Boyle’s Law:

At constant temperature, the volume of a definite mass of a gas is inversely proportional to its pressure.

That is, the volume of a given quantity of gas, at a constant temperature decrees with the increasing of pressure and increases with the decreasing pressure.
Let, a Cylinder contain 10 ml of gas, at constant temperature and 1 atm pressure. if the pressure increases to 2 atm then the volume also decrees to 5 ml.
Mathematical Representation of Boyle's Low:

V ∝ 1/P
∴ PV = K

Where, K is a constant whose value depends upon the,
a. Nature of the Gas.
b. Mass of the Gas.
For a given mass of a gas at constant temperature,
P_{1}V_{1} = P_{2}V_{2} 

Where, V_{1} and V_{2} are the volume at P_{1} and P_{2} are the pressure respectively.
Graphical Representation of Boyle's Low:

The relation between pressure and volume can be represented by an arm of rectangular hyperbola given below.
As the value of the constant in equation will change with temperature, there will be a separate curve for each fixed temperature. These curves plotted at different fixed temperature are called isotherms.
Graphical Representation of Boyle's Low 

At Constant temperature a given mass of an Ideal Gases the product of Pressure and Volume is always same. if the product of pressure and volume represents in Y axis and Pressure represents X axis a straight line curve is obtained with parallel to X axis.
This Graph is shows that, at constant temperature the product of pressure and volume is does not depends on its pressure.
Relation between Pressure and density of a gas:

At constant temperature(T) a definite mass of gas has pressure P_{1} at volume V_{1} and pressure P_{2} at volume V_{2}.
According to Boyle’s Law,
P₁ V_{1} = P_{2} V_{2}
or, P_{1}/P_{2} = V_{2}/V_{1}
Again, Let the mass of the Gas = M
Density D_{1} at Pressure P_{1}
and Density D_{2} at Pressure P_{2}
Thus, D_{1} = M/V_{1} and D_{2} = M/V_{2}
or, V_{1} = M/D_{1} and V_{2} = M/D_{2}
Again, P_{1}/P_{2} = (M/D_{2}) × (D_{1}/M)
= D_{1}/D_{2}
P₁ V_{1} = P_{2} V_{2}
or, P_{1}/P_{2} = V_{2}/V_{1}
Again, Let the mass of the Gas = M
Density D_{1} at Pressure P_{1}
and Density D_{2} at Pressure P_{2}
Thus, D_{1} = M/V_{1} and D_{2} = M/V_{2}
or, V_{1} = M/D_{1} and V_{2} = M/D_{2}
Again, P_{1}/P_{2} = (M/D_{2}) × (D_{1}/M)
= D_{1}/D_{2}
P_{1}/P_{2} = D_{1}/D_{2} P/D = Constant(K) P = K × D P ∝ D 

Thus, at constant temperature density of a definite mass of a gas is Proportional to its Pressure.
Relation between Volume and Temperature of a Gas:

At constant pressure a definite mass of gas, with the increasing of temperature, volume also increases and with decreasing temperature, volume also decreases.
That is, the volume of a given mass of gas at constant pressure is directly proportional to its Kelvin temperature.
Charl’s Law:

At constant pressure, each degree rise in temperature of a definite mass of a gas, the volume of the gas expands 1/273.5 of its volume at 0^{0}C.
Mathematical Representation of Charl's Low:

If V_{0} is the volume of the gas at 0^{0}C, then
1^{0}C rise of temperature the volume of the gas rise V_{0}/273.5 ml.
∴ 1^{0}C temperature the volume of the gas,
(V_{0} + V_{0}/273) ml
= V_{0} (1 + 1/273) ml.
At t^{0}C temperature the volume of the gas,
V_{t} = V_{0} (1+ t/273) ml
= V_{0} (273 + t°C)/273 ml.
(V_{0} + V_{0}/273) ml
= V_{0} (1 + 1/273) ml.
At t^{0}C temperature the volume of the gas,
V_{t} = V_{0} (1+ t/273) ml
= V_{0} (273 + t°C)/273 ml.

It is convenient to use of the absolute temperature scale on which temperature is measured Kelvin(K). A reading on this scale is obtained by adding 273 to the Celsius scale value.
Temperature on Kelvin scale is,
T K = 273+t°C 
∴ V_{t} = (V_{0} × T)/273 = (V_{0}/273) T 

Since V_{0}, the volume of a gas at 0°C, has a constant value at a given pressure, the above relation expressed as,
V_{t} = K₂ T
∴ V ∝ T 

Where K_{2} is constant whose value depends on the, Nature, mass and pressure of the gases.
According to the above relation Charl’s Law states as,
At constant pressure the volume of a given mass of gas is directly proportional to its Kelvin temperature.
Graphical Representation of Charl's Low:

A typical variation of Volume of a gas with change in its kelvin temperature a straight line plot was obtained, Called isobars.
The general term isobar, which means at constant pressure, is assigned to these plots.
Graphical Representation of Charl's Low. 
Absolute Temperature or Absolute Zero:

Since volume is directly proportional to its Kelvin temperature, the volume of the gas is theoretically zero at zero Kelvin or  273^{0}C.
However this is indeed hypothetical because all gases liquefies and then solidity before this low temperature reached. In fact, no substance exists as a gas at the temperature near Kelvin zero, through the straight line plots can be extra plotted to zero volume.
The temperature corresponds to zero volume is 273^{0}C.
Relation between temperature and Density of a given Ideal Gas at constant Pressure:
From the Charl’s Law,
V_{1}/V_{2} = T_{1}/T_{2}
Again, the mass of the gas is M.
Density D_{1} and D_{2} at the Volume V_{1} and V₂ respectively.
Then, V_{1} = M/D_{1} and V_{2} = M/D_{2}
Thus, (M/D_{1} )/(M/D_{2} ) = T_{1}/T_{2}
V_{1}/V_{2} = T_{1}/T_{2}
Again, the mass of the gas is M.
Density D_{1} and D_{2} at the Volume V_{1} and V₂ respectively.
Then, V_{1} = M/D_{1} and V_{2} = M/D_{2}
Thus, (M/D_{1} )/(M/D_{2} ) = T_{1}/T_{2}
or, D₂/D_{1} = T_{1}/T_{2} ∴ D ∝ 1/T 

Thus at constant pressure, density of a given mass of gases is inversely proportional to its temperature.
Combination of Boyle’s and Charl’s Law:
From Charl's Law,
V ∝ 1/P When T Constant.
From Charl's Law,
V ∝ T When P Constant.
When all the variables taken into account the variation rule states as,
Then, V ∝ T/P
PV/T = K(Constant)
V ∝ 1/P When T Constant.
From Charl's Law,
V ∝ T When P Constant.
When all the variables taken into account the variation rule states as,
Then, V ∝ T/P
PV/T = K(Constant)
∴ (P_{1}V_{1})/T_{1} = (P_{2}V_{2})/T_{2} = Constant PV = KT 

Thus the product of the pressure and volume of a given mass of gas is proportional to its Kelvin temperature.
 Problem 1:

At 1 atm pressure and 300 K temperature, the volume of the gas is 2000 cm^{3}, then calculate the volume of this gas at 600 K temperature and 2 atm pressure.
 Answer:

From the Combination of Boyle's and Charl's Law is,

(P_{1}V_{1})/T_{1} = (P_{2}V_{2})/T_{2}

Here, P_{1} = 1 atm; V_{1} = 2000 cm^{3} and T_{1} = 300K and P_{2} = 2 atm; V_{2}= ? and T_{2} = 600 K.

∴
1×2000/300 = 2×V₂/600

or, V_{2} = (1×2000×600)/(300×2)

= 2000 cm^{3}