Dalton’s Law of Partial Pressure, Boyle's Law and Charle's Law

Parameter of gas

Gas can be explained in terms of pressure, volume, temperature and entropy. These are called parameter of gas. The relation between these three parameters keeping one constant at a time is known as gas law. The gas law reduced into the simple equation is known as equation of state.

Dalton’s Law of Partial Pressure

It states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressure of the individual gas when there is no chemical interaction between them.

Consider two bulbs A and B separated by a tap key T. The bulb A has a V₁ gas at pressure p₁ and bulb B has V₂ gas at pressure P₂. When the tap key is opened the final volume occupied by each gas is (V₁ + V ₂). Experimentally it is found that total pressure exerted by the mixture of gases is

$$p = \frac {p_1V_1}{V_1 + V_2} + \frac {p_2V_2}{V_1 + V_2} \dots (i)$$

If \( p_1'and p_2'\) are the pressure exerted by first and second gas when they are mixed, then total pressure exerted by first and second gas when they are mixed, then total pressure of the mixture can be written as

$$p = p_1' + p_2'\dots (ii)$$

For first gas,

$$p_1V_1 = p_1' (V_1 + V_2) $$

$$\text{or,} p_1' = \frac {p_1V_1}{V_1 + V_2} $$

For second gas,

$$p_2V_2 = p_2'(V_1 + V_2) $$

$$\text{or,} p_2' = \frac {p_2V_2}{V_1 + V_2} $$

Putting values of and in equation (ii), we get

$$p = \frac {p_1V_1}{V_1 + V_2} + \frac {p_2V_2}{V_1 + V_2} \dots (iii)$$

Equation (iii) shows Dalton’s law of partial pressure.

Gas Lasws

Boyle’s law: Boyle’s law states that at a constant temperature, the volume of the given mass of gas is inversely proportional to the pressure.

\(\text{or,}p\propto \frac{1}{v}\)P a at constant temperature

$$ PV = constant $$

Note: Gas obeys the gas law at high temperature and low pressure.

The graph between pressure and volume at constant temperature is shown below:

Charle’s law (at constant pressure): It states that at a constant pressure, the volume of the given mass of gas is directly proportional to its absolute temperature.

\(\text{i.e.}V \propto T\) at constant pressure

$$\frac {V}{T} = constant, \frac {V_1}{T_1} =\frac {V_2}{T_2} = \dots \dots \dots\frac {V_n}{T_n} $$

Charle’s law (at constant volume) : Charle’s law at constant volume states that at constant volume, the pressure of the given mass is directly proportional to its absolute temperature.

\(\text{i.e.} P \propto T\) at constant volume

$$\frac {P}{T} = constant$$

$$\frac {P_1}{T_1} =\frac {P_2}{T_2} = \dots \dots \dots\frac {P_n}{T_n} $$

Volume Coefficient or Coefficient of Volume Expansion

Volume coefficient is defined as the ratio of the change in volume to the original temperature per degree rise in temperature in constant pressure.

$$\text{i.e.} \gamma_p = \frac{V - V_o}{V_o \times \theta}$$

$$\text{or,} \gamma_p \times V_o \times \theta = V - V_o$$

$$\text{or,} V_o (1 + \gamma_p \times \theta) = V $$

Experimentally, \(\gamma_p =\frac {1}{273 k}\)

$$V = V_o(1 + \frac{\theta}{273})$$

$$\text{or,} V = V_o(\frac{273 +\theta}{273})$$

$$273 + \theta = T $$

$$273 + 0 = T_o$$

$$\text{or,} V = V_o(\frac{T}{T_o} $$

$$\text{or,} \frac{V}{T} = \frac{V_o}{T_o} $$

$$\text{or,} V \propto T $$

Hence, Charle’s law at constant pressure is verified.

Pressure coefficient

The pressure coefficient is defined as the ratio of the change in pressure to original pressure per degree rise in temperature at constant volume.

$$\text{i.e.} \gamma_v = \frac{p - p_o}{p_o \times \theta}$$

$$\text{or,} \gamma_p \times V_o \times \theta = V - V_o$$

$$\text{or,} P_o (1 + \gamma_v \times \theta) = P$$

Experimentally, \(\gamma_v =\frac {1}{273 ^oC}\)

$$p = p_o(1 + \frac{\theta}{273})$$

$$\text{or,} p= p_o(\frac{273 +\theta}{273})$$

$$273 + \theta = T $$

$$273 + 0 = T_o$$

$$\text{or,} p =p_o(\frac{T}{T_o} $$

$$\text{or,} \frac{p}{T} = \frac{p_o}{T_o} $$

$$\text{or,} p\propto T $$

Hence, Charle’s pressure law is verified.

Hence, Charle’s law is verified.

Equality of Pressure and Volume Coefficient of Gas (different figure numbering from book)

Let us consider an ideal gas is enclosed piston having pressure and volume ‘P₁’ and ‘V₁’ respectively at temperature 0as shown in the figure. Now, if the gas is expanded to the volume ‘V₂’ by changing its temperature from 0 to such that pressure of the gas is kept constant.

So, at constant pressure

From Charle’s law of volume

We can write

$$V_2 = V_1[1 + \gamma_p(\theta - 0)]\dots(i)$$

Again, if the volume of the gas is compressed to V₁ by changing its pressure to P₂ such that the temperature of the gas is kept constant as shown in the figure

From Boyle’s law

At constant temperature,

$$p_1V_2 = p_2V_1$$

using equation (i)

$$p_1V_1[1 + \gamma_p(\theta - 0)] = p_2V_1$$

$$\therefore p_2 = p_1[1 + \gamma_p(\theta - 0)]\dots(ii)$$

Again , from Charle’s law of pressure

$$ p_2 = p_1[1 + \gamma_v(\theta - 0)]\dots(iii)$$

From equation (ii) and (iii)

$$\gamma_p = \gamma_v$$

Hence, if the gas obeys Charle’s law and Boyle’s law, then the volume coefficient and pressure coefficient of the gas must be equal.