Priyam Study Centre

A Man would do nothing, if he waited until he could do it so well that no one would find fault with what he has done.

Nov 18, 2018

Slater's Rules

Slater's Rules:

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.
Example: 
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

Slater's Rule for calculation of Screening Constant.
Evaluation of the Effective Nuclear Charge
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
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.


Nov 17, 2018

Electron Affinity

Electron Affinity:

Electron Affinity of an element is defined as the amount of energy released when an electron is added to a neutral gaseous atom in its lowest energy state that is, ground state to convert it into uni-negative gaseous ion.
In simple words, the electron affinity of an atom is defined as the energy liberated when a gaseous atom captures an electron.
A (neutral gaseous Atom) + e (electron) A⁻ (gaseous anion) + Electron Affinity (EA)
Evidently electron affinity is equal in magnitude to the ionisation energy of the species formed. The definition of electron affinity as the exothermicity of the electron capture reaction would lead to the negative sign according to our usual thermodynamic convention. But electron affinity is normally describes with positive sign. Negative electron affinities are known. These negative values indicate that the species concerned does not like to have any more electron.

Measurement of Electron Affinity:

Electron affinities are difficult to obtain. These are often obtained from indirect measurements, by analysis of Born - Haber energy cycles in which one step is electron capture or by the direct study of electron capture from heated filaments.
Calculate the electron affinity of chlorine from the Born -  Haber cycle, given the following date : lattice energy = - 774 kJ mol⁻¹ , Ionization Potential of Na = 495 kJ mol⁻¹, heat of sublimation of Na = 108 kJ mol⁻¹, energy for bond dissociation of chlorine (Cl₂) = 240 kJ mol⁻¹ and heat of formation of NaCl = 410 kJ mol⁻¹.
Born - Haber Cycle for formation of NaCl (S) is:
How to determine Electron Affinity
Born - Haber Cycle Applied to NaCl
From the above Born - Haber cycle we can written as:
- UNaCl - INa + ECl - SNa - (1/2) DCl - ΔHf =0
or,  EClUNaCl + INa + SNa + (1/2) DCl +ΔHf
∴ ECl = -774 + 495 + 108 + 120 + 410
= 359 kJ mol⁻¹
Factors Influencing the magnitude of Electron Affinity:
The magnitude of Electron Affinity (EA) is influenced by following factors such as,

(i) Atomic Size.

(ii) Effective nuclear Charge and

(iii) Electronic Configuration.

Atomic Size:

Larger the atomic radius lesser the tendency of the atom to attract the additional electron towards itself and lesser is the force of attraction exerted by the nucleus on the extra electron being added to the valence shell of the atom.
Thus EA values decreases with increases atomic atomic radius.

Effective Nuclear Charge:

Higher the magnitude of effective nuclear charge (Z*) towards the periphery of an atom greater is the tendency of the atom to attract the additional electron electron towards itself. Again greater the force of attraction exerted by the nucleus on the extra electron being added to the valence shell of the atom. As a result higher the energy released when extra electron is added to form an anion.
Thus the magnitude of Electron Affinity (EA) of he atom increases with increasing Z* value.

Electronic Configuration:

The magnitude of electron affinity depends on the electronic configuration of the atom. The elements having nS², nS² nP³, nS² nP⁶ valence shell configuration posses very low electron affinity value due to having their stable valence shell configuration. 
Examples:
(a) Electron Affinity of Be and Mg are -0.60 eV and - 0.30 eV respectively
These elements posses filled S-Orbital in their valence shell which is stable hence Be and Mg do not prefer to except the additional electron in order to form anion. In such cases energy is required to form anion and consequently the electron affinity of Be and Mg posses negative sign with low magnitude.
From the above discussion it is clear that elements of Group - IIA posses low electron affinity vales with negative sign due to stable nS² valence shell electronic configuration. 
Account for the large decrees in electron affinity from lithium to beryllium despite the increase in nuclear charge.

Atomic number and the electronic distribution of lithium and beryllium are:
Li : 3 : 1S² 2S¹
Be : 4 : 1S² 2S²
Lithium has an incompletely filled 2S sub-shell while beryllium has the sub shell filled. Thus lithium can receive an electron in its 2S sub-shell but for beryllium a still higher energy 2P level has to be made of. A filled shell or sub-shell leads to some extra stability. Hence beryllium resists an extra electron more than does lithium.
(b) The elements of Group - VA (group - 15) having nS² nP³ valence shell configuration also posses low electron affinity values
Why electron affinity of Nitrogen is - 0.10 eV while that of Phosphorus is + 0.70 eV ?

Due to smaller size of Nitrogen atom when extra electron is added to the stable half - filled 2P orbital some amount of energy is required and hence electron affinity of nitrogen is negative. 
On the other hand, due to bigger size of P atom in compare to nitrogen less amount of energy released when the extra electron is added to the stable half - filled 3P orbitals and thus electron affinity of Phosphorus is expressed with positive sign.
(c) Electron Affinity data on noble gases are not available.
Complete lack of evidence for stable noble gas anion is sufficient prof of their low electron affinity. An electron, even if accepted by a noble gas atom, would have to go into an orbital of next quantum shell and presumably the nuclear charge not high enough to hold an electron placed so far out.

Periodic Variations:

In a Group:
In a group when move from top to bottom the size of atom generally increases with increasing atomic number and consequently the magnitude of electron affinity decreases in the same direction.
Exceptions: 
There are some exceptions to this general rule as is evident from the following examples:
(a) Although the elements of the second period of the periodic table are relatively smaller in size than those of the third period elements, but the electron affinity values of elements of second period is smaller than the electron affinity values of third period elements.
This unexpected behavior is explained by saying that the much smaller sizes of the second period elements give a very much higher value of charge densities for the respective negative ions. A high vales of electron density is opposes by the inter - electronic repulsion forces. For example, electron affinity of fluorine is lower than that of Chlorine . 
Thus the magnitude of electron affinity of the elements of group 17 decreases in the order :
Cl〉F〉Br〉I
Electron affinity of Cl is greater than that of F - Explain.
The halogen possess large positive electron affinity values indicating their strong tendency to form anions. This is easily understandable because their electronic configurations are only one electron short of the next noble gas element.
The electron affinity of chlorine is greater than that of F. This is probably due to very small size of F atom which results in a very high charge density and consequently strong repulsion between existing valence shell electrons and the entering electron (new coming electron). On the other hand Cl being bigger size , charge density is small and thus such repulsion is not strong enough. Hence the electron affinity Cl is greater than that of F.
In a Period:
In a period, when we move from left to right Z* value towards the new coming electron gradually increases with the increasing atomic number and hence the electron affinity gradually increases in the same direction.
Exceptions:
Exceptional cases may arise in case of the elements having stable nS², nS²nP³ and extraordinarily stable 1S², nS² nP⁶ valence shell configurations.
For example in case of the elements in 2nd period :
Be ,N and Ne the magnitude of electron affinity decreases with increasing atomic number.
A plot of electron affinities of elements up-to Chlorine against atomic number shown as:

What is Electron Affinity
Electron affinities as Function of Atomic Numbers

Nov 10, 2018

Ionization Potential

Ionization Potential:

If enough energy is supplied to an electron of a particular atom, it can be promoted to successively higher energy levels. If the supplied energy is large enough it may possible to remove the electron completely from the pull of the positive nucleus.
The amount of energy required to remove the most loosely bound electron, that is outermost electron from an isolated gaseous atom of an element in its lowest energy state, that is ground state to produce cation is known as Ionization Potential or Ionization Energy of that element.
M (isolated gaseous atom) + Ionisation Energy M⁺ ( gaseous cation) + e (electron)
It is generally represented as I or IP and it is measured in electron volt (eV) or Kilo calories (Kcal) per gram atom.
One electron volt (eV) being the energy gained by an electron falling through a potential difference of one volt.
∴ 1 eV = (charge of an electron) × (one volt)
= 1.6 × 10⁻¹⁹ coulomb volt = 1.6 × 10⁻¹⁹ Joule = 1.6 × 10⁻¹² erg
Calculate the ionization potential of hydrogen atom in eV.
We know that, the energy corresponding to the transition from n = ∞ to n = 1 gives the ionization potential of the hydrogen atom.
Thus the Ionization Potential of the Hydrogen atom = (2π²me⁴/h²) {(1/1²) - 0}
 = 2.179 × 10⁻¹¹ erg = 2.179 × 10⁻¹⁸ Joule 
= (2.179 × 10⁻¹⁸)/(1.6 × 10⁻¹⁹) eV = 13.6 eV
Calculate the second ionization potential of helium (given ionization potential of hydrogen = 13.6 eV)
The ground state electronic configuration of helium is 1S². The second ionization potential means removal of the second electron from the 1S orbital against the nuclear charge of +2.
So we have: Ionization Potential = (2π²me⁴/){(1/n₁²) - (1/n₂²)} = Z² ×
∴ Second Ionization Potential of Helium = 2² × 13.6 = 54.4 eV

Successive Ionization Potentials:

The electrons are removed in stages one by one from an atom. The energy required to remove the first electron from a gaseous atom is called its first Ionization potential. The energy required to remove the second electron from the cation is called second Ionization potential. Similarly we have third, fourth Ionisation potentials.
Thus, M (g) + IP₁ M⁺ (g) + e
M⁺ (g) + IP₂ M⁺² (g) + e
M⁺² (g) + IP₃ M⁺³ (g) + e
The values show that these increase in order:
IP₁ IP₂IP₃IP₄
The successive increase in their values is due to the fact that it is relatively difficult to remove an electron from a cation having higher positive charge than from a cation having lower positive charge or from neutral atom.

Factor Affecting the Magnitude of Ionisation Potential and its Periodic Variation: 

The following factors influence the magnitude of the Ionization Potentials:
(1) The distance of the electron from the nucleus.
(2) The charge on the nucleus, that is, atomic number.
(3) The screening effect of lower lying inner electrons.
(4) Overall charge on the ionizing species.

The distance of the electron from the nucleus:

The greater the distance of an element from the positive charge nucleus of an element, the weaker will be the attraction and hence the lower the ionization potential of that element. If an atom is raised to an exited state by promoting one electron to a higher energy level, then that electron is further from the nucleus and it is observed that this excited electron is more easily detached than when the electron is in its ground state.
Thus on moving from top to bottom in a group the ionization potential of the elements decreases with the increase of their nucleus electron separation.

What is Ionisation Potential
First Ionization Potential in a Group

The Charge on the Nucleus that is, Atomic Number:

The higher the charge on the nucleus the more difficult it is to remove an electron and hence the higher the value of ionization potential.
Thus the value of ionisation potential generally increases in moving from left to right in a period, since the nuclear charge of the elements (atomic number) also increases in the same direction.
What is Ionization Potential
Variation of Ionization Potential along the 2nd Period

With increasing atomic number electrons are added to orbitals of the same principal quantum number. These electrons add little to the size of the atom, the increasing nuclear charge bring about a contraction in size. In effect therefore ionisation potential steadily increases along a period.

Completely - filled and half - filled orbitals:

According to the Hund's rule atoms having half - filled or completely filled orbitals are comparatively more stable and hence move energy is needed to remove an electron from such atom. The ionization potential of such atoms is therefore relatively higher than expected normally from their position in the periodic table.
A few regulation that are seen in the increasing vale of ionization potential along a period can be explain on the basis of the concept of the half - filled and completely filled orbitals.
Example:
Be and N in the second period and Mg and P in the third period have slightly higher vale of ionization potentials than those normally expected.
This is explained on the basis of extra stability of the completely - filled 2S - orbital in Be (2S²) and 3S - orbital in Mg (3S²) and of half - filled 2P - orbital in N (2S²2P³) and 3P - orbital in P (3S²3P³).

The Screening effect of the lower lying inner electrons:

Electrons provide a screening effect on the nucleus. The outermost electrons are shielded from the nucleus by the inner electrons. The radial distribution functions of the S, P, d orbitals shows that for the same principal quantum number the S - orbital is the most penetrating., next is P - orbital and least penetrating is the d - orbital. Screening efficiency falls off in the order S〉P〉d.
As we move down a group, the number of inner - shells increases and hence the ionisation potential tends to decreases.
Elements of II A Group: BeMgCaSrBa
In first transition series electron filling up processes begins in the 3d level below a filled 4S² level. During ionization process will a 4S electron or a 3d electron be lost first ? Explain with reference to Chromium.

Electronic Configuration of Chromium : 1S² 2S² 2P⁶ 3S² 3p⁶ 3d⁵ 4S¹.
Screening constant (σ) for 4S electron is : 
σ = (2 × 1.0) + (8 × 1.0) + (8 × 0.85) + (5 × 0.85)= 21.05.
∴Effective nuclear charge = (24 - 21.05) = 2.95.
Screening constant (σ) for 3d electron is : 
σ = (2 × 1.0) + (8 × 1.0) + (8 × 1) + (4 × 0.35)= 19.4.
∴Effective nuclear charge = (24 - 19.4) = 4.60.
Hence a 3d electron is more tightly held than a 4S electron. So during ionization the 4S electron lost first.

Overall Charge on the Ionizing Species:

An increasing in the overall charge on the ionizing species (M⁺, M⁺², M⁺³, etc) will enormously influence the ionization potential since electron withdrawal from a positive charged species is more difficult than from a neutral atom.
The first ionization potentials of the elements very with their positions in the periodic table. In each of the table the noble gas has the highest value and the alkali metals the lowest value for the ionization potential.
What is Ionization Potential
First Ionization Potentials as Functions of Atomic Number
Again the Ionization Potentials of an elements in a group gradually decreases with increas-ing atomic number.
As for example, LiNaKRbCs
Arrange the following with the increasing order of Ionization Potentials Li, Be, B, C, N, O, F, Ne.
LiBBeCONFNe


Nov 8, 2018

Formulation of Kinetic Gas Equation

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. 

That is, 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. 
Postulates of Kinetic Theory of Gases and Derivation of the 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.

Postulates of Kinetic Theory of Gases and Derivation of the Kinetic Gas Equation

Change of Momentum along X, Y and Z Direction
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 mCy²/l) + (2 mCz²/l
= 2m/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) + .......
Thus Rate of Change of Momentum = 2mN/l{(N₁C₁² + N₂C₂² + N₃C₃²)/N}
 = 2 mNCrms²/l
Where Crms = 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
or, PV(1/3)mNCrms² (Here l³ = Volume of the Cube Containing gas Molecules)
The other from of the equation is P = (1/3)(mN/V)Crms² = (1/3)dCrms²
Where (mN/V) is the density(d) of the gas molecules.
Postulates of Kinetic Theory of Gases and Derivation of the Kinetic Gas Equation
The Kinetic Gas Equation
These 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.
From the given data, we have N = 10²³, m = 10⁻²² gm = 10⁻²⁵ Kg, V = 1 dm³ = 10⁻³ m³ and Root Mean Square(C) = 10⁵ cm sec⁻¹ = 10³ m sec⁻¹.
Therefore, from Kinetic Gas Equation, PV = (1/3)mNC² = (1/3)(mN/V)C²
Putting the value We have P = (1/3)(10⁻²⁵ Kg × 10²³/10⁻³ m³) × (10³ m sec⁻¹)²
or, P =  0.333 × 10⁷ Pa

Expression of Root Mean Square Velocity:

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
RT = (1/3)MCrms²
or, Crms² = 3RT/M
or, Crms = √(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 27°C.
We know that Crms = √(3RT/M).
Here, M = 32 gm mol⁻¹ and T = 27°C = (273+27)K = 300 K.
Thus Crms = √{(3 × 8.314 × 10⁷ erg mol⁻¹K⁻¹ × 300 K)/(32 gm mol⁻¹)} 
= 48356 cm sec⁻¹
Calculate the rms speed of ammonia in N.T.P.
Here V = 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 Crms = √(3PV/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)mCrms².
Again from Kinetic Gas Equation PV = (1/3)mNCrms²
or, PV = (2/3)N{(1/2)mCrms²}
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 = (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) = (3/2)RT.
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 27°C.
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 27°C is (0.5 × 900) cal = 450 cal