Transcript 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
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
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





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.
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)
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)
*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

P (v-b') = RT
PV/RT=mP+1
*Which can be written as?
P (v-mRt) =RT
OR
Where b'=mRT




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
pv=RT, since an ideal gas comprises a system
equation of state,
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




**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 +vab=0
At tcr we can write
Pcr=r
VCR=3b, Pcr=a/27b2
Thus Tcr 8a/27br,
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).










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,
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







1.6 The Thermodynamic
Properties of Ideal Gases
&Mixtures of Ideal Gases
1.6.1The Isothermal Free EnergyPressure 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:
: At constant T∂G =V ∂P
∂
THUS:
G (P2‚T) =G (P1‚T) + R T ln
p2/p1





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 
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
Pi = Ni

P=Σ Pi=RT/VΣ

Thus:
RT/V
Thus:
Ni
Hence:
Pi/P=Ni/Σ Ni=Xi
6) (

*Therefore:
Pi=Xi P

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








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 

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 
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 
For the case of mixing gases such 
that:
V mix = V1+V2+…… = 
V
(Pint) 1= (Pint) 2 =……=P mix= P 
1.7The Thermodynamic
Treatment of Imperfect
Gases
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 energypressure equation of the pressure ensures linearity
between G and the logarithm of this function in any state
for any gas.
garithm of this function in any state for any gas
alled fugacity, f, and is partially defined by the equation:
dG = RT d ln (T)
F/p — 1 as P—
G=G+RT ln (T)
e energy of the gas in its attendant state which now defined as that
=1 and the temperature of concern.
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














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
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






*(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=TdSPdV
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













