Derivation of Gas Laws from Kinetic Theory of Gases

Boyle's Law from Kinetic theory of gases

From kinetic theory of gases, we have

$$ P = \frac {1}{3}\frac {M}{V} \bar c^2 $$

$$ \text{or,}PV = \frac {1}{3} M \bar c^2 $$

Since \(\bar C^2 \propto T\), if the temperature of the gas is kept constant, for given mass of a gas (i.e. M = constant),

$$PV = constant$$

$$\text{i.e.} p \propto \frac{1}{V}$$

It means pressure of the given mass of a gas is inversely proportional to it's volume at constant temperatures which is Boyle's law.

Charle's Law from Kinetic theory of gases

From kinetic theory of gases, we have

$$ P = \frac {1}{3}\frac {M}{V} \bar c^2 $$

$$ \text{or,}V = \frac {1}{3} \frac {M}{P} \bar c^2 $$

Since \(\bar C^2 \propto T\), if the pressure of the gas is kept constant, for given mass of a gas, we have

$$v \propto T$$

It means volume of the given mass of a gas is directly proportional to it's absolute temperature at a constant pressure which is Charle's law.

Law of pressure from Kinetic theory of gases

here,

$$ P = \frac {1}{3}\frac {M}{V} \bar c^2 $$

Since \(\bar C^2 \propto T\) and at constant volume of a gas, the above equation becomes

$$P \propto T$$

It means pressure of the gas is directly proportional to its absolute temperature at constant volume which is law of increase of pressure.

Perfect Gas Equation from Kinetic theory of gases

From kinetic theory of gases, we have

$$ P = \frac {1}{3}\frac {M}{V} \bar c^2 $$

$$ \text{or,}V = \frac {1}{3} \frac {M}{P} \bar c^2 \dots (i)$$

Since \(\bar C^2 \propto T\) so, equation (i) becomes,

$$PV \propto T$$

$$\text{or,} PV = RT $$

Where R is the constant foe one mole of an ideal gas. This equation is the perfect gas equation.

Dalton's Law of Partial Pressure from Kinetic theory of gases

Consider a mixture of gases in a vessel of volume V in which no chemical interaction takes place. . If p₁, p₂, . . .. .. . are the partial pressure of the respective gases, then from the kinetic theory of gases. we have,

$$ P_1 = \frac {1}{3}\frac {n_1m_1}{V} \bar c_1^2, P_2 = \frac {1}{3}\frac {n_2m_2}{V} \bar c_2^2, \dots and so on$$

and so on

where, n₁,n₂, n₃, . . . . . are the number of molecules of the respective gases,

m₁, m₂, m₃ . . . . . are the mass of molecules of the respective gases,

and c₁, c₂, c₃. . . . . are the root mean square velocity of the respective gases,

According to Dalton's law, the total pressure exerted by a mixture of pressure is equal to this sum of the pressure exerted by each gas. So total pressure

$$ p = p_1 + p_2 + p_3 + \dots$$

$$ = \frac {1}{3}\frac {n_1m_1}{V} \bar c_1^2 + \frac {1}{3}\frac {n_2m_2}{V} \bar c_2^2 + \dots+ \frac {1}{3}\frac {n_nm_n}{V} \bar c_n^2 $$

$$=\frac {1}{3V} [n_1m_1c_1^2 + n_2m_2c_2^2 + \dots n_nm_nc_n^2]$$

If the temperature of the mixture of gases is same, their kinetic energy per molecule must be equal.

That is

$$\frac {1}{2}m_1c_1^2 = \frac {1}{2}m_2c_2^2 = \dots = \frac {1}{2}m_3c_3^2 = \frac {1}{2}m_nc_n^2$$

So, we have

$$p = \frac {1}{3V} m \bar c^2 [n_1 +n_2 +\dots +n_n]$$

$$\text{or,} p =\frac {m \bar c^2}{V} n$$

where, n₁,n₂, n₃, . . . . . is the total numbers of molecules of the mixture of gases.

$$ \therefore P = \frac {1}{3}\frac {M}{V} \bar c^2 $$

which is the total pressure exerted by a gas having n number of molecules each mass m and mean square velocity \(\bar c^2\).

Thus above equation proves the Dalton's law of partial pressure.