- If the thermal energy is much greater than the forces of attraction then we have the matter in a gaseous state.

- Molecules in the gaseous state move very large speeds and the forces of attraction between them are not sufficient to bind the molecules at one place with the result the molecules move practically independent of one another.

- There exist no boundary surface and therefore gases tend to fill completely any available space, that is they do not possess fixed volume.

### 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 decreases with the increase of pressure and increases with the decreasing pressure.

- Let, a Cylinder contains 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.

#### Representation of Boyle's law

V ∝ 1/P

∴ PV = K

where, K is a constant whose value depends upon the,

(i)Nature of the gas.

(ii)Mass of the gas.

For a given mass of a gas at a constant temperature,

P₁V₁ = P₂V₂

where V₁ and V₂ are the volumes at pressure P₁ and P₂ respectively.

∴ PV = K

where, K is a constant whose value depends upon the,

(i)Nature of the gas.

(ii)Mass of the gas.

For a given mass of a gas at a constant temperature,

P₁V₁ = P₂V₂

where V₁ and V₂ are the volumes at pressure P₁ and P₂ respectively.

#### Graphical representation of Boyle's law

- The relation between pressure and volume can be represented by an arm of a rectangular hyperbola given below.

- As the value of the constant in the equation will change with temperature, there will be a separate curve for each fixed temperature. These curves plotted at different fixed temperatures are called isotherms.

Boyle's law |

- At constant temperature, a given mass of ideal gases the product of pressure and volume is always the same. If the product of pressure and volume represents in Y-axis and pressure represents X-axis a straight line curve is obtained parallel to X-axis. This graph shows that at a constant temperature the product of pressure and volume does not depend on its pressure.

### Pressure density relationship of gases

- At constant temperature(T) a definite mass of gas has pressure P₁ at volume V₁ and pressure P₂ at volume V₂.

According to Boyle’s law,

P₁ V₁ = P₂ V₂

or, P₁/P₂ = V₂/V₁

Again, let the mass of the gas = M

Density D₁ at pressure P₁ and density D₂ at pressure P₂

Thus, D₁ = M/V₁ and D₂ = M/V₂

or, V₁ = M/D₁ and V₂ = M/D₂

Again, P₁/P₂ = (M/D₂) × (D₁/M) = D₁/D₂

or, P₁/P₂ = D₁/D₂

or, P/D = Constant(K)

or, P = K × D

∴ P ∝ D

P₁ V₁ = P₂ V₂

or, P₁/P₂ = V₂/V₁

Again, let the mass of the gas = M

Density D₁ at pressure P₁ and density D₂ at pressure P₂

Thus, D₁ = M/V₁ and D₂ = M/V₂

or, V₁ = M/D₁ and V₂ = M/D₂

Again, P₁/P₂ = (M/D₂) × (D₁/M) = D₁/D₂

or, P₁/P₂ = D₁/D₂

or, P/D = Constant(K)

or, P = K × D

∴ P ∝ D

- Thus, at constant temperature density of a definite mass of a gas is proportional to its pressure.

### Volume temperature relationship of gases

- At constant pressure a definite mass of gas, with the increasing temperature, the volume also increases and with decreasing temperature, the 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 0⁰C.

### Representation of Charl's law

- If V₀ is the volume of the gas at 0⁰C, then 1⁰C rise of temperature the volume of the gas rise V₀/273.5 ml.

∴ 1⁰C temperature the volume of the gas, (V₀ + V₀/273) ml = V₀ (1 + 1/273) ml.

At t⁰C temperature the volume of the gas,

Vt = V₀ (1+ t/273) ml

= V₀ (273 + t°C)/273 ml

At t⁰C temperature the volume of the gas,

Vt = V₀ (1+ t/273) ml

= V₀ (273 + t°C)/273 ml

- It is convenient to use the absolute temperature scale on which temperature is measured Kelvin(K). Reading on this scale is obtained by adding 273 to the celsius scale value.

The temperature on the Kelvin scale is,

T K = 273 + t°C

∴ Vt = (V₀ × T)/273 = (V₀/273) T

T K = 273 + t°C

∴ Vt = (V₀ × T)/273 = (V₀/273) T

- Since V₀, the volume of a gas at 0°C, has a constant value at a given pressure, the above relation expressed as,

Vt = K₂ T

∴ V ∝ T

∴ V ∝ T

- Where K₂ 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 law

- A typical variation of volume of gas with a 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.

Char's law |

### Absolute temperature

- Since volume is directly proportional to its Kelvin temperature, the volume of the gas is theoretically zero at zero Kelvin or - 273⁰C.

- However, this is indeed hypothetical because all gases liquefy 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 -273⁰C.

### Density temperature relationship of gases

From Charl’s law, V₁/V₂ = T₁/T₂

Again, the mass of the gas is M.

Density D₁ and D₂ at the volume V₁ and V₂ respectively.

Then, V₁ = M/D₁ and V₂ = M/D₂

Thus, (M/D₁ )/(M/D₂ ) = T₁/T₂

or, D₂/D₁ = T₁/T₂

∴ D ∝ 1/T

Again, the mass of the gas is M.

Density D₁ and D₂ at the volume V₁ and V₂ respectively.

Then, V₁ = M/D₁ and V₂ = M/D₂

Thus, (M/D₁ )/(M/D₂ ) = T₁/T₂

or, D₂/D₁ = T₁/T₂

∴ D ∝ 1/T

- Thus at constant pressure, the 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 Boyl's law, V ∝ T when P constant.

From Boyl's law, V ∝ T when P constant.

- When all the variables are taken into account the variation rule states as,

Then, V ∝ T/P

or, PV/T = K(constant)

∴ (P₁V₁)/T₁ = (P₂V₂)/T₂ = constant

∴ PV = KT

or, PV/T = K(constant)

∴ (P₁V₁)/T₁ = (P₂V₂)/T₂ = constant

∴ PV = KT

- Thus the product of the pressure and volume of a given mass of gas is proportional to its kelvin temperature.

### Questions answers

Problem- At 1 atm pressure and 300 K temperature, the volume of the gas is 2000 cm³, then calculate the volume of this gas at 600 K temperature and 2 atm pressure.

From the combination of Boyle's and Charl's law is,

(P₁V₁)/T₁ = (P₂V₂)/T₂

(P₁V₁)/T₁ = (P₂V₂)/T₂

- Here, P₁ = 1 atm; V₁ = 2000 cm³ and T₁ = 300K and P₂ = 2 atm; V₂= ? and T₂ = 600 K

∴ 1×2000/300 = 2×V₂/600

or, V₂ = (1×2000×600)/(300×2)

= 2000 cm³

or, V₂ = (1×2000×600)/(300×2)

= 2000 cm³