**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**.

**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 a constant

**temperature**,

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

- Where V

_{1}and V

_{2}are the volumes 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 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

__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.__

**Ideal Gases****The relation between Pressure and density of a gas:**

- At constant temperature(

**T**) a definite mass of gas has pressure

**P**at volume

_{1}**V**and pressure

_{1}**P**at volume

_{2}**V**.

_{2}According to

P₁ V

or, P

Again, Let the mass of the Gas =

Density

and Density

Thus, D

or, V

Again, P

= D

**,**__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**at Pressure_{1}**P**_{1}and Density

**D**at Pressure_{2}**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 × DP ∝ D |

- Thus, at

**density of a definite mass of a gas is**

__constant temperature__**to its Pressure.**

__Proportional__**The relation between Volume and Temperature of a Gas:**

- 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**Mathematical Representation of Charl's Low:**

- If

**V**is the volume of the gas at

_{0}**0**, then

^{0}C**1**rise of temperature the volume of the gas rise

^{0}C**V**ml.

_{0}/273.5∴

(V

= V

At

V

= V

**1**temperature the volume of the gas,^{0}C(V

_{0}+ V_{0}/273) ml= V

_{0}(1 + 1/273) ml.At

**t**temperature the volume of the gas,^{0}CV

_{t}= V_{0}(1+ t/273) ml= V

_{0}(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 |

∴ V_{t} = (V_{0} × T)/273 = (V_{0}/273) T |

- Since

**V**, the volume of a gas at 0°C, has a constant value at a given pressure, the above relation expressed as,

_{0}**V**

_{t}= K₂ T∴ V ∝ T |

- Where

**K**is constant whose value depends on Nature, mass, and pressure of the gases. According to the above relation,

_{2}**states as, At constant pressure, the volume of a given mass of gas is directly proportional to its**

__Charl’s Law__**.**

__Kelvin temperature__**Graphical Representation of Charl's Low:**

- 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 or Absolute Zero:__

__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**. 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

^{0}C**-273**.

^{0}C**The relation between temperature and Density of a given Ideal Gas at Constant Pressure:**

From

Again, the mass of the gas is

Density

Then, V

Thus, (M/D

**,**__Charl’s Law__**V**_{1}/V_{2}= T_{1}/T_{2}Again, the mass of the gas is

**M**.Density

**D**and_{1}**D**at the Volume_{2}**V**and_{1}**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, the density of a given mass of gases is inversely proportional to its temperature.

**Combination of Boyle’s and Charl’s Law:**

From

From

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

Then, V ∝ T/P

PV/T = K(Constant)

**,**__Charl's Law__**V ∝ 1/P**When**T**Constant.From

**,**__Charl'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)

∴ (P _{1}V_{1})/T_{1} = (P_{2}V_{2})/T_{2} = ConstantPV = 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**, then calculate the volume of this gas at

^{3}**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**= (1×2000×600)/(300×2)

_{2}**= 2000 cm**

^{3}