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Notes on State of Matter

193 / June-06-2021 23:09:49 By- srijan /

Gaseous State

Postulates of kinetic theory of gases :

The kinetic and molecular behaviour of gases is theoretically explain by the assumption of kinetic theory of gases. The main points of kinetic theory of gases are :
1. All gases are made up of large number of tiny particles called molecules.
2. The molecules of particular gas are similar to all respects such as shape, size, volume, energy but the molecules of different gases are different in all respect.
3. The gas molecules are in constant rapid motion in all direction colliding with each other or against the inert wall of the container.
4. The collision between the gas molecules is perfectly elastic. Thus, there is no change in momentum when they collide with each other or against the inner wall of the container.
5. The presence exerted by a gas is due to the bombardment of gas molecules on the inner wall of container or bombardment due to their collision.
6. The average kinetic energy of the gas molecules is directly proportional to the absolute temperature.
7. The volume occupied by a gas molecule is negligible as compared to the total volume of gas.
8. There is no intermolecular force of interaction (attraction or repulsion) between the gas molecules.
9. There is negligible effect of gravity on the movement of gas molecules.

GAS LAWS:

1. Boyle's Law: It state's that," the volume of given amount of gas is inversely proportional to its pressure at constant temperature".

i.e. V ∝ 1/P [at constant temperature in kelvin scale.]
V= K/P [K is proportionality constant]
therefore, PV= K
therefore, PV= constant.
Thus, the product of pressure and volume of gas is always constant at constant temperature.

Let, P1 and V1 be the initial pressure and volume of gas respectively. Then from Boyle's law,
P1V1= constant <------- (i)

If volume is changed to V2 and corresponding pressure to P2 in final condition, then from boyle's law
P2V2= constant <------- (ii)

therefore from equation (i) and equation (ii)
P1V1=P2V2

a. P against V graph:


When pressure of gas (P) against volume of gas (V) is plotted a hyperbolic curve is obtained which indicates that the pressure of gas is inversely proportional to its volume at constant temperature.

b. P against 1/V graph:


When pressure of gas (P) against 1/V is plotted as a straight line passing from the origin is obtained which indicates that the pressure of gas is directly proportional to the inverse of volume at constant temperature.

c. P against PV graph:


When pressure of gas (P) against PV is plotted, a straight line parallel to the X-axis is obtained which indicates that the pressure of gas is independent of PV at constant temperature.

2. Charle's law: The volume of given amount of gas at 0 degree celsius is increases or decreases by 1/273 of its volume at 0 degree celsius for every 1 degree celsius rise or fall in temperature at constant pressure.

Let, V0 be the volume of gas at 0 degree celsius, then from charle's law,
The volume of gas at:
1 degree celsius= [V0+V0 x (1/273)] = V0 [1+(1/273)]
2 degree celsius= [V0+V0 x (2/273)] = V0 [1+(2/273)]
3 degree celsius= [V0+V0 x (3/273)] = V0 [1+(3/273)]
t degree celsius= [V0+V0 x (t/273)] = V0 [1+(t/273)]
-t degree celsius= [V0+V0 x (-t/273)] = V0 [1-(t/273)]
-1 degree celsius= [V0+V0 x (-1/273)] = V0 [1-(1/273)]
-273 degree celsius= [V0+V0 x (-273/273)] = V0 [1-(273/273)]= 0

If V1 and V2 are the volume of gas at t1 degree celsius and t2 degree celsius respectively. Then from charle's law:
V1= V0 [1 + (t1/273)] <------- (i)
V2= V0 [1 + (t2/273)] <------- (i)

Dividing equation (i) by equation (ii) we get,
(V1/V2) = [(273 + t1)/(273 + t2)]
V1/V2 = T1/T2
where T1 = 273 + t1 and T2= 273 + t2
i.e. V ∝ T [directly proportional]
Thus, the volume of gas is directly proportional to the absolute temperature at constant pressure.

# Graphical Representation of Charle's law:



This graph indicates that the volume of given amount of gas is directly proportional to absolute temperature at constant pressure.
absolute zero= 0 kelvin or -2730C.

Absolute zero temperature:
At -2730C or 0 kelvin there is no any kinetic energy and volume of the gas. This temperature is known as absolute zero temperature. This is the lowest theoretical value of temperature explain by charle's law.
The volume of gas at t0C is given as:
Vt= V0[1 + (t/273)]
Then, the volume of gas at -2730C is given as:
V-273= V0 [1 - (273/273)] = 0
 

Ideal Gas Equation:

The gas which obeys all the gas laws and ideal gas equation (PV= nRT) at all temperature and pressure is called ideal gas.
If we give high temperature and low pressure to real gas then it shows the behaviour of ideal gas because the volume is maximum.

Derivation of Ideal Gas Equation:
1. According to Boyle's law, the volume of given amount of gas inversely proportional to its pressure at constant temperature.
i.e. V ∝ 1/P <-------- (i) (inversely proportional)
2. According to Charle's law, the volume of given amount of gas is directly proportional to the absolute temperature at constant pressure.
i.e. V ∝ T <----------- (ii) (directly proportional)
3. According to Avogadro's hypothesis, the volume of gas is directly proportional to the number of moles of gas.
i.e. V ∝ n <----------- (iii) (directly proportional)
Combining equation (i), equation (ii), and equation (iii) we get
PV ∝ nT
PV= nRT where,
P= pressure of gas.
V= volume of gas.
n= number of moles of gas.
R= gas constant
T= absolute temperature.

Combined gas equation
From boyle's law: The volume of given amount of gas is inversely proportional to its pressure at constant temperature.
i.e. V ∝ 1/P <------- (i)
From charle's law: The volume of given amount of gas is directly proportional to the absolute temperature at constant pressure.
i.e. V ∝ T <--------- (ii)
Combining equation (i) and equation (ii)
PV ∝ T
PV= kT
PV/T = k
PV/T = constant <-------- (iii)
Let, P1, V1, T1 be the initial pressure, volume and temperature of gas respectively. Then, from equation (iii) we get,
P1V1/T1 = constant <-------- (iv)
If P2, V2 and T2 be the final pressure, volume and temperature of gas respectively. Then from equation (iii) we get,
P2, V2/T2 = constant <-------- (v)
Now from equation (iv) and (v)
P1V1/T1 = P2, V2/T2

Dalton's Law of Partial Pressure:

Statement: It states that, " the total pressure exerted by certain volume of mixture of non-reacting gases is equal to the sum of the partial pressure exerted by individual component gases at constant temperature."
Derivation:


Let us consider a vessel contains n1 moles of gas 1, n2 moles of gas 2 and n3 moles of gas 3. Then, the total number of moles of gas in the vessel is given as,
nt= n1 + n2 + n3 <-------- (i)
multiplying both sides of equation by RT/V we get,
ntRT/V= n1RT/V + n2RT/V + n3RT/V
We know from ideal gas equation,
PV= nRT
P1= n1RT/V
P2= n2RT/V
P3= n3RT/V

From equation (iii) and equation (iv) we get,
Pt= P1 + P2 + P3

# Application of Dalton's Law of Partial Pressure :
i. It is used to determine a partial pressure exerted by the water vapour in the moist gas.
ii. It is used to determine the pressure exerted by the pure gas when it is collected over the water.
 

Grahm's Law of diffusion:

The spontaneous of mixing of non-reacting gases is called diffusion.
Statement: It states that " the rate of diffusion of gases are inversely proportional to the square root of their density."
i.e. r ∝ 1/√d
Let, r1 and r2 be the rate of diffusion of two gases. Then from grahm's law of diffusion,
r1∝ 1/√d1 or, r1= k/√d1 <------- (i)
and r2∝ 1/√d2 or, r2= k/√d2 <------- (ii)
Dividing equation (i) by equation (ii) we get,
r1/r2 = √d2/d1 <--------- (iii)

We know that,
Molecular mass (M) = 2 x vapour density (d)
d1= M1/2 and d2= M2/2
Substituting the value of d1 and d2
we get,
r1/r2 = √m2/m1 <------- (iv)

Thus, the rate of diffusion of gases are inversely proportional to the square root of their molecular masses.
Rate of diffusion (r)= Volume of gas diffuse (V) / Time taken (t)
therefore, r1= V1/t1 and r2= V2/t2
Substituting the value of r1 and r2 into equation (iii) we get, V1/t1/V2/t2 = √d2/d1
(V1t2)/(t1V2) = √d2/d1 <-------- (v)

But, volume of gas (V)= mass (m)/density (d)
therefore, V1= m1/d1 and V2= m2/d2

Substituting the value of V1 and V2 in equation (v) we get,
m1/d1t1/m2/d2t2 = √d2/d1

therefore, m1/d2t2/m2/d1t1 = √d2/d1