Gaseous State- Boyle’s and Charle’s Law

The most mobile state of matter with no fixed shape and volume is gaseous state.

Characteristics of gases

1. A gas has neither fixed shape nor fixed volume. It takes the shape and volume in which the gas is kept.
2. The distance between the gas molecules are large and hence the molecular force of attraction is weak.
3. Gaseous molecules can be compressed by applying pressure and can expand indefinitely.
4. They have low density and can diffuse through each other to form a homogeneous mixture.

Some physical quantities in gaseous state

i. Mass: Mass of gaseous substance is usually expressed in gram.

ii. Mole: The quantity of matter can be expressed in terms of mole (n).

n = \frac{Mass(gm)}{Molar \hspace{1mm}mass(gm/mol)}

Molar mass is the mass of 1 mole of a substance.
1 mol of gas = Molecular mass in gram
= 22.4 liter volume at NTP
= 6.023 x 10²³ number of molecules

iii. Volume: Volume of the gas is the volume of the container. At normal temperature (273K) and pressure (760 mm Hg), the volume occupied by one mole of gas is 22.4 L.
1 liter=1000 ml=1dm³=1000cm³

iv. Pressure: When gas molecules collide with each other and with the wall of the container, pressure is exerted. Gas pressure is the force exerted by the gaseous molecules per unit area on the wall of the container. The pressure created by atmospheric air is called atmospheric pressure. Atmospheric pressure is measured by a barometer and gas pressure by a manometer.
1 atm= 760 mm Hg=760 torr=101325 Nm^-2 =1.01325 bar

Gas laws

The laws which show the relationship between various quantities of gases (P, V, T and n) are called gas laws.

1. Boyle’s law (PV relation)

Statement: At constant temperature, the volume of a given mass of gas is inversely proportional to its pressure. Mathematically, let P be the pressure and V be the volume of a given mass of gas, then at a constant temperature,

V ∝ 1/P
or, V=k/P   k=proportionality constant
or, PV = k………(i)

Boyle’s law also states that “At constant temperature, the product of pressure and volume at a given mass of a gas is constant.

A certain mass of gas having pressure P₁ and volume V₁ is changed to pressure P₂ and volume V₂ at a constant temperature. Applying Boyle’s law,

P₁V₁ = k ——(ii)
P₂V₂ = k ——(iii)

From equation (ii) and (iii), we get

P₁V₁=P₂V₂ ——(iv)

Graphical representation of Boyle’s law

Significance of Boyle’s law

• It shows the relationship between pressure and the density of the gas. At high altitude, gaseous pressure is low, the density of the gas is also low. This results in less availability of oxygen gas for breathing at high altitude. So, mountaineers carry oxygen cylinder for breathing purpose.
• It shows how the volume of a given mass of gas varies with pressure.

Charle’s Law

Statement – At constant pressure, the volume of a given mass of a gas increases or decreases by 1/273^th of its original volume at 0 °C for every 1 °C rise or fall in temperature.

Let V_o be the volume of gas at 0 °C, then

Volume at 1 °C (V₁) = V_o + V_o /273 = V_o (1+ 1/273)
Volume at 2 °C (V₂) = V_o + V_o x 2/273 = V_o (1+ 2/273)
Volume at t °C (V_t) = V_o + V_o x t/273 = V_o (1+ t/273)

= V_{o}\frac{(t + 273)}{273}\\ V_{t} = \frac{V_{o}}{273}\times T \, ----(i)

Here, T = (t+273) is the temperature in the Kelvin scale and V_o /273 is constant.

So, equation (i) becomes V ∝ T ——(ii)

Hence, Charle’s law can also be stated as,” At constant pressure, the volume of a given mass of a gas is directly proportional to the temperature in Kelvin state (absolute scale)”.

or,\:V = constant\; T\\ or,\:\frac{V}{T}= constant

Let V₁ and V₂ be the volume of gas at temperature T₁ and T₂, then,

\frac{V_{1}}{T_{1}}= constant\ and \ \frac{V_{2}}{T_{2}}= constant \\ \\so, \frac{V_{1}}{T_{1}}= \frac{V_{2}}{T_{2}}

Graphical representation of Charle’s Law

Significance of Charle’s law

• It gives the relation of volume and temperature at constant pressure.
• It introduced the concept of the Kelvin scale of temperature.
• Hot air balloons are made on the basis of Charle’s law. At high temperature, air expands, becomes less denser and rises up.

Solved Examples

1. 600 ml of chlorine gas at 27 °C was cooled to -23 °C at the same pressure, calculate the contraction in volume.
solution:

From Charle’s law,

\frac{V_{1}}{T_{1}}= \frac{V_{2}}{T_{2}}\\ or,\; \frac{600}{300}= \frac{V_{2}}{250}\\ \therefore V_{2} = 500\;ml\\ Hence, contraction\; in\; volume= 600-500= 100\;ml

Equation of state and combined gas equation

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

V ∝ 1/P —–(i)

According to Charle’s law, the volume of a given mass of a gas is directly proportional to the temperature in Kelvin scale at constant pressure.

V ∝ T —–(ii)

Combining both laws,

V\; \alpha\; \frac{T}{P}\\\\ or,\; V= \frac{KT}{P}\\\\ or,\; PV = KT

where K is constant and its value depends upon the amount of substance present.

a. If 1 gm of gas is taken, then K is called specific gas constant. Since 1 gm of different gas occupies different volume, its value is different for different gas.

b. If one mole a gas is taken, its value is same for all gases and called universal gas constant and denoted by R.

PV = RT

For n mole of gas, PV=nRT —–(iv)

This equation (iv) is called the equation of state or ideal gas equation.

From equation (iv),

\frac{PV}{T}= nR\\\\ \therefore \frac{PV}{T}= K

Let P₁, V₁ and T₁ be the initial and P₂, V₂ and T₂ be the final pressure, volume and temperature of a given mass of gas respectively, then

\frac{P_{1}V_{1}}{T_{1}}= K\; and\; \frac{P_{2}V_{2}}{T_{2}}= K

Combining these two equations,

\frac{P_{1}V_{1}}{T_{1}}= \frac{P_{2}V_{2}}{T_{2}}

This is called combined gas equation.

Universal gas constant (R)

The ideal gas equation is

PV=nRT \\ or,\; R= \frac{PV}{nT}\\ =\frac{pressure\times volume}{no.\;of\; moles \times temp.}\\ = \frac{force/area \times volume}{no.\; of\; moles \times temp.}=\frac{force/length^2 \times length^3}{no.\; of\; moles \times temp.}\\ =\frac{force\times length}{no.\; of\; moles \times temp.}\\ = energy\;K^{-1}\;mol^{-1}

Universal gas constant (R) is the amount of energy required to raise the temperature of one mole of an ideal gas by one degree on the Kelvin scale.

The Value of R = 0.0821 L atm K^-1 mol^-1
= 1.987 cal K^-1 mol_-1
= 8.314 x 10⁷ erg K^-1 mol^-1
= 8.314 J K^-1 mol^-1

Some Important Questions

1. Why gases do not settle on the bottom of the container?
2. Why mountaineers carry oxygen cylinder with them?
3. Why does the size of the air balloon expand as it moves to high altitude?
4. Draw volume vs temperature graph and find the temperature corresponding to zero volume.
5. State Dalton’s law of partial pressure.
6. Why does hydrogen diffuse faster than oxygen?
7. Calculate the value of R in Latmmol-1K-1.
8. How is the density of a gas is related to its molecular mass?
9. What is aqueous tension? Write the formula to calculate the pressure of dry gas?
10. What is an ideal gas? Under what condition it behaves like an ideal gas?
11. What are the reasons for the deviation of real gas from ideal behaviour?
12. Derive PV = nRT
13. What are the main postulates of the Kinetic theory of gases?
14. State and explain Graham’s law of diffusion.
15. Starting from P₁V₁=P₂V₂, show that P₁/d₁=P₂/d₂.

Some Important Numericals

1. A sample of gas occupies 50 cc volume at 730 mmHg and 20°C. What volume will the gas occupy if pressure is changed to 650 mmHg keeping the temperature constant? (56.2 cc)
2. A vessel of 1500 cc capacity contains a certain mass of gas at 15°C and 1 atm pressure. The gas was transferred completely into another vessel of 2500 cc capacity at the same temperature. What will be the pressure of the gas in the new vessel? (456 mm)
3. 273 ml of gas is taken in a vessel at 0°C. What will be the volume of gas at 1oC and – 1°C at the same pressure? (274 and 272 ml)
4. Two litres of gas is heated from 10 to 50°C at the same pressure. What will be the resulting volume? (2.28 lit)
5. 150 cc of nitrogen gas at 25°C is cooled to 10°C. Find the new volume of gas at constant pressure. (142.44 lit)