VI+1)/2 G (A) + (VI+VI+1)/2

*∆P*∆P . .The graphical integration of figure 6 is shown table 2 and The graphical integration of figure 6 is shown table 2 and

figure 9 (8.8) is schematic representation of the isothermal figure 9 (8.8) is schematic representation of the isothermal G-P variation of Vander Waals gas at a temperature lower G-P variation of Vander Waals gas at a temperature lower

than the critical temperaturethan the critical temperature..

Since the state of the lower free energy at the same Since the state of the lower free energy at the same pressure is the most stable states A,B,C,D,,L,M,N,Oas the pressure is the most stable states A,B,C,D,,L,M,N,Oas the

stable statesstable states..The A, B, C, and States represent the stable states of The A, B, C, and States represent the stable states of

gases as phase, the D-L pointsgases as phase, the D-L pointsRepresent the gas- liquid transformation equilibrium of Represent the gas- liquid transformation equilibrium of

Vander waals gas, the K, M, V, and O states represents the Vander waals gas, the K, M, V, and O states represents the states liquid states of the concerned Vander waals gasstates liquid states of the concerned Vander waals gas..

))Figure 10Figure 10 ( (represent the P-V isothermals for van der Waals represent the P-V isothermals for van der Waals carbon dioxide for whichcarbon dioxide for which::

a=3.59 lit 2/atom, b=0.0427 lit/mole, Tcr=3042k, Pcr=73.0 a=3.59 lit 2/atom, b=0.0427 lit/mole, Tcr=3042k, Pcr=73.0 atmatm,,

Vcr= 95.7*10^-3 cm3 /mole as Zcr=0.280Vcr= 95.7*10^-3 cm3 /mole as Zcr=0.280

)*)*Figure11Figure11 ( (ref ref 8.108.10 represents the ( G–p )variation for van der Waals Co2 at several represents the ( G–p )variation for van der Waals Co2 at several temperaturestemperatures

)*)*Figure 12Figure 12 ( (ref ref 8.108.10 shows a companion between the variation with temperatures of shows a companion between the variation with temperatures of the vapor pressure of van der Waals liquid Co2 and the actual vapor pressure of the vapor pressure of van der Waals liquid Co2 and the actual vapor pressure of

liquid Co2liquid Co2 * *For van der waals liquids we have Δ H (evap) =Hv-He=Uv-U2+P(Vv-V2)For van der waals liquids we have Δ H (evap) =Hv-He=Uv-U2+P(Vv-V2) Since dU=TdS-PdVSince dU=TdS-PdV

ThusThus ) ) əU/əVəU/əV((t =T(əP/əV)t- Pt =T(əP/əV)t- P By using Maxwell relation: (əS/əV) t= (əp/əV) vBy using Maxwell relation: (əS/əV) t= (əp/əV) v Thus: (əU/əV) t= (əP/əT) v-PThus: (əU/əV) t= (əP/əT) v-PApplying van der Waals equation yield: (əU/əV) t=T(R/ V-P) –P=a/V^2Applying van der Waals equation yield: (əU/əV) t=T(R/ V-P) –P=a/V^2Integration givenIntegration given : : ΔU=-a/v +constantΔU=-a/v +constantThus ΔH(evap)=-a/Vv +a/VL +P (Vv-Vl)Thus ΔH(evap)=-a/Vv +a/VL +P (Vv-Vl)-= -= a (1/Vv -1/Vl) +P (Vv-Vl)a (1/Vv -1/Vl) +P (Vv-Vl)Hence: ΔHevap=0 at T=T criticalHence: ΔHevap=0 at T=T critical