**
**Gas Laws :

The quantitative relationship between volume, pressure, temperature and the rate of diffusion for a given quantity of gas
are termed as ’Gas Laws’.

These Laws are

* Boyle’s law

* Charles’ law

* Pressure-temperature law

* Avogadro’s law

* Graham’s law of diffusion

**
**Boyle’s Law

Robert Boyle, in 1662, published a mathematical statement on the relation between the volume and pressure of gas at constant
temperature called **Boyle’s Law**. This law states that "At constant temperature, volume of a definite mass of a
dry gas is inversely proportional to its pressure."

It can be mathematically expressed as :

The magnitude of constant depends on temperature, mass and nature of a gas.

Boyle’s law can be useful in calculating the volume of a gas at any required pressure if the volume at some other
pressure is known by using the following equation.

P_{1}V_{1} = P_{2}V_{2} = K

If 10 dm^{3} of nitrogen gas originally at 20.26 KPa is allowed to reach a pressure of 40.52 KPa while keeping
temperature constant, final volume can be calculated as follows :

P_{1} = 20.26 KPa P_{2} = 40.52 KPa

V_{1} = 10 dm^{3} V_{2} = ?

According to Boyle's Law

**
** = 5 dm3

It can be seen that since volume and pressure are inversely proportional, on increasing pressure, the volume decreases.

**
**Example :

The volume of a certain gas at constant temperature was found to be 14 liters when the pressure was 1.2 kg/ cm^{2}
If the pressure is decreased by 30% find the final volume of the gas.

V_{1} = 14 liters P_{1} = 1.2 kg/cm2

V_{2} = ? P_{2} = P_{1} - 30%

According to Boyle's law

P_{1}V_{1} = P_{2}V_{2 }

**
** The final volume of the gas is 20 liters.

Charle's Law

Jacques Charles, in 1787, formulated the relationship between the volume and temperature of a given mass of a dry gas at
constant pressure called Charles' Law which states that "At constant pressure, the volume of a fixed mass of dry gas is directly
proportional to the absolute temperature;

It can be mathematically expressed as

The magnitude of constant depends on pressure, mass and nature of a gas.

Charles' law is useful for calculating the volume of a gas at any required temperature if the volume at some other temperature
is known by using the following equation.

V_{1} = T_{1}

V_{2} = T2

** Example : **

A gas at constant pressure is kept at 100^{0}C. On decreasing the temperature to 50^{0}C, the gas occupies
a volume of 800 ml. Find the initial volume of the gas.

V_{1} = ? T_{1} = 100^{0}C = 100 + 273 K = 373 K

V_{2} = 800 ml T_{2} = 50^{0}C = 50 + 273 K = 323 K

According to Charles' Law

**
**Initial volume of the gas was 923.8 ml.

Pressure Temperature Law

The Pressure and Temperature law is similar to Charles' law. It states that "For a given mass of a dry gas, the pressure
is directly proportional to the absolute temperature, if the volume is kept constant."

It can be mathematically expressed as

P µ T, if V is constant

P = constant ´ T

This law can be useful in calculating the pressure of a gas at any required temperature if the pressure at some other
temperature is known by using the following equation :

If 25 dm^{3} of a gas at 36.5 KPa is cooled from 298 K to 136 K, keeping the volume constant, the final pressure
can be calculated as follows :

P_{1} = 36.5 KPa / P_{2} = ?

T_{1} = 298 K T_{2} = 136 K

**
**= 16.66 KPa

**
**Gay Lussac Law

In 1805, Gay Lussac observed a phenomenon based on his experiments on different gases. He framed his observations in the
Law of combining volumes of Gases which states that "when gases react they do so in volumes which bear simple whole number
ratios to one another and to the volume of the products, if gaseous, when measured at the same conditions of pressure and
temperature.

**
**Example

1. One volume of H_{2} combines with one volume of Br_{2} to form two volumes of HBr gas

**
**Ratio of reactions and products 1 : 1 : 2

2. One volume of N_{2} combines with 3 volumes of H_{2} to produce two volumes of NH_{3} gas

**
****
**Ratio of reactions and products 1 : 3 : 2

**
**Example

500 cm^{3} of Nitric oxide (NO) reacted with 300 cm^{3} of O_{2} to form nitrogen dioxide (NO2).
What would be the composition of final mass?

\ 2 vol of NO requires 1 vol O2

2 cm^{3} of NO requires 1 cm^{3} O2

\ 500 cm^{3} will require 250 cm^{3} of O2

Now

2 vol of NO yields 2 vol of NO2

\ 2 cm^{3} of NO yields 2 cm^{3} of NO2

\ 500 cm^{3} NO yields 500 cm^{3} of NO2

The resulting mixture consists of

1. Unused O_{2} = (300 - 250) = 50cm3

2. Product NO_{2} formed = 500 cm3

**
**Avogadro’s Law

**Avogadro’s Law** states that "equal volumes of different gases

under the same conditions of temperature and pressure contain an equal number of molecules."

**
**Avogadro’s Number (NA)

It is defined as the number of molecules present in 22.4 dm^{3} of gas ,

or gram-molecule weight of any substance
at N. T. P.

i.e. NA_{ }= 6.023 ´ 1023

e.g.

1) 22.4 dm^{3} of H_{2} gas at N.T.P. contains 6.023 ´ 10^{23 }H_{2} molecules.

2) 1 gm molecule of N_{2} (i.e. 28 gms of nitrogen) contains 6.023 ´ 10^{23 }nitrogen molecules.

**
**Graham’s Law of Diffusion

The phenomenon of **diffusion of a gas** can be defined as the tendency of a gas to spread uniformly throughout the
space available to it.

The relation between its density and the rate of diffusion of a gas can be represented by **Graham’s law** which
states that "The rate of diffusion of a gas is inversely proportional to the square root of its density under given conditions
of temperature and pressure."

Vapor density of a gas can be calculated if 0.08 dm^{3} of gas diffuses in the same time as 0.002 dm3 of chlorine
having vapor density 35.5.

r_{1 }= rate of diffusion of gas = 0.08 dm^{3}/t

r_{2} = rate of diffusion of chlorine = 0.002 dm^{3}/t

D_{1} = ?

D_{2} = 35.5

**
**General or Ideal Gas Equation

**The General or Ideal Gas Equation **is obtained by combining relations such as Boyle’s Law, Charles’ Law
and Avogadro’s Law.