Introduction to Boyle's Law

Applications of Boyle's law

a)As we know from Boyle's law, PV = constant. With the increase in altitude, the pressure of the gas decreases which causes the decrease in the density of gas decreases. So, mountain climbers may face a problem called altitude sickness. To recover this problem, they carry oxygen gas cylinder with them.

b)From Boyle's law, pressure and volume have an inverse relation. With the increase in altitude, the pressure decreases which in turn increases the volume of the gas. Therefore, the size of air balloon goes on increasing as it ascends the height.

Charle’s law

• Jacques Charles, a French chemist introduced this law in 1787 AD.
• He gave the relationship between volume and temperature at constant pressure.

Charle’s law states that “ The volume of a given mass of given mass increase or decrease by (\(\frac{1}{273}\) )^th of its volume at 0⁰C for every ⁰C rise or fall in temperature”.

Let V_o be the volume of the gas at 0⁰C,

Volume of the gas at 1⁰C

= V_o + V_o \(\frac{1}{273}\)

= \( \frac {V_o ( 1 + 273)}{273}\) at 1⁰C

Volume of the gas at 3⁰C

= V_o + V_o \(\frac{3}{273}\)

= \( \frac {V_o (3 + 273)}{273}\) at 3⁰C

Volume of the gas at t⁰C

= V_o + V_o \(\frac{t} {273}\)

= \( \frac {V_o (t + 273)}{273}\) at t⁰C

Let 'V' be the final temperature of the gas at 't⁰C',

Then, V = \( \frac {V_o (t + 273)}{273}\) -------------(i)

Equation (i) is the mathematical form of Charle's law.

Verification of Charle's law

Rearranging equation (i)

V = V₀ + V₀ \( \frac{t}{273}\)

or, V = \( \frac{V_o}{273}\) t + V₀ -(ii)

Equation (ii) is in the form of y = mx + c. So, a graph of 'V' vs 't' gives a straight line with slope \( \frac{V_o}{273}\) and y-intercept 'V₀'

Alternative form of Charle's law

Charle’s law states that “ The volume of a given mass of given mass increase or decrease by (\(\frac{1}{273}\) )^th of its volume at 0⁰C for every ⁰C rise or fall in temperature”.

Mathematically,

V = \( \frac {V_o (t + 273)}{273}\) -------------(i)

Let, V₁ and V₂ be the volume of given mass of gas at t₁⁰C and t₂⁰C respectively. Then, from equation (i)

V₁ =\( \frac {V_o (t_1 + 273)}{273}\) ----(ii)

V₂ =\( \frac {V_o (t_2 + 273)}{273}\) ----(iii)

Dividing euqation (ii) by (iii)

\( \frac {V_1}{V_2}\) = \( \frac {273 + t_1}{273 + t_2}\)

Since, 273 + t₁ = T₁ and 273 + t₂ = T₂

\( \frac {V_1}{V_2}\) =\( \frac {T_1}{T_2}\)

∴ V∝ T at constant pressure

From above expression, Charles law can also be regarded as " pressure remaining constant, volume of a given mass of gas is directly proportional to the absolute temperature"

Verification,

V∝ T

or, V = kT

This equation is the form of y = mx. So, a graph of 'V' vs "T' gives a straight line passing through origin as shown in the figure below.

Absolute zero ( Kelvin zero) and Absolute scale of temperature

From Charle's law

V =\( \frac {V_o (t + 273)}{273}\)

when, t = -273⁰C

V = V_o - V₀ = 0

At -273⁰C, the volume of the gas becomes theoretically 0. All the gases become solid or liquid before this temperature is reached. This hypothetical temperature is known as absolute zero. It is denoted by OA, OK, etc.

The temperature scale based on the absolute zero as its starting point is known as Kelvin scale of temperature or absolute scale of temperature.

Relation between density and temperature of gas

From definition of density

Density (D) = \( \frac{Mass(M)}{Volume(V)}\)

From Charles law,

V = kT, where '' is a proportionality constant

or, \( \frac {M}{D}\) = kT

or, \( \frac {M}{k}\) = DT

For a given mass of gas, the value of 'm' is constant. So, \( \frac{M}{k}\) also gives another constant value

Then, TD = constant

Hence, T₁D₁ = T₂D₂............T_nD_n at constant pressure