COLLISION FREQUENCYAND MEAN FREE PATH
3.10 IDEAL GAS AND REAL GAS
1. Ideal Gas A Gas Which Obeys Boyle’s law, Charles law, etc. under all conditions of temperatureandpressureisknownasanidealgasoraperfect gas. An idealgasmay also be defined as
Agas which obeys the gas equation (PV = nRT) under all conditions of temperature and pressure is calledanideal gas.
Wefind that there is nogaswhich obeys thegaslaws or the gasequation under all conditions of temperature andpressure.Thegasesarefoundtoobey thegaslaws fairly wellifthe pressure is low or the temperature is high.
2.RealGas temperature.
All gasesarerealgases.They show more and more deviations from the gas laws as the pressure is increased or the temperature is decreased.
Deviationof Real Gases from Ideal Behaviour
A realgasisonewhich obeys thegaslaws fairly well under low pressure or high
According to Boyle’slaw,PV=constantatconstanttemperature. This means that the productPV should remain constantatall pressure providedtemperatureis kept constant.Thus, a plotof PVvsP should be a horizontal straight line paralleltothe X-axis.However,inactual practice, itisfound that nogas gives a straight-line plot. Twotypesof curvesaregenerally obtained.
(a) For gases likeCOandCH4, the productPVfirst decreases withincreaseof pressure, reaches a minimumvalue and then beginstoincrease(seeFig. 3.17.)
(b)Forgases like hydrogen and helium, the productPVcontinuouslyincreaseswithincreasesof pressure (seeFig. 3.17.)
O
<3 45
i03 40
55 35
.£ 30
< Ideal Gas
25
j> 20 CO CH4
150 200 400 600 800 1000
Pressure(Atmospheres)
Fig.3.17Plots ofPVvsPfor differentgases at0°C (273K)
3.10.1 Deviation of Real Gases fromBoyle’s and Charles’Law in Terms of Compressibility Factor,z
Theeffect oftemperatureand pressure on the behaviour of agasmay be studiedin termsof quantity
‘z’calledcompressibility factor,which is defined as
Foran ideal gas,z=1atalltemperatureand pressures.In caseof real gases, the factorzvaries from values less than 1to values greater than 1, with changes of temperature and pressure. For example, the plots of compressibility factor zvs. pressure P at different constant temperatures for nitrogengasareshownin(Fig 3.18).Fromthese plots,it is observedthat as thetemperatureincreases, the minimum in the curves shift upwards.Ultimately,atemperatureisreachedatwhich the valueofz remainscloseto1 overanappreciable rangeof pressure. For example, in case of nitrogen gas,at 50°C,the valueofzremainscloseto1uptonearly 100 atmospheres.This temperature atwhicha real gas behaves likeanideal gasover anappreciable pressure range is called Boyletemperature or Boyle point becauseatthistemperatureBoyle'slaw is obeyedover arangeof pressures.Obviously, above the Boyletemperature, agasshows positive deviations only.
z= PV nRT
-50°e
i.a
S Q *C
* £ 1.61.4 50"C
100*c
1 2
IdealGas
1.0 d 0.6
0.6 200 400eooaooiooo Pressure
^ifnj
Fig.3.18Plot o f Z v s P forN2differenttemperatures
Deviations from Charles Gay-Lussac’s Law According to Charles Gay-Lussac’s law,the coefficient of volume expansion of every gas should be same [i.e. 1/273 of its volume at 0°C]
independent of pressure. However,experiments have shown that thisis true onlyatlow pressures.
Appreciable deviationsareobservedathigh pressure.
Example 19 Calculate the volume of 10moles of methane at100atm pressure and0°C.Atthis temperature andpressure,Z=0.75.
Solution:
z - PV
t)RT V = ZnRT
P
0.75 x 10 x 0.0821X273
= 1.68 litres 100
3.10.2 Causes ofDeviation of Real Gas from Ideal Behaviour
Thegaslaws are derivedfrom kinetic theory ofgaseswhich is baseduponcertainassumptions.Thus theremustbe somethingwrongwithcertainassumptions.A carefulstudy shows thatathigh pressure orlowtemperature,the followingtwoassumptions of kinetic theory ofgasesbecome invalid.
1.The volume occupied by the gas molecules is negligibleascomparedtothe total volume of the gas.
2.The forces of attractionorrepulsionbetween the gas moleculesarenegligible.
The abovetwoassumptions are valid onlyifthe pressure is low or the temperature is high so that the distance between the molecules is large.However,ifthe pressure is high or thetemperatureislow, thegasmolecules come close together.
Under these conditions:
1.Theforces of attraction or repulsion between the molecules arenotnegligible.
2.The volumeof thegasissosmall that the volume occupied by thegasmoleculescannotbe neglected.
3.10.3 Derivationofvander Waals Equation for RealGases
van der Waals obtained the equation for real gases by applying the corrections for volume and pressure asexplained below:
1.Correction fortheVolume Supposeuisthe actual volume occupied by one mole of the gas molecules.Then since the gas molecules are in motion, it has been found that the effective volume occupied by the gas moleculesisfour times the actual volume i.e., equalto4U.Letitbe represented byb.The volumebisalso calledco-volumeorexcluded volume. Thus, thefree space available for the movement of the gas moleculesis(V-b).Hence, in the real gas equation,Vshould be replaced by (V-b)[see Fig.3.19].
Actual volume of gasmolecules
•t*
••t (if)
fa) <b>
Fig.3.19(a) Gas molecules aredispersed inthe wholespaceofthe container (b) Actual volume of the gasmolecules
2. Correction forthe Forces of Attraction, i.e. Pressure Correction Agasmolecule lying inthe interior of the gas such as molecule AinFig.3.20 is attracted by all other gas molecules surroundingit. Hence,thenetforceofattractionexertedonsuch a molecule by the other moleculesis zero. However,for amoleculelying near the wall of thecontainer,suchasmolecule BinFig.3.20, the molecules lying inside the bulk of thegasexertsomenetinward pull.Thus,theeffect of such an inward pull is “dragging back” of the molecule. Consequently, the pressure with which the gas molecule strikes the wallof the vesselisless than the pressure that would have been exertedif there werenosuch inward pull.
MoleculeA Molecule3B Fig.3.20intermolecular forces of attraction
Thus, the ideal pressure would begreaterthan the observed pressure by a factorpwhere pisthe
“correctionfactor duetothe inward pull.This inward pull on thegasmolecules lying near the wall depends upon
(a) Number of molecules surrounding a molecule.
(b) Total numberof molecules
Eachof these factors in turn depends upon the density (p)of thegas.Hence,correctionfactor,poc
P2
Butfor a given mass of thegas,
1 V whereVisthe volumeof thegas.
1
or P =7TTa
V2 where aisaconstant, depending upon thenatureof thegas. Thus,
correctedpressure =P+—d
V2
Putting the corrected values of volume andpressureinthe idealgas equation,PV=RT,for1mole of thegas,the equationismodifiedto
(V-b)= R T V2J
This equationisknownas vander Waals equation. Theconstantsaandbarecalledvander Waals constantwhose values depend upon thenatureof thegasandareindependent of thetemperatureand pressure.
Fornmolesof thegas,0oc —ft and effective volume=nb.Hence, thevander Waals equation takes the form V
Fig.3.21Johannes Diderickvan der waals(1837-1923)was a Dutch physicist.He received the Nobel prize in 1910 for his work on the properties of gases and liquids.
3.10.4 Significance ofvan Der Waals Constants a and b
It is foundthat the values of afor the easily liquefiablegasesaregreaterthan those for the so-called
permanentgaseslike H2and He.Moreover,the valueof aincreaseswith the easeof liquefaction of the gas.We know that a gaswhich is more easily liquefiable has greater intermolecular forces of attraction.Hence,‘a’, isa measureof the intermolecular forces of attraction inagas,
*b'isthe‘effective volume’of the gas molecules.The constancy in the value ofbfor any gas over a widerangeof temperature and pressure confirms thatthe gas moleculesareincompressible.
3.10.5 BehaviourofReal GasesUsingvan Der WaalsEquation
1.A LowPressure Atextremely low pressure,Vis very large.Hence,the correction term— v2 is
very small. Similarly, thecorrection term bis also very small as compared to V. Thus, both the correctiontermscanbe neglectedsothat thevander Waals equation reducestoPV=RT.Thisiswhy atextremely low pressure,thegases obey the idealgas equation.
2. At ModeratePressure As the pressure is increased, the volume decreases and hence the factora/V2increases. Thus,thefactora/V2can nolonger be neglected.However, if the pressureis not toohigh,the volumeV isstill sufficiently largesothatbcanbe neglectedincomparison withV. Thus, thevander Waals equation reducesto
p ^V2 V= RT
PV+ —V = RT
PV = RT-—
or
or V
Thus,PVisless thanRTby a factor Asthe pressureincreases, Vdecreases,sothat the factor—
V V
increases.Thus,PVdecreasesasthe pressureisincreased.This explains why a dipinthe plots of PV vs P of realgasesisobtained (seeFig. 3.17).
3.At HighPressure Asthe pressureisincreased furthersothat itisfairly high,Vissosmall that bcan nolonger be neglected in comparison with V.Although under these conditions, the factora/V isquite large butsinceP is very high soa/V2canbe neglectedincomparison withP.Thus, the van der Waals equation reducesto
P(V-b)= RT PV-Pb = RT
PV = RT+P b
or or
Thus, PVis greater than RT by a factor Pb.Now as the pressure is increased, the factor Pb increasesmoreand more.This explains why after the minima in the curves, the product PV increases continuously as the pressure is increased more and more (Fig.3.17).
4.AtHighTemperature Atany given pressure,if thetemperatureissufficiently high, Vis very large so that as in case 1,the van der Waals equation reducestoPV=RT.Hence,athightemperature, realgasbehaves like an idealgas.
3.10.6 Exceptional Behaviour of Hydrogen and Helium
In case of hydrogen and helium, their molecules have very small masses.Hence, in the case of these gases, intermolecular forces of attraction are extremely small even at high pressures.In other words, thefactora/V isnegligible at all pressures.Hence,the van der Waals equation is applicable in the following formatallpressures and ordinarytemperatures.
P(V-b)=RT or PV=RT+Pb
This explains why hydrogen and helium show positive deviation only which increases with increase in die value ofP.
Example20 Two moles of ammonia gas are enclosedin a vessel of 5-litre capacity at 27°C.
Calculate thepressure exerted by the gas.assuming that
(a) The gas behaves like an ideal gas (using ideal gas equation) ( b) The gas behaves like a real gas (usingvander Waals’equation)
Giventhat for ammonia,a =4.17atmlitre2mol~2(or4.17atmdm6mol~2) and b =0.037litre moC1(or0.037dm3moC1)
Solution: Wearegiven that n=2moles
V=5litresordm3
T=27°C=(27 + 273) K=300K
a=4.17 atmlitre2mol-2or atmdm6mole2
b=0.037litre mol-1ordm3mol-1
Weknow that
R=0.0821litreatmK 1mol1atmordm3atmK 1mol1
(a) If the gas behaves like an ideal gas,wehave,
PV =itRT
P = ttRT V
2 x 0.0821 x 300
= P.85atm
5
(b)Ifthegasbehaves like a realgas,weapply,vander Waals equation,i.e.
2\
£7/?
P + {V-nb) -nRT V2
nRT an2
P =
V2 V-nb
2 x 0.0821 x 300 4.17 x (2)2
<5}2
5-2 x 0.031
= P.33atm
Example 21 Calculate the pressureexerted by one mole of carbon dioxide gas in a1.32dm vesselat 48°Cusing thevander Waalsequation.Thevander Waals’constantsare a =3.59dm6atm
mot~2and b=0.0427dm3moC1
Solution: The van der Waals equation can be written as
nRT n2a V ->?b Substituting the values for different quantities, we have
( l mol)( 0.08206 dm3atm fC~lmoPJ )(321 K ) (1 mol)2( 3.5P dm6atm mol”2>
(1.32 dm3)-(1 mol) (0.0427 dm3mol~J)
= 20.62atm-2.06atm= 18.56 atm
P =
V2
P =
(1.32 dm3)2
Example 22 Calculate from thevander Waalsequation the temperature at which 3 moles of SO
wouldoccupya volume of 0.01 m3at 1519875 Nm2pressure (a =0.679Nm4mot 2,b=5.64 x 105 m3mor1,R=8.314JFT1mol-1).
Solution: The van der Waals equation may be writtenas
1 an2
T = — P +
nR (V-n b)
Y2 )
Substituting the values whichare in SI units,
1 0.679 x 3
(0.01-3 x 5.64 x icr5>
7"= 15198754
(0.01)2 j
3 x 8.314
1 X 1580985 x 0.0098308 24.942
= 623.1 K
3.10.7 Kammerlingh-Onnes(Virial) Equation ofStateforRealGases
It is an empirical equation which expresses PV as a power series of pressure P at any given temperature.Itis writtenas
PV=A+BP+CP2+DP3+...
wherePisthe pressureinatmospheres andVisthe molar volumeinlitres.The coefficientsA,B, Cetc.,areknown respectively as the first,second,third,etc.,virialcoefficients.The following points may be noted about the above equation:
1.The virial coefficients are differentfor differentgases.
2.Atvery lowpressure,only the first virial coefficientissignificant andit isequaltoRT,i.e.
A=RT
3.Athigher pressures,the other virial coefficients also become important andmustalso be considered.
4.Forany particulargas,the valuesof A,B, C,etc.,areconstantat constanttemperature.Their values change with change of temperature.The first virial coefficient Aisalways positive and increaseswithtemperature.Onthe other hand,the second virial coefficient B is negativeatlow temperatures.With increaseoftemperature,it increasestozeroand then becomes more and more positive.Thetemperature atwhich B=0iscalledBoyletemperature.Atthistemperature, Boyle’slaw is valid over a fairly wide pressurerange.
The otherform in which the Onnes equation is often written is by expressing the compressibility factor (Z) as a power series in 1/V, i.e.
_ PV
f B C
Z = = 1 + — + —+ _
V2 RT V
B C
PV = RT 1 +-+ — + _
V v2
or
Where the coefficients B, C, etc.(called second, third virial coefficient, etc.)for a particular gas depend only on thetemperatureand not on the pressure.
Example 23 Whatis the molar volume ofN2(g)at 500 K and 600 atm according to (a) the perfect gaslaw,and (b) the virialequation? The virial coefficient B of N2(g)at 500 K is 0.0169 litremot“.
Solution:
(a) Accordingtothe perfect gas law,
RT 0.0821X500
V= = 0.0684litre
P 600
(b) B RT
PV=RT 1 + —V =RT4 V B
PV =RT+PB
RT + PB or
V=
or P
( 0.0821X500
V= RT + B = +0.0169
or P 600
-i
= 0.0684 + 0.0169 = 0.0853 litre mol 3.10.8 Boyle’s Temperature
It is the temperature at which agasbehaves like an idealgas.This temperature is given by T - a
Rb
Boyle’s temperature may be derived from the van der Waal’s equation as follows:
/>+ —V2 )(V-b)= RT
Itmay bewrittenin the form
ab
PV= RT-—a + bp4- ...(3.61)
V2
Asbothaandbaresmall, andif the pressure isnot toohigh so thatVisnotsosmall,ab neglected.
Further, Vin the correctionterma/Vmay be replaced byRT/P.
Then Eq.(3.61) reducesto
canbe V2
PV= RT- aP fbP RT
= RT+ P b-—
RT
Sincethegasbehaves ideallyatBoyle’s temperature,PV=RT
Hencethe second term on RHS should be zero.
SincePhas a finite value,
b— a-= o or T = a RT Rb
orBoyle’s temperature
Tb a . . .(3.62) Rb
3.10.9 SecondVirial Coefficient usingvan der Waals Equation The general equation ofstateas given by Onnes is
PV = RT 1 + —V + T v --4...
The second virial coefficient B of this equation canbe obtained by comparing itwith van der Waals equation
. . .(3.63)
P (V-b)= R T V2 )
which may berewritten intheform
PV = RT - Pbp -P
^V2r- abVa PV = RT + bp~-
V + V2
Asaandbarevery small andif the pressure is not too high,a ^ canbe neglected.Further,Pmay V2
be replaced by R T or
Hence,wegetV
R T a R T+b
V V
PV -RT .M M)
or
Comparing Equation (3.63) and (3.64),weget
B = b — RT £
Thus, the second virial coefficient can be calculated from a knowledge of the van der Waals constantaandbof thegasattemperatureT.
Example 24 T h ev a nd e r Waalsconstant b for CO2is 4.28*102dm3mol1.Assuming the CO2 moleculeto be spherical,calculateits molecular diameter.
Solution: For 1 mole of thegas,
b=4Num
whereNisAvogadro’snumber andUmisthe actual volumeof thegasmolecule.
0.0428
f ) —23 litres
Vm =
4V 4 x 6 . 0 2 2 x 1 0
= 1.777 x 1(T26litres
= 1.777 x 10-23cm3
4 3 4
= — n
But Pm =T3 * ' 3
6 x 1.777 x 10 -23
3.143
K
-24
= 33.92 x 10
a = 3.24 x 10 s cm or 3.24 A ° which gives
3.10.10 OtherEquationsof State
Besidesvander Waals equations and virial equation, a number of other equations have been worked out togive P-V-T relationin caseof realgases.Someof theseare describedas under.In all these gases,the quantity ofgas consideredisonemole.
1.BerthelotEquation Berthelot proposed the following equationtoexplain the behaviour of real gases:
p +-T V A H2 ) < v = RT
Here,the correctiontermfor pressure isa/TV2inplace ofa/V2.
2.DietericiEquation Dietericiproposed the following equation for theP-V-Tbehaviourof real gases:
pea/RTV(V-b)=R T
-a:RTV
P(V - b)= R T e
or
This equation gives satisfactory resultsatmoderate pressure in line with the van der Waals equation.
However,athigher pressures,it gives results ingreater agreementwith experimented data.
3.Radlich-KwongEquation This equation may be written as
P -h a i <V-b)= R T
V(V +b)T2
This equation has been found to be the besttwo-parameterequation of state for real gases.The constantais temperaturedependent.It isbeing frequently used because of its simplicity.
4. ClausiusEquation following equation:
To account for the variation of a with temperature, Clausius used the
P + a {V- b)= R T
T{V + c)2
c isan additional constant.The Clausius equation explains fairly well the behaviour of real gases.