WBCHSE Class 11 Chemistry For Ideal Gas law – Definition, Equation Notes

Avogadro’s Hypothesis

You have already learnt that Avogadro had proposed the hypothesis that equal volumes of all gases contain an equal number of molecules under the same conditions of pressure and temperature.

This means the volume of a gas is directly proportional to the number of molecules (N) at a given temperature and pressure.

V ∝ N at constant temperature and pressure. You also know that 1 mol of any gas at stp contains 6.022 x 10²³ molecules, which means that the number of molecules and the number of moles are related as follows.

Number of moles (n) = \(\frac{\text { number of molecules }(N)}{\text { Avogadro constant }\left(N_{\mathrm{A}}\right)}\)

Thus, Avogadro’s hypothesis can be expressed in terms of the number of moles rather than the number of molecules. Equal volumes of all gases contain the same number of moles under the same conditions of pressure and temperature, or volume is directly proportional to the number of moles at constant temperature and pressure.

V ∝ n at constant temperature and pressure

or V = a constant x n, where n is the amount or number of moles of a substance.

• The above observation was generalised by Amedeo Avogadro in the year 1811 as follows. All gases containing an equal number of moles occupy the same volume at the same temperature and pressure. This means that all the gases with the same value of n, at the same temperature and pressure, will have exactly the same value of V. It also follows that 1 mol of any gas occupies the same volume at the same temperature and pressure.
• This volume is known as molar volume (V_m) of the gas. Experiments show that the volume of 1 mol of any gas at standard temperature and pressure (stp) is 22.7109 L. Standard temperature refers to 0°C or 273.15 K and standard pressure to 1 bar. (According to the earlier convention stp referred to the same temperature, CTQ, but 1 atm pressure, and the molar volume was 22.4 L.) Accordingly the molar volume of any gas under these conditions is 22.7 L.

The total volume of n moles of a gas is V = nV_m.

Ideal Gas Equation

You have learnt that provided the pressure is constant, the volume of a gas is directly proportional to its Kelvin temperature. You have also leamt that the volume of a gas varies inversely with the pressure applied on it, provided the temperature is constant.

And then that the volume of a gas is proportional to the number of moles at constant pressure and temperature. Is there a way in which we can sum up these three relations into one?

Let us consider a particular amount of gas occupying volume V₁ at pressure p₁ and temperature T₁ What volume will it occupy at some other pressure p₂ and temperature T₂? Since the laws we have considered so far do not consider situations in which both the pressure and temperature change, let us take one step at a time. Let the temperature remain constant in the first step and the pressure change from p₁ to p₂ and let the volume change from V₁ to V_x, i.e.,

⇒ \(p_1 V_1 \stackrel{T_1 \text { constant }}{\longrightarrow} p_2 V_x\)

Then, according to Boyle’s law (p₁ V₁ = p₂V₂),

⇒ \(V_x=p_1 V_1 / p_2\) …..(1)

Now let the pressure remain constant at p₂ and let the temperature change from T₁ to T₂ and the volume change from V₁ to V₂, i.e.,

⇒ \(V_x T_1 \stackrel{p_2 \text { constant }}{\longrightarrow} V_2 T_2\)

Then according to Charles’s law \(\left(V_x / T_1=V_2 / T_2\right)\) ……(2)

From equations (1) and (2),

∴ \(p_1 V_1 / p_2=V_2 T_1 / T_2 \quad \text { or } \quad \frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2} \text {. }\)

This is a very useful equation for calculating the value of one of the variables if the other five arc known. It A implies that for a particular amount of gas, the ratio of the product of pressure and volume to the absolute temperature is constant

pV/T = constant, say K. ……. (3)

This constant K depends upon the mass or quantity of the gas we are considering. How do we make Equation (3)independent of the quantity of the gas? According to Avogadro’s hypothesis, the volume of a gas at constant temperature and pressure is directly proportional to the number of moles. In that case K in Equation (3) must also be proportional to the number of moles.

K = nR n = number of moles

⇒ pV/T = nR, where R is a constant independent of the amount of gas.

or pV = nRT.

This equation is called the ideal gas equation. The word ideal is used because it describes the behaviour of an ideal gas. In reality, no gas is ideal but you will read about that later in this chapter. The constant R is independent of the quantity of the gas and is the same for all gases. It is called the universal gas constant and has the value of pV/T for 1 mol of a gas.

The ideal gas equation may be alternatively derived by combining the four measurable variables of a gas as follows:

V ∝ n(p,T are constant) — Avogadro’s law

V ∝ r (n, p are constant) — Charles’s law

V ∝ yp («, T are constant) — Boyle’s law

Combining the above three equation,

⇒ \(V \propto \frac{n T}{p} \quad \text { or } \quad p V \propto n T \quad \text { or } \quad p V=n R T \text {. }\)

The ideal gas equation can be used to find the molar mass and density of a gas. We can express the ideal gas equation in terms of molar mass as follows.

⇒ \(p V=\frac{m}{M} R T{\left[n=\frac{m}{M}=\frac{\text { mass of gas }}{\text { molar mass of gas }}\right]}\)

⇒ \(p=\frac{m}{V} \frac{R T}{M}{\left[\frac{m}{V}=\rho \text {, density }\right]}\)

⇒ \(p=\rho \frac{R T}{M}\)

⇒ \(M=\rho \frac{R T}{n}\)

Universal gas constant: What exactly does the gas constant signify? To understand its nature, let us analyse the different quantities in the ideal gas equation.

pV = nRT

∴ \(R=\frac{p V}{n T}=\frac{\frac{\text { force }}{\text { area }} \times \text { volume }}{n \times \text { temperature }}=\frac{\frac{\text { force }}{\text { length }}{ }^2 \times \text { length }^3}{n \times \text { temperature }^3}\)

= \(\frac{\text { force } \times \text { length }}{n \times \text { temperature }}\)

Force multiplied by length has the dimensions of work or energy. Therefore, R lias the dimensions of (let us say R represents) work done or energy per degree per mole.

Depending upon the units used to express work (or energy), R may have different numerical values. The values of R can be calculated using the gas equation once the values of the pressure, volume and temperature are known.

1. In cgs units

Since p = hpg,

density of Hg = 13.6 g cm^-3, g = 981 cm s^-2

p = 76 cmHg = (76 x 13.6 x 981) dynes cm^-2

V = 22400 mL

T = 273.15 K

n = 1 mol

∴ R = \(\frac{p V}{n T}=\frac{76 \times 13.6 \times 981 \times 22400}{1 \times 273}=8.314 \times 10^7 \mathrm{erg} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\).

2. In SI units

p = 1 bar or 10⁵Pa = 10⁵ Nm^-2

V = 22.7 x 10^-3 m³

T = 273.15K

∴ R = \(\frac{1 \mathrm{~atm} \times 22.4 \mathrm{~L}}{1 \mathrm{~mol} \times 273.15 \mathrm{~K}}=0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)

3. If p i taken as 1 atm and V as 22.4 L

R = \(=\frac{1 \mathrm{~atm} \times 22.4 \mathrm{~L}}{1 \mathrm{~mol} \times 273.15 \mathrm{~K}}=0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)

4. If p is taken as 1 bar and T as 273 K (stp conditions)

We know that the volume occupied by 1 mol of any gas is 22.7109 L.

∴ R = \(\frac{(22.7109 \mathrm{~L})(1 \mathrm{bar})}{(273.15 \mathrm{~K})(1 \mathrm{~mol})}=0.08314 \mathrm{~L} \text { bar } \mathrm{K}^{-1} \mathrm{~mol}^{-1} .\)

Note that the value differs from the value in SI units only in the place of decimal point.

Example 1. 40 mL of oxygen was collected at 10°C and 1 bar pressure. Calculate its volume at 273 K and 1.013 bar.
Solution:

Given

40 mL of oxygen was collected at 10°C and 1 bar pressure.

p₁ = 1 bar

p₂ = 1.013 bar

V₁ = 40mL

V₂ = ?

T₁ = 10C = 283K

T₂ = 273 K

According to the gas equation

⇒ \(\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}\)

∴ \(\quad\frac{1 \times 40}{283}=\frac{1013 \times V_2}{273}\)

∴ \(\quad V_2=\frac{1 \times 40 \times 273}{283 \times 1013}=38.07 \mathrm{~mL}\).

Example 2. Calculate the number of moles of hydrogen contained in 20° L of the gas at 27°C and 70 cmHg pressure. Given that R = 0.0821 L atm K^-1 mol^-1.
Solution:

Given R = 0.0821 L atm K^-1 mol^-1

V = 20L

p = 70 cm = 70/76 atm

T = 27°C = 300 K

Using the ideal gas equation pV = nRT,

n = \(\frac{p V}{R T}=\frac{(70 / 76)(20)}{(0.0821)(300)}=0.747\)

Example 3. A gas has a density of 1.2504 kg m^-3 at 0°C and a pressure lx 10⁵Pa. Calculate its molar mass (given R = 8.314 J K^-1 mol^-1).
Solution:

Given

A gas has a density of 1.2504 kg m^-3 at 0°C and a pressure lx 10⁵Pa.

⇒ \(p V_m=p \cdot \frac{M}{d}=R T\)

or \(M=\frac{d R T}{p}\)

= \(\frac{1.2504 \times 8.314 \times 273.15}{10 \times 10^5} \quad\left(1 \mathrm{~Pa}=1 \mathrm{Nm}^{-2}\right)\)