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Showing posts with label PROPERTIES OF GASES. Show all posts
Showing posts with label PROPERTIES OF GASES. Show all posts

Jan 1, 2019

Van der Waals Equation

Van der Waals Equation :

In 1873, Van der Waals modified the Ideal Gas Equation,
PiVi = RT
By incorporating the size effect and inter molecular attraction effect of the real gases. These above two effects are discussing under the Volume Correction and Pressure Correction of the ideal gas equation.

Volume Correction:

Molecules are assumed to be a hard rigid spheres and in a real gas, the available space for free movement of the molecules becomes less then V
let us take the available space for free movement of 1 mole gas molecules , 
Vi = (V - b)
where V is the molar volume of the gas and b is the volume correction factor
Vi = molar volume of the ideal gas where the gas molecules are regarded as point masses. Let us take r is the radius of the molecule and σ = 2r is the diameter using the molecule as a rigid sphere.
How can we derived Ven der Waals Equation?
Representation of Gas Molecules
Then it can be shown that the volume correction term or effective volume of one mole gas molecules,
b = 4 × N0 (4/3) πr3
When two molecules encounter each other the distance between the centers of the two molecules would be σ. They can not approach beyond this distance. Thus the sphere of radius σ will occupy a space unavailable for a pair of molecules.
This excluded volume = 4/3 π σ3 for a pair of molecules.
Thus the effective volume for a single molecule,
= 1/2 × 4/3 π σ3
The effective volume for Avogadro number of molecules (present in 1 mole gas),
How can we derived Ven der Waals Equation?
The effective volume for Avogadro number of molecules
Thus b measures and helps to calculate the radius ( or diameter) of the gas molecule.
The equation becomes, 
Pi(V - b) = RT

Pressure Correction:

Pressure of the gas is developed due to the wall collision of the gas molecules. But due to inter-molecular attraction, the colliding molecules will experience an inward pull and so pressure exerted by the molecules in real gas will be less than that if there had not been inter molecular attraction as in ideal gas Pi
Thus, Pi 〉P
or, Pi = P + Pa
Where Pa is the pressure correction term originating from attractive forces. Higher the inter molecular attraction in a gas, greater is the magnitude of Pa. Pressure depends on both the frequency of molecular collisions with the walls and the impulse exerted by each collision. Both contributions are diminished by the attractive forces and the strength with which these operate is roughly proportional to the density of the molecules. Therefore the average pressure exerted by the molecules is decreased by Pa is proportional to the square of the density.
Pa ∝ 1/V2  Since, density ∝ 1/V 
Pa = a/V2 
Where a = constant for the gas that measures the attractive force between the molecule. Thus, Pi = P + (a/V2
Using the two corrections we have Van der Waals equation for 1 mole real gas is, 
(P + a/V2 )(V - b) = RT
To convert the equation for n moles volume has to change as it is the only extensive property in the equation. 
Let V be the volume of n moles gas hence, V = v/n. Putting this term in the Van der Waals equation, we have,
what is the Van der Waals equation?
Van der Waals equation for n Moles Real Gas

Calculate the pressure of 2 moles of Nitrogen gas occupying 10 lit volume at 27°C. Given, a = 1.4 atm lit² mol⁻² and b=0.04 lit mol⁻¹ using Van der Waals equation. 
Also calculate the pressure of the gas using ideal gas equation and find the extent of deviation from ideal behavior.

P = 4.904 atm, Pi = 4.92 atm and Deviation=0.325%

Let us consider two hypothetical cases to show the size effect and attraction effect on the pressure of the gas:

(a)For the real gas a=0 (that is no inter molecular attraction ) but b≠0 (size considers):

We have Van der Waals Equation,
(P+ an²/V²)(V - nb) = nRT but a=0
Hence, P = nRT/(V - nb) ) 〉Pi
Since, Pi = nRT/V only.

(b)For the real gas a≠0 (existence of attractive force) but b=0 (no size consider):

It means that the molecular size effect (repulsive interaction) creates higher pressure then that observed by the ideal gas where molecules have no volume. 
We have Van der Waals Equation, 
(P+ an²/V²)(V - nb) = nRT
 but b=0(no size). 
 Hence, P = nRT/V - (an²/V²ㄑPi 
Since, Pi = nRT/V only.
 Thus, inter molecular attraction effect reduces the pressure of the real gases.

Units of a and b:

From the Van der Waals Equation, 
(P+ an²/V²)(V - nb) = nRT 
where, Pa = an²/V²
Where Pa is the pressure correction term, called often the internal pressure of the gas. 
Then, a = Pa × (V²/n²)
Thus the unit of a = atm lit² mol⁻²
Again, nb = Unit of volume (say liter) 
Hence, the unit of b = lit mol⁻¹

Find the Units and Dimensions of  Van dar Walls constant 'a' and 'b'
from Van der walls Equation.

In SI system,unit of 'a' 
= N m⁻² m⁶ mol⁻² 
N m⁴ mol⁻²
In CGS system,unit of 'a' 
= dyne cm⁻² cm⁶ mol⁻² 
= dyne cm⁴ mol⁻²
In SI system,unit of 'b' = m³ mol⁻¹
In CGS system,unit of 'b' = cm³mol⁻¹

Significance of a and b:

a’ term originates from the inter molecular attraction and Pa = an²/V² . 
Thus, ‘a’ is a measure of internal pressure of the gas and so the attractive force between the molecules. Higher the value of ‘a’ grater is the inter molecular attraction and more easily the gas could be liquefied.
Thus, a of CO₂ = 3.95 atm lit² mol⁻² while, a of H₂ = 0.22 atm lit² mol⁻².
Another constant, ‘b’ measures the molecular size and also a measure of repulsive forces. The value of b can also be utilized to calculate the molecular diameter(σ). 
The grater the value of ‘b’ larger the size of the gas molecule.
Thus, b of CO₂ = 0.04 lit mol⁻¹ while, b of H₂ = 0.02 lit mol⁻¹.

Calculation of Boyle Temperature (TB):

Mathematical condition for Boyle’s temperature is,
TB = [d(PV)/dP]T When P0
From the Van der Waals Equation for 1 mole gas,
(P + a/V²)(V - b) = RT
or, P = (RT/V - b) - a/V²
Hence, PV = {RTV/(V - b)} - a/V
Thus, TB = [d(PV)/dP]T 
= [RT/(V-b)-{RTV/(V-b)²}+a/V²][(dV/dP)]T
= [{RT(V - b) - RTV/(V - b)²} + a/V²] [(dV/dP)]T
=[{- RTb/(V - b)²} + a/V²] [(dV/dP)]T
But when T = TB
[d(PV)/dP]T = 0 and [(dV/dP)] ≠ 0
Thus We have, 
RTBb/(V - b)² = a/V²
or, T= (a/Rb) {(V - b)/V}² 
Since P → 0, V is large.
Thus, (V - b)/V ≃ 1 
Hence, TB = a/Rb

Calculate the Boyle temperature of nitrogen gas 
given a=1.4 atm lit² mol⁻² and b=0.04 lit mol⁻¹.

TB = 427 K

Comment on the possibility of defining a Boyle Temperature if a=0 and a=b=0.

When, a = 0, Van der Waals Equation, P (V - b) = RT 
or, PV = RT + Pb 
Differentiating with respect of pressure at constant temperature,
[d(PV)/dP]T = b, but b ≠ 0 hence [d(PV)/dP]T ≠ 0
The gas has no Boyle temperature since at any temperature since at any temperature, [d(PV)/dP]T ≠ 0
Again when a = b = 0, Van der Walls equation becomes PV = RT
or, [d(PV)/dP]T = 0
That is at any temperature this becomes zero and so all the temperature are Boyle Temperatures.

Explanation of Amagat’s Curves:

We have Amagat Curves,
How can we Explain Amagat Curves By Ven der Walls Equation?
(a) Amagat Curve For Different Gases
Van der Waals equation for 1 mole real gas, 
(P + a/V₂ )(V - b) = RT 
or, PV - Pb + (a/V) + (a/V₂) = RT 
Neglecting the small term (a/V₂), 
we get , PV = RT + Pb - (a/V
Using ideal gas equation for small term, a/V = aP/RT and taking Z = (PV/RT),
We have, 
Z = 1 + (1/RT){b - (a/RT} P
This Shows that, Z = (T,P)
This equation can be uses to explain Amagat curve qualitatively at low pressure and moderate pressure region.
(a) For CO₂a’ is very high since the gas is easily liquefied. 
What is the Ven der Waals Equation?
(b) Amagat Curve For CO₂
Thus (a/RT) 〉b in the equation and {b - (a/RT)} = - ve . That is slope of Z vs P curve is ( -) for carbon dioxide at moderate pressure region. That is the value of Z decreases with increase of Pressure and it is also found in the curve of carbon dioxide.
For H₂a’ is very small and It is not easily liquefied.
 so a/RT〈 b and slope of Z vs P curve for H₂ becomes (+) ve and the value of Z increases with pressure.
(b) (i) When T〈 Tв that is T〈 a/Rb or,b〈 a/RT and {b−(a/RT)} = (-)ve. 
That is the value of Z decreases with increases with P at the moderate pressure region of the curve of CO₂ , Z〈 1 and more compressible. 
(ii) When T = Tв = a/Rb or, b = a/RT hence, {b - (a/RT)} = 0 That is Z = 1 the gas shows ideal behavior. The size effect compensates the effect due to inter molecular attraction of the gas. Z runs parallel to P axis up to certain range of pressure at low pressure region. 
(iii) When T 〉TB, that is T 〉a/Rb or, b 〉a/RT and {b - (a/RT} = (+)ve. 
That is value of Z increases with increase of P when T 〉TB. The size effect dominates over the effect due to inter molecular attraction. For hydrogen and helium, 0°C is grater then their TB values and Z vs P slope becomes (+)ve. At very low pressure P→0 and high temperature,the volume is very large and both the size effect and attraction effect becomes negligible that is Pb and (a/RTP are negligible and Z=1 that is the gas obeys ideal behavior.

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. 
C2RMS =(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/l2 mCy2/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 + N3C32 .....)/N
= 2 mNCRMS2
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, × 6l2 = 2 mNCRMS2/l
or, P × l3 = 13 mNCRMS2
or, PV = 13 mNCRMS2
Here l3 = Volume of the Cube Contain Gas Molecules. 
The other form of the equation is PV = 13 mN × mNV  × CRMS2 
or, PV = 13 dCRMS2 Where mNV 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 102gas 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 = 13 mNCRMS2 
or, P = 13 × mNV × CRMS2
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 = 13 mNCRMS2 
RT = 13 mNCRMS2 
or, CRMS2 = 3RTM
or, CRMS = √3RTM 
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 = √3RTM .
Here, M = 32 gm mol⁻¹ and T = 27°C = (273+27)K = 300 K.
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 = √3RTM = (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 =12 CRMS2 .
Again from Kinetic Gas Equation, PV = 13 mNCRMS2
PV = 23 N × 12mCRMS2
or, PV = 23 N Ē
For 1 mole ideal gas PV = RT and N = N₀
Thus RT = 23N₀ Ē
or, Ē = (32)(RN₀)T = 32kT
Where k = RN0 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) = 32RT.
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