Priyam Study Centre

A Page about Chemistry and Related Topics.

Jan 29, 2019

Formulation of Kinetic Gas Equation

After knowing the experimental gas lows, it is of interest to develop a theoretical model based on the structure of gases, which can correlated to the experimental facts. Fortunately such theory has been developed and known as the Kinetic Theory of Gases based upon the certain postulates which are supposed to be applicable to an ideal gas.

Postulates of Kinetic Theory:

1.The gas is composed very small discrete particles, now called molecules. For a gas the mass and size of the molecules are same and different for different gases.
2.The molecules are moving in all directions with verity of speeds. Some are very fast while other are slow.
3.Due to random motion, the molecules are executing collision with the walls of the container (wall collision) and with themselves (inter molecular collision). These collision are perfectly elastic and so there occurs no loss of kinetic energy or momentum of the molecules by this collision.
4.The gas molecules are assumed to be point masses, that is their size is very small in comparison to the distance they travel.
5.There exist no inter molecular attraction specially at law pressure, that is one molecule can exert pressure independent of the influence of other molecules.
6.The pressure exerted by a gas is due to the uniform wall collisions. Higher the frequency of the wall collision greater will be the pressure of the gas.
This explains Boyle's Law since when volume is reduced, wall collision becomes more frequent and pressure is increased.
7.Through the molecular velocity are constantly changing due to inter molecular collision, average kinetic energy(є) of the molecules remains fixed at a given temperature.
This explain the Charl's law that when T is increased , velocity are increased, wall collision become more frequent and pressure(P) is increased when T kept constant or Volume(V) is increases when P kept constant.
Root Mean Square Speed:
Root mean square(RMS) speed is defined as the square root of the average of the squares of speeds.
CRMS2 = (N1C12 + N2C22 + N3C32 + ..)/N

Derivation of the Kinetic Gas Equation:

Let us take a cube of edge length l containing N molecules of a gas of molecular mass m and RMS speed is CRMS at temperature T and Pressure P.
What is the Kinetic theory of gases? How can we formulate kinetic gas equation?
Molecular Velocity and its Components.
Let in a gas molecules, 
N1 have velocity C1
N2 have velocity C2
N3 have velocity C3, and so on.
Let us concentrate our discussion to a single molecule among N1 that have resultant velocity C1 and the component velocities are Cx, Cy and Cz.
So that, C12 = Cx2 + Cy2 + Cz2
The Molecule will collide walls A and B with the Component Velocity Cx and other opposite faces by Cy and Cz.
Change of momentum along X-direction for a single collision,
= mCx - (- mCx) = 2 mCx
Rate of change of momentum of the above type of collision,
= 2 mCx × (Cx/l)
= 2 mCx2/l
Similarly along Y and Z directions, the rate of change of momentum of the molecule are 2 mCy2/l and 2 mCz2/l respectively.
Total rate change of momentum for the molecule,
= 2 mCx2/l + 2 mCx2/l +2 mCz2/l 
= 2 (m/l) (Cx2 + Cy2 + Cz2)
= 2 mC12/l
For similar N1 molecules, it is 2 mN1C12/l
Taking all the molecules of the gas, the total rate of change of momentum,
= (2 mN1C12/l)+(2 mN2C22/l)+(2 mN3C32/l)+ ..
= 2 mN {(N1C12 + N2C22 + N3C13 ..)/N}
= 2 mNCRMS2
Where CRMS2 = Root Mean Square Velocity of the Gases.
According to the Newton's 2nd Low of Motion, rate of change of momentum due to wall collision is equal to force developed within the gas molecules.
That is, P × 6l2 = 2 mNCRMS2/l
or, P × l3 = 1/3 mNCRMS2
∴ PV = 1/3 mNCRMS2
Here, l3 = Volume of the Cube Contain Gas Molecules.
The other form of the equation is,
PV = 1/3 mN × mN/V × CRMS2
∴ PV = 1/3 d CRMS2
Where mN/V is the density(d) of the gas molecules.
What is the Kinetic theory of gases? How can we derived kinetic gas equation?
Kinetic Gas Equation
This equation are also valid for any shape of the gas container.
Calculate the pressure exerted by 1023 gas particles each of mass 10-22 gm in a container of volume 1 dm3. The root mean square speed is 105 cm sec-1.
We have, N = 1023
m = 10-22 gm = 10-25 Kg, 
V = 1 dm3 = 10-3 m3 
and CRMS2 = 105 cm sec-1 = 103 m sec-1
Therefore, from Kinetic Gas Equation,
PV = 1/3 mNCRMS2
or, P = 1/3 × mN/V × CRMS2
Putting the value we have,
P=(1/3)(10-25 Kg×1023/10-3 m3)×(103 m sec-1)2
∴ P = 0.333 × 107 Pa
 
 
Expression of Root Mean Square Velocity(CRMS2) from Kinetic Gas Equation:
Let us apply the kinetic equation for 1 mole Ideal Gas.
In that case mN = mN0 = M and PV = RT.
Hence from Kinetic Gas Equation,
PV = 1/3 mNCRMS2
or, RT = 1/3 mNCRMS2
or, CRMS2 = 3RT/M
CRMS2 = √3RT/M
Thus root mean square velocity depends on the molar mass(M) and temperature(T) of the gas.
Calculate the root mean square speed of oxygen gas at 270C.
We know that, CRMS2 = √3RT/M .
Here, M = 32 gm mol-1 ,and T = 270C = (273+27)K = 300 K. 
CRMS2 = √(3 × 8.314 × 107 erg mol-1K-1 × 300 K)/(32 gm mol-1)
= 48356 cm sec-1
 
 
Calculate the RMS speed of NH3 at N.T.P.
At N.T.PV = 22.4 dm3 mol-1 = 22.4 × 10-3 m3 mol-1
P = 1 atm = 101325 Pa
and M = 17 × 10-3 Kg mol-1
Thus, CRMS2 = √3RT/M 
= √(3 × 101325 × 22.4 × 10-3) /(17 × 10-3)
= 632 m sec-1
Expression of Average Kinetic Energy(Ē):
The average kinetic energy(Ē) is defined as,
 Ē = 1/2 m CRMS2.
Again from Kinetic Gas Equation,
PV = 1/3 mNCRMS2
or, PV = (2/3)  N × (1/2) m CRMS2
or, PV = 2/3 N Ē
For 1 mole ideal gas,
PV = RT and N = N0
Thus RT = 2/3 N0 Ē
or, Ē = (3/2)(R/N0)T 
∴ Ē  = 3/2 kT
Where k = R/N0 and is known as the Boltzmann Constant. Its value is 1.38 × 10-23 JK-1.
The total kinetic energy for 1 mole of the gas is,
ETotal = N0 (Ē) = 3/2RT
Thus Average Kinetic Energy is dependent of T only and it is not dependent on the nature of the gas.
Calculate the kinetic energy of translation of 8.5 gm NH3 at 270C.
We know that total kinetic energy for 1 mole of the gas is,
ETotal = (3/2)RT
= (3/2)(2 cal mol-1 K-1 × 300 K)
= 900 cal mol-1
Again 8.5 gm NH3 = (8.5/17) mol = 0.5 mol
Thus Kinetic Energy of 8.5 gm NH3 at 270C is (0.5 × 900) cal
= 450 cal
Calculate the RMS velocity of oxygen molecules having a kinetic energy of 2 K.cal mol-1. At what temperature the molecules have this value of KE ?
T = 673.9K or 400.90C






Jan 27, 2019

Critical Constants of a Gas

Definitions of Critical Constants:

A gas can be liquefied by lowering temperature and increasing of pressure. But influence of temperature is more important. Most of gases are liquefied at ordinary pressure by suitable lowering of the temperature. But a gas can not be liquefied unless its temperature is below of a certain value depending upon the nature of the gas whatever high the pressure may be applied.
This temperature of the gas is called its Critical Temperature (TC). A gas can only be liquefied when the temperature is kept below TC of the gas.
Critical Temperature (TC):
Critical Temperature (TC) is the maximum temperature at which a gas can be liquefied, that is the temperature above which a liquid cannot exist.
Critical Pressure (PC):
Critical Pressure (PC) is the maximum pressure required to cause liquefaction at the temperature (TC).
Critical Volume (VC):
Critical Volume (VC) is the volume occupied by one mole of a gas at critical temperate (TC) and Critical Pressure (PC).

Andrews Isotherms:

In 1869, Thomas Andrews carried out an experiment in which P - V relations of carbon dioxide gas were measured at different temperatures.
What is the Critical Constant of the gases? How can we Express Critical Constants in Terms of Van der Waals Constants?
Andrews Isotherms.
Following are observed from this graph:
At high temperature, such as T4, the isotherms look like those of an ideal gas.
At low temperatures, the curves have altogether different appearances. Consider, for examples, a typical curve abcd.
As the pressures increases, the volume of the gas decreases (curve a to b).
At point b liquefaction commences and the volume decreases rapidly as the gas converted to a liquid with much higher density. This conversion takes place at constant pressure P
At the point C, liquefaction is complete and thus the cd is evidence of the fact that the liquid cannot be easily compressed. Thus, we note that ab represents the gaseous state, bc, liquid and vapour in equilibrium, and cd shows the liquid state only. 
At still higher temperatures we get a similar type of curve as discussed in the above point, except that the horizontal portion is reduced. The Pressure corresponding to this portion being higher then at lower temperatures.
At temperatures TC the horizontal portion is reduced to a mere point. At temperatures higher then TC there is no indication of qualification at all. Thus for every gas, there is a limit of temperature above which it cannot be liquefied, no matter what the pressure is.
Continuity of State:
It appears from the Amagat curve at T there is discontinuity or break during the transformation of gas to liquid. But it is not so.
The continuity of the states from the gas to liquid can be explains from the following isotherm pqrs at T
What is the Critical Constant of the gases? How can we Express Critical Constants in Terms of Van der Waals Constants?
Continuity of State.
The gas at A is heated to B at constant volume (V) along AB.
Then the gas is gradually cooled at constant P along BC, the volume is reduced considerably. The gas is again cooled at constant V until the point D is reached. No where in the process liquid would appear.
At D, the system is highly compressed gas. But from the curve, this point is the representation of the liquid state.
Hence there is hardly a distinction between the liquid state and the gaseous state. There are no line of separation between the two phases. This is known as the principle of continuity of state.
Critical Phenomena and Van der Walls Equation:
For one mole of a gas the Van der Waals equation, 
(P + a/V2)(V - b) = RT
V3 - (b + RT/P)V2 + (a/P)V - ab/P = 0 
This equation has three roots in V for given values of a, b, P and T. It found that either all the three roots are real or one is real and the other two are imaginary.
What is the Critical Constant of the gases? How can we Express Critical Constants in Terms of Van der Waals Constants?
Critical Phenomena and Van der Walls Equation.
The main characteristics of the above isotherm are as follows:
(a) At higher temperature and higher volume region, the isotherms looks much like the isotherms for a Ideal gas.
(b) At the temperature lower than TC the isotherm exhibits a maximum and a minimum. For certain values of pressure, the equation gives three roots of volume as V1, V2 and V3 at pressure P1. The section AB and ED of the Van der walls curve at T1 can be realized experimentally. ED represents Supersaturated or Super cooled vapour and AB represents super heated liquid. Both these states are meta-stable. These states are unstable in the seance that slight disturbances are sufficient to cause the system to revert spontaneously into the stable state with the two phases present in equilibrium.
(c) The section BCD of the Van der Walls isotherms cannot be realized experimentally. In this region the slope of the P - V curve is positive. Increasing or decreasing the volume of such a system would increase or decrease of pressure. The line BCD represents the meta- stable state.
 
 
Expression of Critical Constants in Terms of Van der Waals Constants:
Again with increase of temperature the minimum and maximum points come close to each other and at the critical point (C) both maximum and minimum coalesce. The slope and curvature both becomes zero at this point.
Thus the mathematical condition of critical point is,
(dP/dV)T = 0 and (d2P/dV2)T = 0
Van der Waals equation for 1 mole gas is, 
(P + a/V2)(V - b) = RT 
or, P = RT/(V - b) - a/V2
Differentiating Van der Waals equation with respect to V at constant T,
We get Slope,
(dP/dV)T = - {RT/(V - b)2} + 2a/V3
 And Curvature,
(d2P/dV2)T = {2RT/(V - b)3} - 6a/V4
Hence at the critical point,
- {RTC/(VC - b)2} + 2a/VC3 = 0  
or, RTC/(VC - b)2 = 2a/VC3 
and {2RTC/(VC - b)3} - 6a/VC4 = 0 
or, 2RTC/(VC -b)3 = 6a/VC4 
Thus, (VC - b)/2 = VC/3
VC = 3b
Putting the value of VC = 3b in 
RTC/(VC - b)2 = 2a/VC3
 We have, RTC/4b2 = 2a/27b3
TC = 8a/27Rb
Again the Van der Walls equation at the critical state,
PC = RTC/(VC - b) - a/VC2 
Putting the value of VC and TC,
PC = a/27b2
The Critical constants for water are 647 K, 22.09 MPa and 0.0566 dm3mol-1. Calculate the value of Van der Waals constants a and b.
We have TC = 647 K 
PC = 22.09 Mpa = 22.09 × 103 kPa 
and VC = 0.0566 dm3 mol-1
Thus, b = Vc/3 = (0.0566 dm3 mol-1)/3
∴ b = 0.0189 dm3 mol-1
a = 3 PC VC2 
= 3 (22.09 × 103 kPa)(0.0566 dm3 mol-1)2
∴ a = 213.3 kPa mol-2
Compressibility Factor at the Critical State (ZC):
The Critical coefficient is defined as, 
RTC/PCVC 
Thus the value of Critical Coefficient,
= {R × (8a/27Rb)}/{(a/27b2) × 3b}
= 8/3
= 2.66
Thus the value of Compressibility factor at the critical state (ZC)
= PCVC/RTC
= 3/8
= 0.375
Van der Waals constants in terms of critical constants:
Van der Waals constants can be determined from critical constants TC and PC of the gas. VC in the expression is avoided due to difficulty in its determination. 
We have, b = VC/
but, PCVC/RTC = 3/8
or, VC = (3/8) × (RTC/PC)
∴ b = (1/8)(RTC/PC)
Again, a = PC × 27b2 = 3 × PC × (3b)2
= 3 PCVC2
= 3PC × (3RTC/8PC)2
∴ a = (27/64)(R2TC2/PC)
Calculate Van der Waals constants for Ethylene. (TC = 280.8 K and PC = 50 atm)
a = 0.057 lit mol-1 and b = 4.47 lit2 atm mol-2
 

 

Argon has (TC = - 122°C, PC = 48 atm). What is the radius of the Argon atom?
1.47 × 10-8 cm
Critical Constant Values of some Commonly used Substance
Gas TC (0K) PC (atm.) VC (c.c.)
Oxygen 154.28 49.7 74.4
Hydrogen 33.2 12.8 69.7
Nitrogen 125.97 33.5 90.0
Ammonia 405.5 111.3 72.0
Carbon
dioxide
304.15 72.9 94.2
Helium 5.2 2.25 61.55
Methane 190.3 45.6 98.8






Jan 26, 2019

Slater's Rules

Slater's Rule for Effective Nuclear Charge:

Slater proposed some set of empirical rules to calculate the screening constant (σ) of various electrons present in different orbitals of an atom or an ion.
Once we get the value of screening constant it is easy enough to find the effective nuclear charge (Z*).

Screening Effect:

The Valence electron in a multi - electron atom is attracted by the nucleus and repelled by the electrons of inner-shells.
What is the Slater's rule ?
Shielding Effect
The combined effect of this attractive and repulsive force acting on the valence electron is that the Valence - electron experiences less attraction from the nucleus. This is known as the screening effect.
Slater's Rules to Calculate the Screening Constant (σ):
Rules for an electron in the nS, nP level:
(i) The first to do is to write out the electronic configuration of the atom or the ion in the following order and grouping.
(1S) (2S, 2P) (3S, 3P) (3d) (4S, 4P) (4d) (4f) (5S, 5P) etc.
It may be noted that so far as the screening effect is concerned the S and P electrons belonging to the same principal quantum shell have the same effect as advocated by Slater.
Na atom :
(1S)² (2S, 2P)⁸ (3S)¹
(ii) An electron in a certain (nS, nP) level is screened only by electrons in the same level and by the electrons of lower energy level.
Electrons lying above (nS, nP) level do not screen any (nS, nP) electron to any extent. Higher energy electrons have no screening effect on any lower energy electrons.
Examples:
Estimation of screening constant of the valence electron of the Sodium atom,
(1S)² (2S, 2P)⁸ (3S)¹ 
(The Valence electron will be excluded from our Calculation).
Estimation of screening constant of the 3d electron of the Sodium atom,
(1S)² (2S, 2P)⁸ (3S)¹
 (The one 2P electron and 3S electrons will be excluded from our Calculation).
(iii) Electrons of an (nS, nP) level shield a valence electrons in the same group by 0.35 each. This is also true for the electrons of the nd or nf that is for the electrons in the same group.
(iv) Electrons belonging to one lower quantum shell, that is (n-1) shell shield the valence electrons by 0.85 each.
(v) Electrons belonging to (n-2) or still lower quantum shell shield the valence electron by 1.0 each. This means the screening effect is complete.
 
 
Examples:
(a) For the valence electron of Na atom:
Screening Constant (σ 
= (2 × 1) + (8 × 0.85) + (0 × 0.35) 
8.8
(a) For the 2P electron of Na atom:
Screening Constant (σ ) 
= (2 × 0.85) + (7 × 0.35) 
4.15
Na+ Ion:
  (1s)2(2s, 2p)8
Screening Constant 
σ  = (2×0.85)+(8×0.35)
= 4.5
Effective Nuclear Charge
Z⋆ = (11 - 4.5) = 6.5
 Mg+2 : (1s)2(2s, 2p)8
Screening Constant 
σ  = (2×0.85)+(8×0.35)
= 4.5
Effective Nuclear Charge
Z⋆ = (12 - 4.5) = 7.5

 K+ (1s)2(2s, 2p)8(3s, 3p)8
Screening Constant 
σ  = (2×1.0)+(8×0.85)+(8×0.35)
= 11.6
Effective Nuclear Charge
Z⋆ = (19 - 11.6) = 7.4

Valence Electron of F ion:
  (1s)2(2s, 2p)7
Screening Constant 
σ  = (2×0.85)+(6×0.35)
= 3.8
Effective Nuclear Charge
Z⋆ = (9 - 3.8) = 5.2

F- ion (1s)2(2s, 2p)8
Screening Constant 
σ  = (2×0.85)+(8×0.35)
= 4.5
Effective Nuclear Charge
Z⋆ = (9 - 4.5) = 4.5
Estimate the screening constant for the outermost 4S electron of Vanadium.
Vanadium has atomic number 23 and the electronic configuration according to the Slater's Rule is :
(1S)² (2S, 2P)⁸ (3S, 3P)⁸ (3d)³ (4S)²
We have consider only one electron of the two 4S electrons.
∴Screening Constant (σ
(2 ×1.0) + (8×1.0) + (8×0.85) + (3×0.85) (1×0.35) 
19.7.
Rules for an electron in the nd, nf level:
The above rules are quite well for estimating the screening constant of S and P orbitals. However when a d or f electron being shielded the (iv) and (v) rule replaced by a new rules for estimation of screening constant.
The replaced rules are:
All electrons below the nd or nf level contribute 1.0 each towards the screening constant.
 
 
Example
Screening Constant for a 3d electron of Vanadium:
Electronic Configuration according to Slater's Rule is:
(1S)² (2S, 2P)⁸ (3S, 3P)⁸ (3d)³ (4S)²
We have consider only two electron of the three 3d electrons.
∴Screening Constant (σ
(2 ×1.0) + (8×1.0) + (8×1.0) + (3×0.35)
 = 18.70
Ion 
Electronic Configuration
 Screening 
Constant
(σ)
Effective 
Nuclear Charge
(Z) 
V(II) 
Ion
  (1s)2
(2s, 2p)8
(3s,3p)8
(3d)3
(2×1.0)+(8×1.0)+(8×1.0)+(3×0.35)
= 19.05
Z=
(23-19.05)
= 3.95
Cu(II) - Ion
(1s)2
(2s, 2p)8
(3s,3p)8
(3d)9
(2×1.0)+(8×1.0)+(8×1.0)+(9×0.35)
= 21.15
Z=
(29-21.15)
= 7.85
Find out the Screening Constant of the 4S and 3d electron of Chromium Atom.
Chromium has atomic number 24 and the electronic configuration according to the Slater's Rule is :
(1S)² (2S, 2P)⁸ (3S, 3P)⁸ (3d)⁵ (4S)¹
∴ Screening Constant (σ) for the 4S electron is :
σ =(2×1.0) + (8×1.0) + (8×1.0) 
(3×0.85) (0×0.35) 
21.05.
And the Screening Constant (σ) for the 3d electron is :
σ =  (2×1.0) + (8×1.0) + 
(8×1.0) + (4×0.35) 
= 19.4.
Calculate the effective nuclear charge of the hydrogen atom.
Hydrogen atom has a single 1S valence electron. There is no other electron to screen it from the nuclear charge of single proton.
Thus, σ = 0 and Z* = 1.0 -0 = 1.0
Thus hydrogen electron sees the full nuclear charge of the nucleus, that is the electron is totally exposed to the proton.