# Chapter 4 States of Matter¶

## Example 4_1 pgno:91¶

In [51]:
P=760;#pressure of 14g of nitrogen in mm of Hg#
V=22.4;#Volume occupied by 14g of Nitrogen in litres#
P1=380;#changed pressure of 14g of nitrogen in mm of Hg#
V1=(P*V)/P1;#changed volume of 14g of nitrogen at 380mm pressure in litres#
print'Volume of 14g of Nitrogen at 380 mm Hg pressure=V1=litres',V1
V2=5.6;#changed volume of 14g of nitrogen in litres#
P2=(P*V)/V2;#Pressure of 14g of nitrogen of volume 5.6litres#
print'\nPressure of 14g of Nitrogen of volume 5.6litres=P2=mm Hg',P2

Volume of 14g of Nitrogen at 380 mm Hg pressure=V1=litres 44.8

Pressure of 14g of Nitrogen of volume 5.6litres=P2=mm Hg 3040.0


## Example 4_2 pgno:93¶

In [52]:
V=5.6;#Volume occupied by 8g of Oxygen at 0 C in litres#
T=273;#Temperature at which 8g of Oxygen occupies 5.6litres in Kelvin#
V1=11.2;#Changed volume of 8g of Oxygen in litres#
T1=(V1*T)/V;#Temperature at which 8g of Oxygen occupies 11.6litres in kelvin#
print'Temperature of 8g of Oxygen Occupying 11.6litres=T1=Kelvin=273degrees',T1

Temperature of 8g of Oxygen Occupying 11.6litres=T1=Kelvin=273degrees 546.0


## Example 4_3 pgno:93¶

In [53]:
P=380.;#pressure of 11g of CO2 at 273K in mm of Hg#
T=273.;#Initial temperature of 11g of CO2 in kelvin#
V=11.2;#Volume occupied by 11g of CO2 in litres at 273K#
P1=760.;#changed pressure of 11g of CO2 at 546K in mm of Hg#
T1=546.;#Final temperature of 11g of CO2 in kelvin#
V1=(P*V*T1)/(T*P1);#changed volume of 11g of CO2 at 760mm pressure in litres#
print'Volume of 11g of CO2 at 760 mm Hg pressure at 546K=V1=litres',V1

Volume of 11g of CO2 at 760 mm Hg pressure at 546K=V1=litres 11.2


## Example 4_4 pgno:96¶

In [54]:
P=570./760.;#pressure of 10L of H2 at 300K in Atmospheres#
T=300.;#Temperature of 10L of H2 in kelvin#
V=10.;#Volume occupied by H2 at 300K in litres#
R=0.082;#Value of R in litre-atmospheres#
n=(P*V)/(R*T);#Number of moles of hydrogen present#
print'Number of moles of Hydrogen present=n=mol',round(n,3)

Number of moles of Hydrogen present=n=mol 0.305


## Example 4_5 pgno:98¶

In [55]:
TP=746.;#Total pressure of gas at 298K in mm of Hg#
PP=24.;#Partial pressure of Water vapour at 298K in mm of Hg#
PPG=TP-PP;#partial pressure of gas at 298K in mm of Hg#
print'partial pressure of dry gas at 298k=mm of Hg',PPG
V=200.;#Volume occupied by gas at 298K in millilitres#
P1=760.;
V1=(PPG*V)/P1;#Volume of dry gas at a pressure of 760mm Hg#
print'\nVolume of dry gas at pressure 760mm of Hg=V1=ml',V1

partial pressure of dry gas at 298k=mm of Hg 722.0

Volume of dry gas at pressure 760mm of Hg=V1=ml 190.0


## Example 4_6 pgno:98¶

In [56]:
mA=0.11;#amount of gas A in grams#
MWA=44;#Molecular weight of gas A in grams#
nA=mA/MWA;#No. of moles of gas A#
print'No. of moles of gas A=nA=mol',nA
mB=0.17;#amount of gas B in grams#
MWB=34;#Molecular weight of gas B in grams#
nB=mB/MWB;#No. of moles of gas B#
print'\nNo. of moles of gas B=nB=mol',nB
MFA=nA/(nA+nB);#mole fraction of gas A#
print'\nmole fraction of gas A=MFA=',MFA
MFB=nB/(nA+nB);#mole fraction of gas B#
print'\nmole fraction of gas B=MFB=',MFB
PPA=MFA*759;#Partial pressure of gas A#
print'\nPartial pressure of gas A=PPA=mm of Hg',PPA
PPB=MFB*759;#Partial pressure of gas B#
print'\nPartial pressure of gas B=PPB=mm of Hg',PPB

No. of moles of gas A=nA=mol 0.0025

No. of moles of gas B=nB=mol 0.005

mole fraction of gas A=MFA= 0.333333333333

mole fraction of gas B=MFB= 0.666666666667

Partial pressure of gas A=PPA=mm of Hg 253.0

Partial pressure of gas B=PPB=mm of Hg 506.0


## Example 4_7 pgno:99¶

In [57]:
P1=0.3;#pressure of gas A in atm#
VA=50.;#volume of gas A in ml#
V=500.;#Volume of vessel in ml#
PA=P1*VA/V;#Pressure of gas A in vessel#
print'Pressure of gas A in vessel=PA=atm',PA
P2=0.4;#pressure of gas B in atm#
VB=250.;#volume of gas B in ml#
PB=P2*VB/V;#Pressure of gas A in vessel#
print'\nPressure of gas B in vessel=PB=atm',PB
P=PA+PB;#Total pressure in vessel in atm#
print'\nTotal pressure in vessel=P=atm',P

Pressure of gas A in vessel=PA=atm 0.03

Pressure of gas B in vessel=PB=atm 0.2

Total pressure in vessel=P=atm 0.23


## Example 4_8 pgno:100¶

In [58]:
print'The volume of 1 mole of H2 at 1 atm and 273.15K=22.415Litres'
print'\nThe volume at 0.5atm is obtained by considering the constancy of PV'
P1=1;
V1=22.415;
P2=0.5;
V2=(P1*V1)/P2;#Volume at 0.5atm in litres#
print'\nVolume at 0.5atm of H2 at 273.15K=V2=Litres',V2
MH2=2;#molecular weight of H2#
DH2=MH2/V2;#density of H2 in gram per litre#
print'\nDensity of H2=DH2=gram per litre',round(DH2,4)

The volume of 1 mole of H2 at 1 atm and 273.15K=22.415Litres

The volume at 0.5atm is obtained by considering the constancy of PV

Volume at 0.5atm of H2 at 273.15K=V2=Litres 44.83

Density of H2=DH2=gram per litre 0.0446


## Example 4_9 pgno:101¶

In [59]:
print'One mole of H2 Occupies 22.415Litres at 273.15K'
print'\nThe volume at 300K is obtained by considering the constancy of V/T'
T1=273.15;
V1=22.415;
T2=300.15;
V2=(V1/T1)*T2;#Volume at 300K in litres#
print'\nVolume of 1mole of H2 at 300.15K=V2=Litres',V2
MH2=2;#molecular weight of H2#
DH2=MH2/V2;#density of H2 in gram per litre#
print'\nDensity of H2=DH2=gram per litre',DH2

One mole of H2 Occupies 22.415Litres at 273.15K

The volume at 300K is obtained by considering the constancy of V/T

Volume of 1mole of H2 at 300.15K=V2=Litres 24.6306507414

Density of H2=DH2=gram per litre 0.0811996410896


## Example 4_10 pgno:101¶

In [60]:
print'Density is proportional to molecular weight'
print'\nRelative density of O2 with respect to H2 is 32/2=16.'
t1=12.;#time for certain volume of gas to stream through hole in mins#
p2=16.;#Relative density of O2 w.r.t H2#
t2=15.;#time for same volume(as gas) of oxygen to stream through hole in mins#
p1=((t1/t2)**2)*p2;#according to graham's law#
print'\nRealtive density of gas to oxygen=p1=',p1#here the answer given in textbook is wrongly printed the actual answer is the one we got here through execution#

Density is proportional to molecular weight

Relative density of O2 with respect to H2 is 32/2=16.

Realtive density of gas to oxygen=p1= 10.24


## Example 4_11 pgno:108¶

In [61]:
from math import sqrt,pi
T=320;#47C in kelvin#
R=8.31*10**7;#Universal gas constant in erg per degree per mole#
M=32;#molecular weight of O2 in gram per mole#
C2=(3*R*T)/M;#mean square velocity of Oxygen in (cm/sec)^2#
Crms=sqrt(C2);#Root mean square velocity of Oxygen in cm/sec#
print 'Root mean square velocity of Oxygen=Crms=cm/sec',Crms
Cm=sqrt(8*R*T/(pi*M));#mean velocity of Oxygen in cm/sec#
print '\nMean velocity of Oxygen=Cm=cm/sec',Cm
Cmpv=sqrt(2*R*T/M);#mean probable velocity in cm/sec#
print '\nMean probable velocity of Oxygen=Cmpv=cm/sec',round(Cmpv)#in textbook in Cmpv value it is misprinted as 10^9 but it is actually 10^4#

Root mean square velocity of Oxygen=Crms=cm/sec 49929.9509313

Mean velocity of Oxygen=Cm=cm/sec 46001.3491471

Mean probable velocity of Oxygen=Cmpv=cm/sec 40768.0


## Example 4_12 pgno:109¶

In [62]:
P=10**6;#pressure of gas in dyn per cm^2#
p=0.00333;#density of gas in gram per cm^3#
C2=3*P/p;#mean square velocity of gas in (cm/sec)^2#
Crms=(C2)**0.5;#Root mean square velocity of Oxygen in cm/sec#
print 'Root mean square velocity of Gas=Crms=cm/sec',Crms
Cm=Crms/1.085;#mean velocity of Gas in cm/sec#
print '\nMean velocity of Oxygen=Cm=cm/sec',Cm
Cmpv=Cm/1.128;#mean probable velocity in cm/sec#
print '\nMean probable velocity of Oxygen=Cmpv=cm/sec',round(Cmpv)

Root mean square velocity of Gas=Crms=cm/sec 30015.0112594

Mean velocity of Oxygen=Cm=cm/sec 27663.6048474

Mean probable velocity of Oxygen=Cmpv=cm/sec 24524.0


## Example 4_13 pgno:111¶

In [63]:
print'Mean free path is inversely proportional to N which is proportional to pressure'
L1=5.8*10**-6;#mean free path of N2 at NTP in cm#
K=58;#preesure is raised 58 times so factor is 58#
L2=L1/K;#mean free path of N2 at 58atm pressure in cm#
print'\nMean free path of N2 at 58atm pressure=1*10^-7cm'

Mean free path is inversely proportional to N which is proportional to pressure

Mean free path of N2 at 58atm pressure=1*10^-7cm


## Example 4_14 pgno:113¶

In [64]:
from math import pi,sqrt
T=273;#temperature 0C in kelvin#
R=8.31*10**7;#Universal gas constant in erg per degree per mole#
M=28;#molecular weight of N2 in gram per mole#
print'Since 22.4Litres of Nitrogen gas at 0C and 1atm pressure will contain 6.023*10**23Molecules.'
N=2.69*10**19;#no. of molecules in molecules per cm**3#
G=3.78*10**-8;#collision diameter in cm#
K=sqrt(pi*R*T/M);
Z11=2*N**2*G**2*K;#number of collisions per second of Nitrogen at 0C and 1atm#
print'\nNumber of molecular collisions per second of Nitrogen at 0C and 1atm==10.43*10**28molecular collisions per sec per cm**3',Z11

Since 22.4Litres of Nitrogen gas at 0C and 1atm pressure will contain 6.023*10**23Molecules.

Number of molecular collisions per second of Nitrogen at 0C and 1atm==10.43*10**28molecular collisions per sec per cm**3 1.0432680598e+29


## Example 4_15 pgno:115¶

In [65]:
from math import sqrt,pi
T=273.;#temperature 0C in kelvin#
R=8.31*10**7;#Universal gas constant in erg per degree per mole#
M=28.;#molecular weight of N2 in gram per mole#
print'Since 22.4Litres of Nitrogen gas at 0C and 1atm pressure will contain 6.023*10**23Molecules.'
N=2.69*10**19;#no. of molecules in molecules per cm**3#
Cm=sqrt(8*R*T/(pi*M));#mean velocity of Nitrogen in cm/sec#
print'\nMean velocity of Nitrogen=Cm=cm/sec',Cm
V=22400.;#volume of nitrogen in cm**3#
p=M/V;#Density of nitrogen in gram per cm**3#
print'\nDensity of Nitrogen=p==1.25*10**-3gram per cm**3',p
n=10.99*10**-5;#Viscosity of N2 in poise#
L=(3*n)/(Cm*p);#mean free path of nitrogen in cm#
print'\nMean free path of Nitrogen=L=5.81*10**-6cm'
G=sqrt(1/(1.414*pi*L*N));#Collision diameter of Nitrogen in cm#
print'\nCollission diameter of Nitrogen=G=3.80*10**-8cm'
K=sqrt(pi*R*T/M);
Z11=2*N**2*G**2*K;#number of collisions per second of Nitrogen at 0C and 1atm#
print'\nNumber of molecular collisions per second of Nitrogen at NTP==10.52*10**28molecular collisions per sec per cm**3',round(Z11,3)

Since 22.4Litres of Nitrogen gas at 0C and 1atm pressure will contain 6.023*10**23Molecules.

Mean velocity of Nitrogen=Cm=cm/sec 45422.6927897

Density of Nitrogen=p==1.25*10**-3gram per cm**3 0.00125

Mean free path of Nitrogen=L=5.81*10**-6cm

Collission diameter of Nitrogen=G=3.80*10**-8cm

Number of molecular collisions per second of Nitrogen at NTP==10.52*10**28molecular collisions per sec per cm**3 1.05226395096e+29


## Example 4_16 pgno:120¶

In [66]:
T1=27;#initial temperature in C#
T2=177;#final temperature in C#
print'NH3 being a nonlinear molecule has 3 translational,3 rotational and 6 vibrational degrees of freedom.'
print'\nCv=3*0.5*R+3*0.5*R+6*R=9*R'
Cv=18;#Molar heat capacity in cal per deg per mol#
dU=Cv*(T2-T1);#Change in internal energy of a mole in cal per mole#
print'\nChange in internal energy of a mole of NH3=dU=cal per mole',dU
print'\nThe actual increase in energy may not be 2700cal per mol\nbecause at the given temperature,none or only some of the vibrational degrees of freedom may be contributing to the total energy.'#answer should come as 2700 not 2400,it was just misprinted#

NH3 being a nonlinear molecule has 3 translational,3 rotational and 6 vibrational degrees of freedom.

Cv=3*0.5*R+3*0.5*R+6*R=9*R

Change in internal energy of a mole of NH3=dU=cal per mole 2700

The actual increase in energy may not be 2700cal per mol
because at the given temperature,none or only some of the vibrational degrees of freedom may be contributing to the total energy.


## Example 4_17 pgno:120¶

In [67]:
print'CH4 being a nonlinear molecule has 3 translational,3 rotational and 9 vibrational degrees of freedom.'
print'\nCv=3*0.5*R+3*0.5*R+9*R=12*R'
Cv=24;#Molar heat capacity in cal per degree per mol#
print'\nActual Cv is 16cal per degree per mol'
R=2;
Cvr=16-R;#Real molar heat capapcity in cal per degree per mole#
print'\nReal molar heat capapcity of a mole of CH4=Cvr=cal per degree per mole',Cvr
print'\nThe Real molar heat capacity at constant volume is 10cal less than the theoretical value.\nSince each vibrational degree of freedom can contribute 2cal,this means that 5 vibrational degrees\nare not contributing at the given temperature.'

CH4 being a nonlinear molecule has 3 translational,3 rotational and 9 vibrational degrees of freedom.

Cv=3*0.5*R+3*0.5*R+9*R=12*R

Actual Cv is 16cal per degree per mol

Real molar heat capapcity of a mole of CH4=Cvr=cal per degree per mole 14

The Real molar heat capacity at constant volume is 10cal less than the theoretical value.
Since each vibrational degree of freedom can contribute 2cal,this means that 5 vibrational degrees
are not contributing at the given temperature.


## Example 4_18 pgno:123¶

In [68]:
a=3.60;#Van der Waals constant for CO2 in L^2 atm per mol^2#
b=4.30*10**-2;#Van der Waals constant for CO2 in litre per mol#
N=300.;#No. of moles of CO2 the cylinder contains#
V=100.;#volume of cylinder in litres#
P=150.;#Maximum pressure the cylinder can withstand in atm#
R=0.082;
K1=P+(N**2*a/V**2);
K2=V-(N*b);
T=(K1*K2)/(N*R);#The lowest temperature at which cylinder can explode in Kelvin#
print'The lowest temperature at which cylinder can explode=T=K',round(T,1)

The lowest temperature at which cylinder can explode=T=K 645.8


## Example 4_19 pgno:137¶

In [69]:
from math import pi
tA=280.;#time of flow for liquid A in seconds#
tB=200.;#time of flow for liquid B in seconds#
pA=1.;#density of liquid A in gram per cm**3#
pB=1.1;#density of liquid B in gram per cm**3#
h=10.;#height of liquid responsible for the flow in cm#
g=980.;#gravity constant in dyns#
V=1.;#volume of liquid in ml#
L=10.;#length of the capillary in cm#
r=0.1;#radius of the capillary in cm#
PA=h*pA*g;#Pressure of liquid A#
PB=h*pB*g;#Pressure of liquid B#
nA=(pi*PA*tA*r**4)/(8*L*V);#Viscosity of Liquid A in centipoise#
print'\nViscosity of Liquid A=nA=centipoise',round(nA,1)
nB=(pi*PB*tB*r**4)/(8*L*V);#Viscosity of Liquid B in centipoise#
print'\nViscosity of Liquid B=nB=centipoise',round(nB,1)

Viscosity of Liquid A=nA=centipoise 10.8

Viscosity of Liquid B=nB=centipoise 8.5


## Example 4_20 pgno:142¶

In [70]:
ST=520.;#surface tension of mercury in dyn per cm#
d=13.6;#density of mercury in gram per cm**3#
g=980.;#gravity constant in dyns#
r=0.1;#radius of the capillary tube in cm#
h=(ST*2)/(g*r*d);#Depression observed in capillary tube in cm#
print'Depression observed in capillary tube=h=cm',round(h,3)#the answer is given wrong in textbook,it should actually be double the one given#

Depression observed in capillary tube=h=cm 0.78


## Example 4_21 pgno:146¶

In [71]:
T=293.;#temperature 20C in kelvin#
R=8.31*10**7;#Universal gas constant in erg per degree per mole#
dr=165.;#Change of surface tension in erg per cm^2 per mol litre#
C=0.06;#concentration we are considering in M#
t=(C*dr)/(R*T);#surface excess concentration of phenol in mol per cm^2#
print'surface excess concentration of phenol=mol per cm^2',t

surface excess concentration of phenol=mol per cm^2 4.06599228694e-10


## Example 4_22 pgno:148¶

In [72]:
M=256;#Molecular weight of acid in grams#
w=2.56*10**-5;#weight of palmitic acid in grams#
N=w/M;#No. of molecules of acid#
A=123.;#Total area occupied in cm^2#
AM=A/N;#Area per molecule in cm^2#
print'Area per molecule=20.4*10^-16cm^2.\nThis is the area of crosssection A for a vertical position.'
V=w/0.81;#Volume of acid in cm^3#
VM=V/N;#Volume of one molecule of acid in cm^3#
L=VM/AM;#Length of molecule in cm#
print'\nLength of a molecule=',L

Area per molecule=20.4*10^-16cm^2.
This is the area of crosssection A for a vertical position.

Length of a molecule= 2.56950717655e-07


## Example 4_23 pgno:150¶

In [73]:
T=298.;#Temperature in kelvin#
ST=0.2;#Lowered surface tension in dyn per cm#
K=1.37*10**-16;#Boltzman's constant#
AM=(K*T)/ST;#Area per molecule on the surface in cm**2#
A=300.;#Total area occupied in cm**2#
N=A/AM;#no. of molecules#
W=3*10**-7;#weight of insoluble substance in grams#
w=W/N;#weight of one molecule in grams#
N1=6.023*10**23;
MW=w*N1;#Molecular weight of substance in grams#
print'Molecular weight of substance=MW==123',MW

Molecular weight of substance=MW==123 122.947499


## Example 4_24 pgno:152¶

In [74]:
MW=461.;#molecular weight of lead iodide in  grams#
CR1=3000.;#initial count rate in count per minute#
CR2=900.;#final count rate in count per minute#
CR=CR1-CR2;
print'The count for lead iodide as absorbed=CR=%fcount per minute',CR
print'\nThe ratio of weights of lead iodide and radio lead iodide in solution is equal to that of the same ratio on surface'
print'\nWeight of lead iodide in solution=0.0014grams'
print'\nWeight of radio lead iodide is proportional to the count.\nWeight of lead iodide on the surface=0.0014*21/9.\nMolecular weight of lead iodide=461.\nArea of the surface per gram=4266cm**2 g**-1'

The count for lead iodide as absorbed=CR=%fcount per minute 2100.0

The ratio of weights of lead iodide and radio lead iodide in solution is equal to that of the same ratio on surface

Weight of lead iodide in solution=0.0014grams

Weight of lead iodide on the surface=0.0014*21/9.
Area of the surface per gram=4266cm**2 g**-1


## Example 4_25 pgno:154¶

In [75]:
print'From the linear plot of the langmuir isotherm the intercept Xm=1/0.19=5.26milligrams'
Xm=5.26*10**-3;
print'\nThis is the weight of N2 that forms a unimolecular layer of 2g charcoal.'
MW=28;#molecular weight of N2#
N=Xm*6.023*10**23/MW;#No. of molecules of N2#
TA=N*16*10**-16;#Total area in cm**2#
A=TA/2;#Area per gram in cm**2#
print'\nArea of N2 per gram=cm**2',round(A)

From the linear plot of the langmuir isotherm the intercept Xm=1/0.19=5.26milligrams

This is the weight of N2 that forms a unimolecular layer of 2g charcoal.

Area of N2 per gram=cm**2 90517.0


## Example 4_26 pgno:155¶

In [76]:
print'From the linear plot of the langmuir isotherm the intercept=0.35*10**-3 and slope=9.47*10**-2'
print'\nVolume is the inverse of summation of intercept and slope and that is 10.52cc'
Vm=10.52;#volume in cc#
m=Vm/22400;#No. of moles of N2#
N=m*6.023*10**23;#No. of molecules of N2#
TA=N*16*10**-16;#Total area in cm**2#
A=TA/17.5;#Area per gram in cm**2#
print'\nArea of N2 per gram=cm**2',round(A)

From the linear plot of the langmuir isotherm the intercept=0.35*10**-3 and slope=9.47*10**-2

Volume is the inverse of summation of intercept and slope and that is 10.52cc

Area of N2 per gram=cm**2 25862.0


## Example 4_28 pgno:161¶

In [77]:
d=13.56;#density of mercury in gr per cm**3#
V=1/d;#Volume of mercury in cm**3#
D=0.07*10**-4;#Diameter of the globule in cm#
print'One globule of mercury occpies a volume of 1.33*3.14*r**3cm**3\nSurface area of one globule=4*3.14*r**2'
Vg=1.33*3.14*r**3;#volume of one globule in cm**3#
y=0.0738/Vg;#No of globules in 0.0738 volume#
print'\nNo. of globules in 0.0738cm**3=y=',y
SA=y*4*3.14*r**2;#Surface area of y globules in cm**2#
print'\nSurface area of y globules=SA=cm**2',round(SA)

One globule of mercury occpies a volume of 1.33*3.14*r**3cm**3
Surface area of one globule=4*3.14*r**2

No. of globules in 0.0738cm**3=y= 4.12164838569e+14

Surface area of y globules=SA=cm**2 63416.0


## Example 4_29 pgno:168¶

In [78]:
from math import pi,log
T=300;#temperature in kelvin#
R=8.31*10**7;#Universal gas constant in ergs#
r=2*10**-5;#average radius of particles in cm#
h=4.15*10**-3;#Vertical seperation in cm#
p=1.21;#density of latex in gram per cm**3#
g=980;#gravity constant in dynes#
print'Since the dispersion medium is water its density is p1=1.\nwhen the system equilibrated average number of colloidal particles seen in the field is halved so N0/N=2.\nthe required expression is derived based on kinetic energy of particles'
p1=1;#density of water#
L=log(2)*0.75*(R*T)/(pi*g*h*(p-p1)*r**3);
print'\nValue of Avagadro number is L==6.037*10**23molecules per mol.',round(L,2)

Since the dispersion medium is water its density is p1=1.
when the system equilibrated average number of colloidal particles seen in the field is halved so N0/N=2.
the required expression is derived based on kinetic energy of particles

Value of Avagadro number is L==6.037*10**23molecules per mol. 6.03775763968e+23


## Example 4_30 pgno:169¶

In [79]:
Vf=117.857*10**-9;#Volume falling within field of view in ml#
AN=9.66;#Average no. of particles per count#
N=AN/Vf;#No. of particles per count#
print'No .of particles per ml=N=',N
W=1.5*10**-7;#Weight of iron oxide per ml#
w=W/N;#weight of one particle in grams#
print'\nWeight of one particle=w=1.83*10**-15grams'
d=5.2;#density of iron oxide in g per cm**3#
V=w/d;#Volume of one particle in cm**3#
print'\nVolume of one particle=V=3.52*10**-16cm**3'

No .of particles per ml=N= 81963735.7136

Weight of one particle=w=1.83*10**-15grams

Volume of one particle=V=3.52*10**-16cm**3
Diameter of particle= 2e-05 5.2


## Example 4_31 pgno:177¶

In [80]:
from math import pi
D=78.;#dielectric constant#
V=0.009;#viscosity of suspension in dyn sec per cm**2#
P=3.2;#potential gradient in volt per cm#
t=38.2;
d=0.033;#standard distance between electrodes in cm#
u=90000.;
EP=(4*pi*V*u*d)/(D*t*P);#Electrokinetic potential of the catalyst in volts#
print'Electrokinetic potential of the catalyst=EP=volts',round(EP,4)

Electrokinetic potential of the catalyst=EP=volts 0.0352


## Example 4_32 pgno:178¶

In [81]:
from math import pi
D=80.36;#dielectric constant#
V=0.01;#viscosity of suspension in dyn sec per cm**2#
P=10;#potential gradient in volt per cm#
v=3*10**-3;#observed velocity in cm per sec#
u=90000;
EP=(4*pi*V*u*v)/(D*P);#Zeta potential of the catalyst in volts#
print'Zeta potential of the catalyst=EP=volts',EP
print'\nThe effective thickness of the double layer can be taken to be 1cm'
e=9*10**-4;
r=0.5*10**-4;
N=4*pi*e*r**2;#total no. of charges carried by a particle#
print'\nTotal no. of charges carried by a particle=N=',N

Zeta potential of the catalyst=EP=volts 0.0422215040552

The effective thickness of the double layer can be taken to be 1cm

Total no. of charges carried by a particle=N= 2.82743338823e-11


## Example 4_33 pgno:179¶

In [82]:
from math import pi
D=80.36;#dielectric constant#
V=0.01;#viscosity of suspension in dyn sec per cm**2#
P=100;#potential gradient in volt per cm#
u=90000;
EP=0.03;#Electrokinetic potential of the catalyst in volts#
FW=(P*EP*D*3600)/(4*pi*V*u);#amount of flow of water in cm per hour#
print'Amount of flow of water through diaphragm=FW=cm per hour',round(FW,1)

Amount of flow of water through diaphragm=FW=cm per hour 76.7


## Example 4_34 pgno:179¶

In [83]:
D=80.36;#dielectric constant#
V=0.01;#viscosity of suspension in dyn sec per cm^2#
P=40;#potential gradient in volt per cm#
u=90000;
ZP=0.05;#Zeta potential of the catalyst in volts#
FW=(P*ZP*D*3600*r**2)/(4*V*u);#amount of flow of water in cc per hour#
print'Amount of flow of water through diaphragm=FW=cc per hour',FW

Amount of flow of water through diaphragm=FW=cc per hour 0.4018


## Example 4_35 pgno:189¶

In [84]:
from math import log
C1=5*10**-4;
C2=2.5*10**-4;
x=(C1**2)/(C2+(2*C1));#The amount of Na+ transported at equilibrium in M#
print'The amount of Na+ transported at equilibrium=x=2*10**-4M'
NaR=C2+x;#Na+ on RHS#
print'\nNa+ on RHS=4.5*10**-4M.\nCl- on RHS=x=2*10**-4M.\nNa+ on LHS=Cl- on LHS=C1-x=3*10**-4M'
print'\nE is in volts and F in coulombs.for homogeneity of units either R should be expressed in coulombs or F in calories.F is 96500C or 23060cal.'
F=23060;
n=1;
T=300;#temperature in kelvin#
R=2;
K=1.5;
E=(R*T*log(K))/(n*F);#Emf across membrane in volts#
print'\nEmf across membrane=E=volts',round(E,4)

The amount of Na+ transported at equilibrium=x=2*10**-4M

Na+ on RHS=4.5*10**-4M.
Cl- on RHS=x=2*10**-4M.
Na+ on LHS=Cl- on LHS=C1-x=3*10**-4M

E is in volts and F in coulombs.for homogeneity of units either R should be expressed in coulombs or F in calories.F is 96500C or 23060cal.

Emf across membrane=E=volts 0.0105


## Example 4_36 pgno:190¶

In [85]:
C1=0.04;
C2=0.02;
x=(C1**2)/(C2+(2*C1));#The amount of Na+ transported at equilibrium in M#
print'The amount of Na+ transported at equilibrium=x=M',x
NaR=C2+x;#Na+ on RHS#
print'\nNa+ on RHS=NaR=M',NaR
NaL=C1-x;#Na+ on LHS#
print'\nNa+ on LHS=NaL=M',NaL

The amount of Na+ transported at equilibrium=x=M 0.016

Na+ on RHS=NaR=M 0.036

Na+ on LHS=NaL=M 0.024


## Example 4_37 pgno:210¶

In [86]:
from math import pi,sin
MW=58.45;#Molar weight of NaCl in grams#
d=2.17;#density of NaCl in g/cc#
MV=MW/d;#Molar volume in cc#
print'Molar Volume of NaCl=MV=cc',MV
print'\nThis must contain 6.023*10^23 NaCl units.\nIn one unit cell of NaCl there are 8 corner Na+ ions and 6 on the face centres.\nThe number of Na+ ions in one unit cell is 8*1/8+6*1/2=4.\nThere are 12Cl- ions on the edges,and one in the centre.\nThe number of Cl- ions are 12*1/4+1=4 '
print'\nThe volume of the unit cell containing 4 NaCl units=179*10^-24cc'
a=5.63*10**-8;#unit cell length in cm#
print'\nUnit cell length of NaCl crystal=a=5.63*10^-8cm'
Id=a/2;#Interionic distance in cm#
print'\nInterionic distance in the crystal=Id=2.815*10^-8cm'
Q=5.9;#glancing angle in degrees#
L=2*Id*sin(Q*pi/180)*10**8;#wavelenth in angstrums#
print'\nWavelength of the Xrays used=L=Angstrums',round(L,2)

Molar Volume of NaCl=MV=cc 26.935483871

This must contain 6.023*10^23 NaCl units.
In one unit cell of NaCl there are 8 corner Na+ ions and 6 on the face centres.
The number of Na+ ions in one unit cell is 8*1/8+6*1/2=4.
There are 12Cl- ions on the edges,and one in the centre.
The number of Cl- ions are 12*1/4+1=4

The volume of the unit cell containing 4 NaCl units=179*10^-24cc

Unit cell length of NaCl crystal=a=5.63*10^-8cm

Interionic distance in the crystal=Id=2.815*10^-8cm

Wavelength of the Xrays used=L=Angstrums 0.58


## Example 4_39 pgno:215¶

In [87]:
N=2.;#No. of atoms or unit cells in BCC structure#
a=4.291*10**-8;#Unit cell edge length in cm#
Na=23.;#weight of Na#
p=(N*Na)/(L*a**3);#density of Na at room temperature in g/cm**3#
print'Density of Na at room temperature=p=g/cm**3',p
P=p*1.0398;#density of Na at -195degrees temperature in g/cm**3#
print'\nDensity of Na at -195degrees temperature=P=g/cm**3',P
a1=5.35*10**-8;#unit cell edge length in cm#
N1=(P*L*a1**3)/(Na);#No. of unitcells at -195degrees#
print'\nNumber of unitcells at -195degrees=N1=',round(N1)
print'\nAt -195degrees temperature Na have 4 unitcells which means it assumes an FCC structure.'

Density of Na at room temperature=p=g/cm**3 0.966650377571

Density of Na at -195degrees temperature=P=g/cm**3 1.0051230626

Number of unitcells at -195degrees=N1= 4.0

At -195degrees temperature Na have 4 unitcells which means it assumes an FCC structure.