- 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*

*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 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.

####
*Representation of Boyle's law*

V ∝ 1/P

∴ PV = K

∴ 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 P1 and P2 are the pressure respectively.

P1V1 = P2V2

Where V1 and V2 are the volumes at P1 and P2 are the pressure 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 temperature are called isotherms.

Graphical Representation of Boyle's Low |

- 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 with parallel to X-axis. This Graph shows that at a constant temperature the product of pressure and volume do not depend on its pressure.

###
*Relation between pressure and density*

*Relation between pressure and density*

- 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

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.

###
*Relation between volume and temperature*

*Relation between volume and temperature*

- At constant pressure a definite mass of gas, with the increasing of 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*

*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.

###
*Representation of Charl's law*

*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 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*

*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.

Graphical Representation of Charl's Low. |

###
*Absolute temperature*

*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.

###
*Relation between temperature and density*

*Relation between temperature and density*

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 PV/T = K(constant)

∴ (P1V1)/T1 = (P2V2)/T2 = constant

∴ PV = KT

∴ (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*

*Questions answers*

*Problem*- 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,

(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