# Gaseous State

### Postulates of Kinetic Molecular Theory Of Gases

1. Every gas consists of extremely minute particles called molecules.
2. The molecules inside a container are in random motion with high velocities. They collide with each other and also with wall of container.
3. The molecular collision are perfectly elastic. Hence, there is not any loss of energy during collision.
4. At low pressure , the distance between the molecules are very large thus force of attraction between them is negligible.
5. Due to continuous bombardment of the moving molecules , pressure is exerted on the wall of container.
6. The average Kinetic Energy of molecules is directly proportional to the absolute temperature of the gas.

### Derivation of $$PV= \frac {1}{3} mnu^2$$

Let us consider a gas is enclosed in a cubical vessel of length ‘l’ and volume ‘V’. According to Kinetic Theory of Gases , gases molecule are in random motion .If ‘v’ be the velocity of a gas molecule then it can be resolved into three components $$v_x, v_y and v_z$$ along x-axis , y-axis and z-axis respectively.
$$v^2=v_x^2+v_y^2+v_z^2$$​

Suppose the motion of one molecule is striking at face B along x-axis.
Momentum of molecule before collision= $$mv_x$$
The molecule rebounds to face A after striking face B
Momentum of molecule after collision= $$-mv_x​$$ (opposite direction)
Change in momentum = $$mv_x– (-mv_x​)$$ = $$2mv_x$$

To strike face B again the molecule have to travel 2l distance. Number of collision per second at face A is given by, $$\frac {v_x}{2l}​​$$

Rate of change in momentum= No. of collision per second × change in momentum per collision
$$= \frac {v_x}{2l}​​ × 2mv_x = \frac{mv_x^2}{l}$$
This is the momentum transferred pre second to the face A.
The total momentum transferred per second due to collision of one molecule on both faces along x-axis $$= \frac{2mv_x^2}{l}$$

Similarly , the rate of change in momentum along y-axis and z-axis will be $$\frac {2mv_y^2}{l}​​$$ and $$\frac {2mv_z^2}{l}$$ respectively.
Hence total change in momentum by single molecule per second is given by
$$= \frac{2mv_x^2}{l} + \frac{2mv_y^2}{l} + \frac{2mv_z^2}{l}$$
$$= \frac{2m}{l}​ (v_x^2 + v_y^2 + v_z^2 )$$
$$= \frac{2mv^2}{l}$$

If there are ‘n’ number of gases molecule, then total change in momentum per second of all the gases molecule will be $$\frac{2mn}{l} \frac{(v_1^2+v_2^2+v_3^2+....v_n^2)}{n}$$
$$= \frac{2mnu^2}{l}$$
Since, Force is the change in momentum per second, force= $$\frac {2mnu^2}{l}$$

$$Pressure = \frac{Force}{Area}$$
$$or, P= \frac{2mnu^2}{l} × \frac{1}{6l^2}$$
$$or, PV= \frac{1}{2} mnu^2$$

### Some Important Terms

Average velocity
It is the arithmetic mean of the velocity possessed by different molecules of gas at a particular temperature.
If $$v_1,v_2,v_3 .......,v_n$$ ​ are the velocities of individual gases molecules and ‘n’ is the total no. of gases molecules present in the gas. Then ,
$$V_avg = ( \frac {v_1+v_2+v_3+......+v_n}{n} )$$
From Maxwell’s Equation ,
$$V_avg = \sqrt{\frac{8RT}{\pi M}}$$

Root Mean Square Velocity(u)
It is the square root of the mean of the square of all the individual velocities of gases molecules.
$$u = \sqrt{( \frac {v_1^2+v_2^2+v_3^2+......+v_n^2}{n} )}$$
From Maxwell’s Equation ,
$$u = \sqrt{\frac{3 RT}{M}}$$

Most Probable Velocity(α)
It is the velocity possessed by maximum number of gases molecules.
From Maxwell’s Equation ,
$$α= \sqrt{\frac{2 RT}{M}}$$

### Derivation of gases law from Kinetic Gas Equation

Boyles Law

We know that,
K.E α T
or, K.E = kT , where k=proportional constant
or, $$\frac{1}{2}​ mu^2 = kT$$
For ‘n’ no. of gases molecule,
or, $$\frac{1}{2}​ mnu^2= kT$$…….(1)

From kinetic gas equation,
$$PV= \frac {1}{3}​ mnu^2$$ …….(2)

Dividing equation 1 by 2 ,
$$PV = \frac{1}{3}​ mnu^2$$
$$or, PV = \frac{2}{3}​( \frac {1}{2} mnu^2)$$
$$or, PV = \frac{2}{3}​ K.E$$
$$or, PV = \frac{2}{3}​ kT$$
When T is constant, the whole terms on RHS becomes constant,
At T constant,
PV= constant
This is the Boyle’s Law.

Similarly ,
$$PV = \frac{2}{3}​ kT$$
$$or V = \frac{2}{3P} ​k T$$
When Pressure=P is constant,
VαT
This is Charle’s Law.

Graham’s Law of Diffuion

We know that,
$$PV= \frac {1}{3}​ mnu^2$$
If m_1​ and m_2 are the masses and u_1​ and u_2​ are the velocities of the molecules of gases A and B,
$$P_1 V_1​= \frac {1}{3}​ m_1n_1u_1 ^2$$
$$P_2 V_2​= \frac {1}{3}​ m_2n_2u_2 ^2$$

For same pressure and volume,
$$\frac {1}{3}​ m_1n_1u_1 ^2 ​=\frac {1}{3} m_2n_2u_2 ^2$$
$$\frac{u_1}{u_2}​​= \sqrt{\frac {m_2 n_2}{m_1 n_1}}$$
If $$n_1​ = n_2​ = N$$ (Avogardo’s Number)
$$\frac{u_1}{u_2}​​ = \sqrt{\frac {m_2 N}{m_1 N}}$$
If $$M_1​ and M_2$$​ are molecular masses of gases A and B
$$\frac{u_1}{u_2} ​​ = \sqrt{\frac {M_2}{M_1}}$$
The rate of diffusion(r) is proportional to velocity of molecules so,
$$\frac{r_1}{r_2}​​ = \frac{u_1}{u_2}​​ = \sqrt{\frac {M_2}{M_1}}$$
This is Graham’s Law of Diffusion

### Maxwell’s Distribution of Molecular Velocities

When any two gases molecules collide, one transfer KE to another molecule. The velocity of molecule which gains energy increases and the velocity of another molecule decreases. Hence, the velocities of molecules are changing. For large number of molecules , a fraction of molecules will have same particular velocity. Maxwell’s derived the equation for distribution of molecular velocities from the laws of probability.

$$\frac{dn_c}{n}​​= 4 \pi (\frac {M}{2 \pi RT})^ \frac{3}{2} e^{\frac{-MC^2}{2RT}}​. c^2 dc$$
where,
$$dn_c$$​ = no. of molecules having velocities between c and c+dc
n= total no. of molecules
M= Molecular weight
T= absolute temperature

#### Derivation of Average Velocity

We know that,
$$V_avg = ( \frac{C_1+C_2+C_3+......+C_n}{n} )$$
This can be written as,
$$V_avg = \frac{\int\limits_0^∞ c. dn_c}{n}$$
Substituting the value of $$\frac{dn_c}{n}$$​​ from Maxwell’s Equation,
$$V_avg = \int\limits_0^∞​ 4 \pi (\frac{M}{2 \pi RT})^ \frac{3}{2} e^ \frac{-MC^2}{2RT} c^3 dc$$
$$= 4 \pi (\frac{M}{2 \pi RT})^ \frac{3}{2} ​ \int\limits_0^∞ ​e^ \frac{-MC^2}{2RT} c^3 dc$$
The integral part $$\int\limits_0^∞ ​ e^ \frac{-MC^2}{2RT}​ c^3$$ dc can be calculated as
$$\int\limits_0^∞ ​e^ \frac{-MC^2}{2RT} c^3 dc = \int\limits_0^∞ e^{-ac^2} c^3 dc$$ where $$a= \frac{M}{2RT}​$$

Putting $$ac^2 = t$$
Differentiating with respect to c we get,
$$\frac{d(ac^2)}{dc} ​= \frac{dt}{dc}$$
$$or, 2ac = \frac{dt}{dc} ​$$
$$or, dc = \frac{dt}{2ac}$$

Now the expression becomes,
$$\int\limits_0^∞ e^{-ac^2} c^3 dc = \int\limits_0^∞ e^-t ​(\frac{t}{a}) \frac{dt}{2a}$$
$$= \frac{1}{2a^2} \int\limits_0^∞ e^{-t} t.dt$$
$$= \frac{1}{2a^2} \int\limits_0^∞ e^{-t} t^{(2-1)} .dt$$
Since, $$\int\limits_0^∞ e^{-x} x^{(n-1)} .dx= \Gamma n$$
$$\frac{1}{2a^2} \Gamma 2$$
$$\frac{1}{2a^2} .1$$
$$= \frac{1}{2} (\frac{2RT}{M})^2$$

Hence, $$V = 4 \pi (\frac{M}{2 \pi RT})^ \frac{3}{2} × \frac{1}{2} (\frac{2 RT}{M})^2$$
$$= \frac{2}{\sqrt{\pi}} (\frac{2RT}{M})^ \frac{1}{2}​$$
$$= (\frac{8RT}{\pi M})^ \frac{1}{2}​$$
$$= \sqrt{\frac{8RT}{\pi M}}$$

#### Derivation of Root Mean Square Velocity

We know that,
$$u= \sqrt{\frac{C_1^2+C_2^2+C_3^2+.......+C_n^2}{n}}$$
$$= ( \bar C ^2)^ \frac{1}{2}$$
This can be written as,

$$u = (\int\limits_0^∞ \frac{c^2. dn_c}{n})^{\frac{1}{2}}$$
Substituting the value of $$\frac{dn_c}{n}$$ from Maxwell’s Equation,
$$u = (\int\limits_0^∞ 4 \pi (\frac{M}{2RT})^ \frac{3}{2} e^ \frac{-MC^2}{2RT} c^4 dc ) ^ \frac{1}{2}$$
$$= ( 4 \pi (\frac{M}{2RT})^ \frac{3}{2} \int\limits_0^∞ ​e^ \frac{-MC^2}{2RT} c^4 dc ) ^ \frac{1}{2}$$
The integral part $$\int\limits_0^∞ ​ e^ \frac{-MC^2}{2RT} c^4 dc$$ can be calculated as
$$\int\limits_0^∞ ​e^ \frac{-MC^2}{2RT} c^4 dc = \int\limits_0^∞ e^{-ac^2} c^4 dc$$where $$a= \frac{M}{2RT}$$

Putting $$ac^2 = t$$
Differentiating with respect to c we get,
$$\frac{d(ac^2)}{dc} = \frac{dt}{dc}$$
$$or, 2ac = \frac{dt}{dc}$$
$$or, dc = \frac{dt}{2ac}$$

Now the expression becomes,
$$\int\limits_0^∞ e^{-ac^2} c^4 dc = \int\limits_0^∞ e^{-t} (\frac{t}{a}) ^ \frac{3}{2}) \frac{dt}{2a}$$
$$= \frac{1}{2a^ \frac {5}{2}} \int\limits_0^∞ e^{-t} t^{\frac{3}{2}} ​.dt$$
$$= \frac{1}{2a^ \frac{5}{2}} \int\limits_0^∞ e^{-t} t^{(\frac{5}{2}-1)} .dt$$
Since, $$\int\limits_0^∞ e^{-x} x^{(n-1)} .dx= \Gamma n$$
$$=\frac{1}{2a^ \frac{5}{2}} \Gamma \frac{5}{2}$$
$$=\frac{1}{2a^ \frac{5}{2}}. \frac{3}{2}. \frac{1}{2} \sqrt{\pi}$$
$$= \frac{3}{8} \sqrt{\pi}(\frac{2RT}{M})^{\frac{5}{2}}$$

Hence, $$u = ( 4 \pi (\frac{M}{2 \pi RT})^ \frac{3}{2} × \frac{3}{8} × \sqrt{\pi} (\frac{2RT}{M})^{\frac{5}{2}})^{\frac{1}{2}}$$
$$= (\frac{3×2RT}{2M})^ \frac{1}{2}​$$
$$= \sqrt{\frac{3RT}{M}}$$

#### Derivation of Most Probable Speed

From Maxwell’s Distribution Law,
$$\frac{1}{N} \frac{d_{nc}}{dc} ​​=4 \pi (\frac{M}{2 \pi RT})^{\frac{3}{2}} e^{\frac{-MC^{2}}{2RT}} c^{2}$$
Since, Most probable speed is the velocity possed by maximum number of molecules, By using condition of Maxima and Minima,
$$\frac{dy}{dx} ​= 0$$
From the curve,
$$\frac{d[4 \pi (\frac{M}{2 \pi RT})^{\frac{3}{2}} e^{\frac{-MC^{2}}{2RT}} c^{2}]}{dc} ​= 0$$
$$or, 4 \pi (\frac{M}{2 \pi RT})^{\frac{3}{2}} \frac{d(e^{\frac{-MC^{2}}{2RT}}c^{2})}{dc} =0$$
$$or, \frac{d(e^{\frac{-MC^{2}}{2RT}}c^{2})}{dc} ​= 0$$
$$or, e^{\frac{-MC^{2}}{2RT}} \frac{d(c)^{2}}{dc} + c^{2} \frac{d(e^{\frac{-MC^{2}}{2RT}})}{dc}​ = 0$$
$$or, e^{\frac{-MC^{2}}{2RT}} . 2c + c^{2} e^{\frac{-MC^{2}}{2RT}} (\frac{-2MC}{2RT}) = 0$$
$$or, c . e^{\frac{-MC^{2}}{2RT}} (2- \frac{MC^{2}}{RT}) =0$$
To satisfy above condition,
Either,
c=0 (not possible for most probable speed)
Or,
$$e^{\frac{-MC^{2}}{2RT}} =0$$ (Only possible when c= \infinity which is not valid for most probable speed)
Or,
$$2- \frac{MC^{2}}{RT} = 0$$
$$or, \alpha = \sqrt{\frac{2RT}{M}}$$

#### Variation of Maxwell’s Curve

1. For same gas at different temperature
2. For different gas at same temperature

#### Vander Waal’s Equation

The gas which obeys PV=nRT equation at all pressure and temperature are ideal gas. Almost all the gases obeys ideal gas equation only at low pressure and high temperature. But, at high pressure and low temperature , the real gas show deviation from ideal gas behaviour. This makes clear that, some of the postulates of Kinetic Theory are incorrect at high pressure and low temperature.They are:-

1. Volume of individual gases molecule is negligiable as compared to total volume of gas.
This is only applicable for low pressure and high temperature. But at high pressure and low temperature, the total volume of gas decreases and hence, volume of individual gas molecure can not be neglected.
2. There is no any force of attraction between gases molecules.
This is only applicable for low pressure and high temperature. But at high pressure and low temperature , the molecules of gas comes closer to each other , thus force of attraction between them exist.

Derivation of Vander Waal’s Equation for Real Gases

Since , the idea gas equation PV=nRT could not hold for real gases. Vander Waal’s did correction in the equation.They are:-

1. Volume correction
The volume of gas is the free space in the container in which molecules can move. Volume of an ideal gas molecule is zero as whole container has free space for movement of molecule.
If b be the effective volume of a molecule per mole then,
Volume of real gas= Volume of container- Total volume of gas molecule
=(V-nb) [for n moles of gas]
2. Pressure correction
Every molecures of gas is attracted by another molecure of gas on all side. Thus this attractive force cancel out. But molecules which is about to strike on the vall of the container is only attracted by molecules on one side only. Hence, that molecule experience inward pull.
The decreased pressure is due to inter moleculer attractive force which depends upon;
• The number of molecules striking on the wall of container which is directly proportional to density of gas ; \frac{n}{V}
• The number of molecules present near the colliding molecules which is also proportional to $$\frac{n}{V}$$
Hence , decreased pressure; $$p \propto \frac{n}{V}​ * \frac{n}{V}$$
$$p \propto \frac{n^2}{V^2}$$
$$p= \frac{an^2}{V^2}$$​ ,where a=Vander Waal’s Constant

Thus ,
Actual pressure(P)= Ideal Pressure – Decreased pressure(p)
P_ideal = P + p
$$=P+ \frac{an^2}{V^2}$$

Substutiting the corrected pressure and volume in ideal gas equation,PV=nRT, we get
$$(P+ \frac{an^2}{V^2}​ )(V-nb) = nRT$$

At extremely low pressure
V becomes very large. Hence all the terms pb , $$\frac{a}{V}$$​ and $$\frac{an^2}{V^2}$$​ are negligiably small. Hence, PV=RT. Thus at low pressure real gases behave ideally.

Units of constant
In the Vander’s Waal’s equation , $$\frac{an^2}{V^2}$$​ is added to P and nb is added to V. hence, they should have same dimension.
$$P= \frac{an^2}{V^2}$$
$$a= \frac{PV^2}{n^2} ​= \frac{( Nm^{-2}) (m^{3})^2}{mole^2} ​= Nm^{4} mole^{-2}$$

V=nb
$$b= \frac{V}{n}​= \frac{m^3}{mole} ​= m^{3} mole^{-1}$$

#### Critical Phenomenon

1. Critical Temperature
The temperature above which a gas cannot be liquified no matter how much greater pressure is applied is know as Critical Temperature.
$$T_{c}​ = \frac{8a}{27Rb}$$
2. Critical Pressure
The minimum pressure required to liquify the gas at it’s critical temperature is known as Critical Pressure.
$$P_{c}​ = \frac{a}{27b^{2}}$$
3. Critical Volume
The volume occupied by a mole of gas at critical temperature and critical pressure is known as Critical Volume.
$$V_{c}​ =3b$$

#### Expression for Critical Phenomenon

We know that,
(P+ \frac {an^{2}}{V^{2}}​) (V-b)= nRT
For 1 mole of gas,
(P+ \frac {a}{V^{2}}​) (V-b)= RT
$$or, ( \frac{PV^{2}+a}{V^{2}} +a​)(V-b)=RT$$
$$or, PV^{3}- PV^{2}b+aV-ab= V^{2} RT$$
$$or, PV^{3}- PV^{2}b - V^{2} RT +aV-ab = 0$$
$$or, V^{3}- V^{2}b - \frac{V^{2} RT}{P} + \frac{aV}{P}​– \frac{ab}{P}​ = 0$$
$$or, V^{3}- V^{2}(b + \frac{RT}{P}PRT​) + \frac{a}{P} – \frac{ab}{P}​= 0$$
At critical point,
$$V^{3} - V^{2} (b + \frac{RT_{c}}{P_{c}}​​) + \frac{a}{P_{c}}​ V – \frac{ab}{P_{c}} ​= 0$$ …………………(1)
Also at critical point,
$$V= V_{c}$$
$$or, V- V_{c}​ = 0$$
$$or, (V- V_{c})^{3} = 0$$
$$or, V^{3}- 3 V^{2}V_{c} + 3 V_{c} ^{2} V - V_{c}^{3} = 0$$ …………………(2)
Comparing eqn (1) and (2) we get,
$$3 V_{c}​ = \frac{RT_{c}}{P_{c}} + b$$ …………………(3)
$$3 V_{c}​ = \frac{a}{P_{c}}$$…………………(4)
$$V_{c}^{3}​ = \frac{ab}{P_{c}}$$​…………………(5)
Dividing eqn (5) by (4), we get
$$V_{c}​ = 3b$$
Putting the value of V_{c}​ in eqn (4),
$$P_{c}​ = \frac{a}{27b^{2}}$$
And, Putting the value of V_{c}​ and P_{c}​ in eqn (3)
$$T_{c}​ = \frac{8a}{27Rb}$$

#### Some Important Terms

1. Mean Free Path
The distance travelled by the molecule before the collision is known as free path.
The mean distance travelled by a molecule between two succesive collision is called mean free path. It is denoted by \lambdaλ. If l_1, l_2,…, l_n are the free path for a molecule of a gas, then it’s mean free path,
\lambdaλ = \frac{l_1 + l_2 + ......+ l_n}{n}nl1​+l2​+......+ln​​
2. Collision diameter
The distance between the centre of molecules at their closest approach is known as collision diameter. It is denoted by \sigmaσ .
3. Collision Frequency
The number of the molecular collision taking place per second per unit volume of the gas is known as collision frequency.
It is given by ,
N_cNc​ = \frac{\pi V \sigma^{2} N^{2}}{\sqrt{2}}2​πVσ2N2​
where,
V = average velocity in cm/sec
\sigmaσ = collision diameter
N = No. of identical gas molecules per c.c

When temperature increases the average velocity also increases.
V_avg = \sqrt{\frac{8 RT}{\pi M}}πM8RT​​
Thus, collision frequency is affected by temperature.

#### Liquification of Gas

If a gas is cooled below the critical temperature and subjected to adquate pressure, it gets liquified.
Two conditions essential to liquify gas:-

1. Low Temperature
2. High Pressure