7.1 Gaseous State of Matter

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🎯 Learning outcomes

By the end of this chapter, Students should be able to:

  1. List the postulates of kinetic molecular theory.
  2. State and explain Gas laws, related equations, and related numerical problems.
  3. Explain Boyle’s law, Charles’ law, Avogadro’s law, combined gas law, Dalton’s law, Graham’s law
  4. State and use the general gas equation PV = nRT in calculations.
  5. Explain the meaning of the universal gas constant and its significance.
  6. Distinguish between real gas and ideal gas.
  7. Explain qualitatively in terms of intermolecular forces and molecular size: The conditions necessary for a gas to approach ideal behavior are:
  8. Explain the cause of deviation of real gas from the gas laws.
1The Different States of Matter

Chemistry is defined as – the scientific study of MATTER, especially its composition, properties and transformation.

Matter – anything that occupies space, possesses mass and can be felt by one of the five senses.

On the basis of physical state, matter can be generally classified into solids, liquids and gases.

Solid Liquid Gas
fixed shape and size, not fixed shape but fixed size, nor fixed shape or fixed size,
could not be compressed could be compressed to a minimal extent could be compressed to a large extent
fixed position of the constituent particles moving constituent particles but movement relatively slow & within the body of liquid very fast-moving constituent particles in random motion
strong force of attraction between constituent particles moderate force of attraction between constituent particles force of attraction between constituent particles negligible
2Kinetic Theory of Gas and its Postulates
  1. Every gas is composed of tiny particles called molecules. Molecules of one kind of gas are identical in all respects.
  2. The gas molecules are incompressible, rigid spheres.
  3. The volume occupied by the individual particles of a gas is negligible compared to the volume of the gas itself.
  4. The particles of an ideal gas exert no attractive or repulsive forces on each other or on their surroundings. It is due to the high kinetic energy of the molecules that overcomes weak forces between them.
  5. Gravitational force or acceleration due to gravity is also negligible for the gas molecules.
  6. Gas particles are in a constant state of random motion and move in straight lines until they collide with another body.
  7. The collisions exhibited by gas particles are completely elastic; when two molecules collide, total kinetic energy is conserved.
  8. Pressure is exerted by the gases due to the continuous bombardment of the gas molecules with the wall of the container.
  9. The average kinetic energy of gas molecules is directly proportional to absolute temperature only; this implies that all molecular motion ceases if the temperature is reduced to absolute zero.
3Boyle’s Law

At constant temperature, the volume of a given mass of a gas is inversely proportional to its pressure.

Let P and V be the pressure and volume of a given mass of gas at a constant temperature, then

V ∝ 1/P
Or, V = K/P   (where K is the proportionality constant)
Or, VP = K

Let us consider a certain mass of gas that has pressure P1 and volume V1, and when the pressure changes to P2, the volume changes to V2. Then, according to Boyle’s law,

V1P1 = K …………… (1)
P2V2 = K ………….. (2)

From equations 1 and 2,

P1V1 = P1V2 at constant temperature for a given mass of a gas.

Significance of Boyle’s law

At high altitudes, atmospheric pressure is low; therefore, the same mass of gas (oxygen) occupies more volume. Therefore, while inhaling (breathing in), a person doesn’t get a sufficient amount of oxygen. For this reason, mountaineers carry an oxygen cylinder.

Graphical representation of Boyle’s law

A plot of volume versus 1/pressure of a constant mass of a gas at constant temperature gives a slanted line.

A plot of volume versus pressure of a constant mass of a gas at constant pressure gives a parabolic curve.

A plot of PV versus V and PV versus P gives a straight horizontal line.

(Refer to textbooks or classwork copies for the graphs)

4Charles’s Law

Statement

The volume of a given mass of gas is directly proportional to the temperature in absolute scale at a constant pressure.

Let a given mass of a gas occupy volume V at absolute temperature T, then according to Charles’s law,

V ∝ T

Explanation

If V1 and T1 are the volume and temperature of a fixed mass of gas, and if the temperature is changed to T2, so that the volume becomes V2. Then

V1 ∝ T1 or, V1 = kT1 or V1/T1 = k …….. (1)
And V2 ∝ T2 or, V2 = kT2 or, V2/T2 = k ……… (2)

Then, from equations (1) and (2),

V1/T1 = V2/T2

Graphical representation and concept of absolute zero

(Refer to textbooks or classwork copies for the graphs and explanation)

5Combined Gas Equation

(Refer to textbooks or classwork copies)

1) Derive the combined gas equation.

Numericals

2) Calculate the pressure required to compress 2000ml of a gas at 700mm pressure and 20°C into a container of 100ml capacity at a temperature of -150°C. (ans = 5877.13 mm or 24.63 atm)
Solution

Given, V1 = 2000ml

P1 = 700mmHg

T1 = 20°C = 20+273K = 293K

V2 = 100 ml

P2 = ?

T2 = -150°C = -150+273K = 123K

We know,

P1V1/T1 = P2V2/T2
3) A spherical balloon of 21 cm diameter is to be filled with hydrogen gas at NTP from a cylinder containing gas at 20 atm and 27°C. If the cylinder can hold 2.66 liters of water vapour at NTP, calculate the number of balloons that can be filled up.
Solution

Given,

For cylinder, at NTP V1 = ?   P1 = 1 atm   T1 = 273 K

Px = 20 atm   Tx = 27°C = 300 K   Vx = 2.66 l

We know that, P1V1/T1 = PxVx/Tx

V1 = 48.43l liters

Out of the total 48.431 liters of gas, 2.66 liters will be left in the cylinder,

So, remaining gas available to fill balloons = 48.431-2.66 = 45.771 liters

= 45771 cc

Volume of balloon in NTP,

V2 = 4/3 πr³ = 4/3 π 10.5³ = 4849 cm³

P2 = 1 atm   T2 = 273 K

No of balloons that can be filled = total volume of gas in the cylinder / volume of one balloon

= 45771/4849 = 9.44

6Ideal Gas Equation, Universal Gas Constant & Deviation from Ideality

Ideal gas and ideal gas equation

Derivation of the ideal gas equation

(Refer to textbooks or classwork copies)

Universal gas constant and its significance

Concept of ideal gases

(Refer to textbooks or classwork copies)

Deviation of real gas from ideality

(Refer to textbooks or classwork copies)

4) 525 cc of a gaseous compound at 28°C and 730 mm Hg pressure was found to be 0.9 g. What will be the volume of 2g of the gas at 30°C and 760 mm Hg pressure? [Given R = 0.0821 liter atm K⁻¹mol⁻¹]
Solution

Given, V1 = 525 cc = 0525/1000 l = 0.525 l

T1 = 28+273 K = 301K

P1 = 730 mm Hg = 730/760 atm

m1 = 0.9 g

From the ideal gas equation,

P1V1 = m1/M RT1

m2 = 2 g

T = 30°C = 30+273 K = 303 K

P = 760 mmHg = 1 atm

V = ?

Again, from the ideal gas equation,

PV = m/MRT

Ans: V = 1.13 ltr

7Dalton’s Law of Partial Pressure

It states, “The total pressure of a mixture of non-reacting gases is the sum of the partial pressure of the constituent gases.” The partial pressure of a component gas is defined as the pressure of the gas if it were occupying the whole mixture volume alone.

Let us consider three gases, A, B, and C, which are non-reacting gases and are mixed in a vessel. The mixture exerts a total pressure of PT. Further, suppose the partial pressures of the three gases are P1, P2, and P3, respectively. Then, according to Dalton’s law of partial pressure,

PT = P1 + P2 + P3

Application of Dalton’s law of partial pressure

Numericals

1) Two vessels of capacity 1.5 liters and 2 liters contain hydrogen gas and oxygen gas, respectively, under a pressure of 750 mm and 100 mm. The gases are mixed together in a 5-litre vessel. What will be the final pressure of the mixture?
Solution

For Oxygen: before mixing VO = 2 ltr

PO = 100 mm

After mixing, V1 = 5 liter

partial pressure of O2, p1 = ?

At constant temperature, applying Boyle’s law, VOPO = V1p1

For hydrogen: before mixing Vh = 1.5 ltr   Ph = 750 mm

After mixing, V3 = 5 litres, partial pressure of H2, p2 = ?

At constant temperature, applying Boyle’s law, VHPH = V1p2

From Dalton’s law of partial pressure,

Total pressure P = p2 + p1

8Graham’s Law of Diffusion

Statement

Under similar conditions of temperature and pressure, the rate of diffusion of gases is inversely proportional to the square root of their densities.

Mathematical expression and explanation

Considering r, d, the rate of diffusion and density of a gas,

r ∝ 1/√d

Similarly, if r1 and r2 are the rates of diffusion and d1 and d2 are the densities of two gases,

r1 ∝ 1/√d1 or, r1 = k/√d1 …………….(1)
r2 ∝ 1/√d2 or r2 = k/√d2 …………….(2)

From equations (1) and (2),

r1/r2 = √(d2/d1) …………………….. (3)

From Avogadro’s hypothesis, we know that the molecular mass of a gas = 2 x its vapour density,

Or, d = M/2   So correspondingly, d1 = M1/2 and d2 = M2/2

Where M1 and M2 are the molecules of the two gases.

Putting these values in equation (3), we get

r1/r2 = √(M2/M1) ……………………..(4)

This is another expression of Graham’s law of diffusion.

Numericals

2) How long will it take 500 ml of hydrogen gas to diffuse through a partition if 250 ml of oxygen diffuses in 50 minutes under similar conditions? (ans = 25 minutes)
Solution

Given, volume of H2 (V1) = 500 ml

t1 = ?

M1 = 2 amu

Volume of O2 (V2) = 250 ml

t2 = 50 minutes

M2 = 32

(M1/M2) = (r2/r1)²

3) The rate of diffusion of a saturated hydrocarbon (CnH2n+2) gas is 1.206 times that of SO2 gas under identical conditions. Find the molecular mass and the value of ‘n’ for the gas. (Molecular mass of SO2 = 64) (Ans: molecular mass = 44, n=3)
Solution

Suppose the rate of diffusion of SO2 = x

Rate of diffusion of hydrocarbon = 1.206 x

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