Chemistry 2nd Paper · Classes 11-12 · Chapter 1 · 1.5

Kinetic Theory of Gases.

Take one mole of hydrogen and one mole of nitrogen at the same temperature. They carry exactly the same kinetic energy, yet the hydrogen molecules rush about nearly 3.7 times faster. How can both be true? The answer is in the kinetic theory of gases.

1. The six postulates

Why does a gas push on its walls, spread out and respond to heat? The kinetic molecular theory of gases explains the basic reason behind this behaviour, and the contributions of Maxwell and Boltzmann are especially notable. The theory stands on six postulates.

Postulate 1, structure: a gas is made of countless tiny particles, atoms or molecules, that move very fast in straight lines in every possible direction. Postulate 2, volume: the total volume of the molecules is negligible compared with the volume of the container, so most of the space is empty.

Postulate 3, attraction and repulsion: there is none between gas molecules; they are free of each other's influence. Postulate 4, collisions: molecules collide with one another and with the walls perfectly elastically, so kinetic energy is not changed into any other form. At a fixed temperature the total kinetic energy of the molecules therefore stays constant.

Postulate 5, pressure: the force the molecules exert per unit area of the container wall through ceaseless elastic collisions is the pressure of the gas. Postulate 6, kinetic energy: the average kinetic energy of the molecules is proportional to the Kelvin temperature of the gas.

2. The kinetic equation and collisions

From the postulates Clausius established the kinetic equation of an ideal gas: PV = (1/3) m N c². Here m is the mass of each molecule, N the number of molecules in the sample and c the root-mean-square speed.

When two molecules come close, a repulsive force becomes active and pushes them in opposite directions; this is a molecular collision. The mean of the distances between successive collisions is the mean free path, l: for four successive distances a, b, c, d, l = (a + b + c + d)/4. The minimum distance between the centres of two molecules during a collision is the collision diameter, written σ (sigma).

3. The three molecular speeds

The molecules do not all move at one speed, so the speed of gas molecules is expressed in three ways: (1) the root-mean-square speed, or RMS speed, (2) the average speed, (3) the most probable speed.

The RMS speed is the square root of the mean of the squares of the speeds: c = √[(c₁² + c₂² + ... + c_N²)/N]. It also equals c = √(3RT/M). The derivation is short: for one mole PV = (1/3) M c² and PV = RT, so c² = 3RT/M. In a calculation take M in kg mol⁻¹ and R = 8.314 J mol⁻¹ K⁻¹, and c comes out in m s⁻¹.

The average speed is the arithmetic mean of the various speeds: c̄ = (c₁ + c₂ + ... + c_N)/N = √(8RT/πM). The RMS speed is slightly greater than the average speed. For three molecules at 2, 3 and 4 cm s⁻¹ the average speed is 3 cm s⁻¹, while the RMS speed is √[(4 + 9 + 16)/3] = 3.109 cm s⁻¹.

Maxwell used the theory of probability to prove that at any instant most molecules share one particular speed. That is the most probable speed, α: α = √(2RT/M).

All three speeds are multiples of √(T/M); only the multiplier differs: α = 1.29×10⁴ √(T/M), c̄ = 1.45×10⁴ √(T/M), c = 1.58×10⁴ √(T/M) cm s⁻¹. So α : c̄ : c = 1.0 : 1.12 : 1.22, and the order is α < c̄ < c. Tap the three speeds from smallest to largest.

Now drag the handles. A lighter gas or a higher temperature lengthens all three bars together, while their ratio stays the same.

Why does the kinetic energy need the RMS speed? Taking every molecule at speed c gives a total kinetic energy of (1/2) m N c², and that is the true total. The average speed or the most probable speed does not give kinetic energy directly.

Molecules moving in a box. Raise the temperature, change the gas, and read the three speeds.

4. Working out RMS speeds

Problem 1.56. A gas of density 1.25 g L⁻¹ at STP has M = 1.25 × 22.4 = 28 g mol⁻¹. At 27°C, c = √(3 × 8.314 × 300/28×10⁻³) = 516.9 m s⁻¹.

Problem 1.57. The RMS speed of CO₂ at 21°C is c = √(3 × 8.314 × 294/44×10⁻³) = 408.24 m s⁻¹; in SI units R = 8.314 J mol⁻¹ K⁻¹ and M is in kg mol⁻¹. Problem 1.58. In CGS units R = 8.32×10⁷ erg mol⁻¹ K⁻¹ and M is in g mol⁻¹. For NH₃ at 25°C, c = √(3 × 8.32×10⁷ × 298/17) = 6.6146×10⁴ cm s⁻¹ = 66146 cm s⁻¹.

Problem 1.59. The density is 1.15 g L⁻¹ at STP, so M = 25.76 g mol⁻¹; at 27°C and 700 mm Hg, c = 5.389×10² m s⁻¹. Problem 1.60. The density is 1.75 g L⁻¹ at 25°C and the molar volume at SATP is 24.789 L, so M = 43.38 g mol⁻¹ and c = 413.932 m s⁻¹.

Problem 1.61. If c = 49330 cm s⁻¹ at 0°C, then M = 3RT/c² = 0.02798 kg mol⁻¹ = 27.98 g mol⁻¹. Problem 1.62. Cl₂ at 20°C has an RMS speed of 320.826 m s⁻¹; for CO₂ to have the same speed, T₂ = 181.57 K = −91.43°C.

At one temperature the ratio of the RMS speeds of two gases is c₁/c₂ = √(M₂/M₁). At 25°C hydrogen has √(28/2) = 3.74 times the RMS speed of nitrogen (the book prints 3.73). That settles the puzzle at the top: the lighter gas is faster.

The RMS speed of O₂, c = √(3RT/0.032), is 461.3 m s⁻¹ at 273 K (STP), 481.9 m s⁻¹ at 298 K (SATP), 483.6 m s⁻¹ at 300 K (27°C) and about 485.97 m s⁻¹ at 303 K (30°C). More heat means more speed, but slowly, because the speed goes with √T.

Two more: N₂ at 33°C (306 K), c = √(3 × 8.314 × 306/0.028) = 522 m s⁻¹; CH₄ at STP, c = √(3 × 8.314 × 273/0.016) = 652 m s⁻¹. Putting M in g mol⁻¹ gives a wrong answer, so convert to kg first.

To find a temperature, turn the formula round: from c = √(3RT/M), T = c²M/3R. For a gas of M = 32 g mol⁻¹ with c = 505.5 m s⁻¹, T = (505.5)² × 0.032/(3 × 8.314) = 327.8 K ≈ 54.8°C. And the temperature at which Cl₂ matches the speed of CO₂ at −91.43°C is T(Cl₂) = 181.7 × 71/44 = 293.2 K ≈ 20°C.

5. Kinetic energy

The average kinetic energy of one molecule is E_K = 3RT/(2N_A). For one mole of gas E_K = 3RT/2. For n moles E_K = 3nRT/2.

By E_K = 3nRT/2 the total kinetic energy depends on the amount n and the Kelvin temperature, not on M. So at one temperature equal moles of all gases have equal kinetic energy, whatever the gas. Light molecules are fast, heavy ones slow, and the product of mass and speed squared comes out the same.

Since k = R/N_A = 1.38×10⁻²³ J K⁻¹ (1.38×10⁻¹⁶ erg K⁻¹), the average kinetic energy of one molecule is E_K = (3/2)kT. At 298 K one CO₂ molecule has E_K = 3 × 8.314 × 298/(2 × 6.022×10²³) = 6.17×10⁻²¹ J (= 6.17×10⁻¹⁴ erg).

Problem 1.63. The kinetic energy of 1 g of CO₂ at 25°C = (3/2) × (1/44) × 8.314 × 298 = 84.46 J. Problem 1.64. The kinetic energy of 16 g of O₂ = 0.5 mol at 27°C is E_K = (3/2) × 0.5 × 8.314 × 300 = 1870.65 J.

The kinetic energy of 1 mol of gas is (3/2) × 8.314 × 290 = 3616.59 J for N₂ at 17°C and 3716.24 J for CO₂ at 25°C. At 30°C, 1 mol each of He and Xe have equal kinetic energy, 3.778×10³ J. When a mass is given, find the moles first: 2 g of He at 25°C is 0.5 mol, so E_K = (3/2) × 0.5 × 8.314 × 298 = 1858 J.

Put a gas, a temperature and an amount into the calculator below. The speed and the kinetic energy come with the working. The presets are from the book.

Now six statements, true or false. Answer each from the postulates.

6. Check yourself

Three questions in the exam's style. Start with the first.

The second puts the postulates to work.

In the last one, watch the units.

The whole note on seven cards.

Source: Chemistry 2nd Paper, Classes 11-12 (NCTB), Chapter 1 “Environmental Chemistry”, section 1.5.