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Behaviors of gases

Behaviors of gases. Supervised by Dr fawzy. 7.1 Introduction.  The application of thermodynamics is the determination of the equilibrium state of a chemical reaction system requires a knowledge of the thermodynamic properties of the reactants in, and the products of the reaction

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Behaviors of gases

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  1. Behaviors of gases Supervised by Dr fawzy

  2. 7.1 Introduction The application of thermodynamics is the determination of the equilibrium state of a chemical reaction system requires a knowledge of the thermodynamic properties of the reactants in, and the products of the reaction The thermodynamic properties of these individual reactants and products are most expressed by means of their equations of state which relate the thermodynamic properties of interest (e.g. free energy, enthalpy, entropy, etc.) are the operational independent variables (pressure, temperature, and composition). This chapter is concerned with the thermodynamic behavior of the simplest states of existence namely gases

  3. 1.2 The P-V-T relationships of gases • For all gases it is experimentally found that : • V PV/RT=1 (limP→0) • This isotherms plotted on a p-v diagram become hyperbolic as P→0 being given by equation (1.) • PV=RT • A gas which obeys equation 1 over a range of states is said to behave ideally in this range of states and the gas is called a perfect gas • The variation of V with P at several temperatures for a typical real gas is shown in (figure

  4. 1.8.1) This figure shows that as the temperature is decreased , the character of the P-V isotherms changes and eventually 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 ; (δP/δV)=0 At temperature below T(cr) two faces can exist. Consider for example one mole of vapor initially A at T (8); if the vapor is compressed isothermally at t(8),the state of vapor move along the isothermal(ABCD)………; at state B the saturated vapor pressure of this liquid at T (8) is reached. Further decrease in the volume of the system causes condensation of the vapor and consequent appearance of the liquid phase.

  5. At state C the system occurs at 100% liquid phase l Vphase, at the range B-C. Liquid phase at state C is in equilibrium with vapor phase at state B, i.e. at this range we have two phases equilibrium. V(c) is the molar volume of the liquid at T (8) and P (8) P (8) is the saturated vapor pressure at T (8)

  6. 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 increase At 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)

  7. *The deviation from ideality and equation state of Real gases. 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) (2) Figure (3). shows the variation of Z values with pressure for same gases at 0’c; these experimental results shows that at pressures less than 10 atm , the value of Z in linearly related to pressure ; i.e. Z=mP+1 PV/RT=mP+1 *Which can be written as? P (v-mRt) =RT OR P (v-b') = RT Where b'=mRT

  8. 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 p< 10 atm is negative, this means that the constant of purely empirical equation 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 imply the need for a correction to the volume term in the ideal gas equation of state, pv=RT, since an ideal gas comprises a system of non interacting ,volume less particles. the z value become independent of the type of species of z is plotted versus the reduced pressure where Pr=P/Pcm ,for fixed value of reduced temperature ,Tr=T/Tcf figure(4.8.4) Thus if two gases have identical values of two reduced variables, then they have approximately equal values of the third, and the two gases are then paid to be in corresponding stat

  9. **1.4 The van der Waals gas • 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 occupy a finite volume and interactions occur among the particles of a real gas, is the Vander waal equation which for 1 mole of the gas is written as: • (p+a/v2)(V-B)=RT • Where a and b are empirical constant for each type of gas • The wan der waal equation can be rewritten as pv3-(bp+Rt) v2 +v-ab=0 • At tcr we can write • Pcr=r • Thus Tcr 8a/27br, VCR=3b, Pcr=a/27b2 • the critical states,van der waals constans,and the values of z at the critical points for several gases are listed in table (1) • the van der waals equation is cubic in v, thus it has three roots • Plotting v against p for different values of t gives the series of isotherms shown in (figure 5.8.6).

  10. Consider the isothermal p-v line shown in figure 6, when the pressure 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 system. 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 physical significance. 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 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 system. 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 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 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 the gas density, or is equal to a/v2 where a is constant,

  11. 1.5 other equations of state for non ideal gases • Other example of empirical equation of state are the dieterici equation: p(v-b')e =RT • The Berthelot equation: (p+a/tv2) (v-b) =RT • The ones or virual, equation: PV/RT=1+BP+CP2+………. • OR PV/RT=1+B'/V+C'/V2+…….. • 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 function if temperature ,it worth noting that as p— 0 and hence v— ∞ then pv/RT—1 • The virial equation converges in the gas phase ,thus the visual equation of state represents the p-v-t relations for real gases over the entire range of preserves and densities • It worth noting that these equations of state have no real fundamental

  12. 1.6 The Thermodynamic Properties of Ideal Gases &Mixtures of Ideal Gases

  13. 1.6.1The Isothermal Free Energy-Pressure Relationship of an Ideal Gas • The Free Energy of an Ideal gas at Temperature (T) and Pressure (p) is given by: • G = G˚ + R T ln(p) (5) • Where G˚ is the standard free energy of ideal gas at temperature T and pressure equals 1atm of P • Equation 5 is derived from the relation: ∂G =V ∂P∂: At constant T • THUS: G (P2‚T) =G (P1‚T) + R T ln p2/p1

  14. 1.6.2 Mixtures of Perfect Gases • Mole Fraction: • (The mole fraction of constituent I in the mixture is defined as the number of mole of constituent I divided by the total number of moles of all constituents in the system). • Xi = Ni /Σ Ni • Based on this definition: Σ Xi = 1

  15. Dalton’s Law of partial pressures • 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 component gases; the contribution made to the total pressure p by each individual gas is called the partial pressure that gas pi • P=Σ pi • Since the component I of the gas mixture behaves ideally and occupies the total volume of the system • Thus: Pi = Ni RT/V • Thus: P=Σ Pi=RT/VΣ Ni • Hence: Pi/P=Ni/Σ Ni=Xi • *Therefore: Pi=Xi P ) 6)

  16. The Partial Moles Quantities: • The partial moles quantity is given by the relation: • Q‾i= (∂Q‾/∂Ni) T, P, Ni ≠ i • .Where Q represents the extensive property of the system Thus, the partial moles free energy of the constituent I of the system is given by: • G‾i= (∂G‾/∂Ni) T, P, Ni≠i = μi • Where μi is the chemical potential of constituent I in the solution • Since ∂ G‾i= RT ∂ ln Pi , and by integrating this equation, we have • G‾i-Gi˚=RT ln Pi • =RT ln Xi P

  17. The Heat of Mixing of perfect gases Since G‾i-Gi˚=RT ln Xi +RT ln P Then (∂ (G‾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 ,xi Thus: -H‾i/T² =-H˚/T²+zero+zero Or: H‾i=H˚ Therefore: Δ H‾μi=Hi-Hi˚=zero Since Gi˚ is a function only of temperature, by definition, thus Hi is independent of pressure and composition *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 particles. Based on the previous analysis we have: ΔHμ=Σ Xi .ΔH-μ=zero

  18. The free energy of mixing of perfect gases • Since ΔGμ =X1(G-μ1-G1 (int)) +X2(G2-Gμ2 (int)) +……. • =RT [X1ln (P mix/Pint) 1 +X2ln (P mix/Pint) 2+……] • If: (Pint) 1= (Pint) 2=……=P mix=P • And: V1+V2+…..=V mix=V • Then: ΔGμˉ=RT Σ Ni lnXi • Or: ΔGμ=RT Σ Xi lnXi

  19. Since the values of Xi are less than unity, the value of ΔGˉ is negative which means that the mixing is spontaneous process. The entropy of mixing of perfect gases Since Δ Hμ ˉ =0 and ΔGμˉ= Δ Hμ ˉ-T ΔSμˉ, Thus: ΔSμˉ=-R Σ Ni ln Xi Or

  20. For the case of mixing gases such that: V mix = V1+V2+…… = V (Pint) 1= (Pint) 2 =……=P mix= P

  21. 1.7The Thermodynamic Treatment of Imperfect Gases

  22. 1.7.1Fugacity of an Imperfect Gas • For ideal gases, the relation between the free energy of the gas and the logarithm of its pressure is G=G˚+RT ln (P) Linearity, this is a direct result of the ideal gas law. • Since the equations of states of real gases deviate from the ideal gas law, the linearity between the free energy and the logarithm of the gas pressure is not valid anymore, so it is important to define a function of the state of the real gas which when used in the free energy-pressure equation of the pressure ensures linearity between G and the logarithm of this function in any state for any gas.

  23. 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.

  24. Consider now that gas obeys the equation of state, V=RT/P –α and since dg=VdP at constant temperature, we can prove that: F/p=e – α P/RT =1- αP/RT =P/Pi (OR) P=√f Prd * Which means that the actual pressure of the gas is the geometric mean of the fugacity and the pressure? It can also be shown that: d ln f/p= -α/RT dp= (V/RT -1/P) dP = (Z-1 / P) dP 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 free energy of anon ideal gas resulting from an Isothermal Pressure change This can be calculated from either of the following equations: dg=VdP Or dg=RT d ln f=RT dln f/p +RT d lnP The integration of these equation can be carried either

  25. 1.7.3 The effect of pressure in the equilibrium state of van der Waals gases. • The isothermal P-V variation of a van der Waals gas at temperature below the critical temperature. • 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 energy of the system at the other states can be calculated using the following equation: • G (i) =G (A) +∫VdP • =G (A) + (VI+VI+1)/2 *∆P. • The graphical integration of figure 6 is shown table 2 and figure 9 (8.8) is schematic representation of the isothermal G-P variation of Vander Waals gas at a temperature lower than the critical temperature.

  26. 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 stable states. The A, B, C, and States represent the stable states of gases as phase, the D-L points Represent the gas- liquid transformation equilibrium of Vander waals gas, the K, M, V, and O states represents the states liquid states of the concerned Vander waals gas. (Figure 10) represent the P-V isothermals for van der Waals carbon dioxide for which: a=3.59 lit 2/atom, b=0.0427 lit/mole, Tcr=3042k, Pcr=73.0 atm, Vcr= 95.7*10^-3 cm3 /mole as Zcr=0.280

  27. *(Figure11) ref 8.10 represents the ( G–p )variation for van der Waals Co2 at several temperatures *(Figure 12) ref 8.10 shows a companion between the variation with temperatures of the vapor pressure of van der Waals liquid Co2 and the actual vapor pressure of liquid Co2 * For van der waals liquids we have Δ H (evap) =Hv-He=Uv-U2+P(Vv-V2) Since dU=TdS-PdV Thus (əU/əV)t =T(əP/əV)t- P By using Maxwell relation: (əS/əV) t= (əp/əV) v Thus: (əU/əV) t= (əP/əT) v-P Applying van der Waals equation yield: (əU/əV) t=T(R/ V-P) –P=a/V^2 Integration given: ΔU=-a/v +constant Thus ΔH(evap)=-a/Vv +a/VL +P (Vv-Vl) =-a (1/Vv -1/Vl) +P (Vv-Vl) Hence: ΔHevap=0 at T=T critical

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