Ideal Gas

Real time intermolecular forces and external forces which act on the gas are not taken into consideration. All the above four relationships combine to form an ideal gas under ideal conditions. Combining all the laws together we get an ideal gas law.

We know from Boyle’s law

V ∝

We know from Charles’ law

V = T

We know from Avogadro’s law

V = n

From this we get,

V ∝

V = R

pV = RnT

R = ------ (1)

R is called gas constant. It is same for all gases. Therefore it is also called as Universal Gas Constant. Equation (1) is called ideal gas equation. Value of R can be calculated if the other values are known. Volume of one mole of an ideal gas under standard temperature and pressure conditions (273.15 K and 1 bar pressure) is 22.710981 L mol^–1. Value of R for one mole of an ideal gas can be calculated under these conditions as follows:

R =

= 8.314 Pa m³ mol⁻¹ K⁻¹

= 8.314 J K⁻¹ mol⁻¹

An Ideal gas equation is a relation between four variables and it describes the state of any gas, therefore, it is also called equation of state. Let us now go back to the ideal gas equation. This is the relationship for the simultaneous variation of the variables. If temperature, volume and pressure of a fixed amount of gas vary from T₁, V₁ and p₁ to T₂, V₂ and p₂ then we can write

R =

n R = and

n R =

= ------ (2)

Equation (2) is a very useful equation. If out of six, values of five variables are known, the value of unknown variable can be calculated from the equation (2). This equation is also known as Combined gas law.

Problems:

At 25 ͦC and 760 mm of Hg pressure a gas occupies 600 ml volume. What will be its pressure at a height where temperature is 10 ͦC and volume of the gas is 640 ml.

Solution:

We have,

T₁ = 25 ͦC

= 25 + 273

= 298 K

T₂ = 10 ͦC

= 10 + 273

= 283 K

V₁ = 600 ml

V₂ = 640 ml

p₁ = 760 mm of Hg

p₂ = ?

p₂ =

p₂ = 676.6 mm Hg

From the ideal gas equation (1),

R =

n =

We know that, n =

Also density d =

Therefore, we get the equation for molar mass

M = ------ (3)

Dalton’s Law of Partial Pressures:

This law states that “the total pressure exerted by a mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases” i.e., the pressures which these gases would exert if they were enclosed separately in the same volume and under the same conditions of temperature.

p_Total = p₁ + p₂ + p₃ +......(at constant T, V)

where,

p_Total the total pressure exerted by the mixture of gases,

p₁ + p₂ + p₃ +...... is the sum of the partial pressures of individual gases.

Gases are generally collected over water and therefore contain some water vapour. The pressure of dry mixture of gases can be calculated by subtracting vapour pressure of water from the pressure of the resulting mixture of gases. Pressure exerted by saturated water vapour is called aqueous tension.

p_Dry_gas = p_Total + Aqueous tension

Let us assume three gases enclosed in volume V at temperature T exert partial pressures p₁, p₂, and p₃, then,

p₁ = as only number of moles of gas n will change

p₂ =

p₃ =

p_Total = p₁ + p₂ + p₃

therefore,

p_Total = n₁ + n₂ + n₃

= n₁ + n₂ + n₃

Let ₁ = be the mole fraction of first gas. Then,

p₁ = ₁ × p_Total

Similarly, p₂ = ₂ × p_Total

p₃ = ₃ × p_Total

Assuming there are i number of gases, then we get a general equation,

p_i = _i × p_Total

Problems:

1. A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5 g neon. If the pressure of the mixture of gases in the cylinder is 25 bars. What is the partial pressure of dioxygen and neon in the mixture?

Solution:

We have,

m₁ = 70.6 g

m₂ = 167.5 g

p_Total = 25 bars

p₁ = ?

p₂ = ?

M = Relative atomic mass × Molar mass constant (1 g/mol)

M₁ = 32 ×1

= 32 g/mol

M₂ = 20.18 ×1

= 20.18 g/mol