Ideal Gas Law and Kinetic Model

pV = nRT; gas laws; kinetic theory; molecular speeds; pressure from molecular collisions

# Ideal Gas Law and Kinetic Model — IB Physics

The ideal gas model connects macroscopic properties (pressure, volume, temperature) with the microscopic behaviour of molecules. It's a beautiful example of how physics bridges scales.


1. Gas Laws

Boyle's Law: pV=kpV = k (constant TT) Charles's Law: V/T=kV/T = k (constant pp) Gay-Lussac's Law: p/T=kp/T = k (constant VV)

2. Ideal Gas Equation

pV=nRT\boxed{pV = nRT}

Or equivalently: pV=NkTpV = NkT

Where:

  • pp = pressure (Pa)
  • VV = volume (m³)
  • nn = moles
  • R=8.31R = 8.31 J mol⁻¹ K⁻¹
  • NN = number of molecules
  • k=1.38×1023k = 1.38 \times 10^{-23} J K⁻¹ (Boltzmann constant)
  • TT = absolute temperature (K)

3. Kinetic Model of an Ideal Gas

Assumptions

  1. Large number of identical molecules in random motion
  2. Volume of molecules negligible compared to container volume
  3. No intermolecular forces except during collisions
  4. Collisions are perfectly elastic
  5. Duration of collisions is negligible

Pressure from Molecules

Pressure arises from molecules hitting container walls: pV=13Nmc2\boxed{pV = \frac{1}{3}Nm\overline{c^2}}

Combining with pV=NkTpV = NkT: 12mc2=32kT\frac{1}{2}m\overline{c^2} = \frac{3}{2}kT

The average KE of a molecule depends only on temperature.

RMS Speed

crms=c2=3kTm=3RTMc_{\text{rms}} = \sqrt{\overline{c^2}} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3RT}{M}}

Where MM = molar mass (kg/mol).

4. Real vs Ideal Gases

Ideal gas behaviour is a good approximation when:

  • Temperature is high (particles have high KE, forces negligible)
  • Pressure is low (particles far apart, volume negligible)

Deviation occurs at high pressure and low temperature (near liquefaction).

Worked Example: Example 1

Problem

A gas occupies 0.02 m³ at 300 K and 100 kPa. Find the number of moles.

n=pV/(RT)=100000(0.02)/(8.31×300)=2000/2493=0.802n = pV/(RT) = 100000(0.02)/(8.31 \times 300) = 2000/2493 = 0.802 mol

Solution

Worked Example: RMS Speed

Problem

Find crmsc_{\text{rms}} for oxygen (M=32×103M = 32 \times 10^{-3} kg/mol) at 300 K.

crms=3RT/M=3(8.31)(300)/0.032=233000=483c_{\text{rms}} = \sqrt{3RT/M} = \sqrt{3(8.31)(300)/0.032} = \sqrt{233000} = 483 m/s

Solution

Worked Example: Combined Gas Law

Problem

Gas at 100 kPa, 2 L, 300 K is compressed to 0.5 L and heated to 600 K. Find pressure.

p2=p1V1T2/(V2T1)=100×2×600/(0.5×300)=800p_2 = p_1 V_1 T_2/(V_2 T_1) = 100 \times 2 \times 600/(0.5 \times 300) = 800 kPa

Solution

6. Practice Questions

    1. State three assumptions of the kinetic theory of ideal gases. (3 marks)
    1. A balloon contains 0.1 mol of helium at 20°C. Find the volume at 101 kPa. (2 marks)
    1. Show that the average KE of a gas molecule is 32kT\frac{3}{2}kT. (3 marks)
    1. Explain why the pressure of a gas increases when it is compressed at constant temperature. (3 marks)

    Answers

    1. Molecules are in random motion; no intermolecular forces except during collisions; collisions are elastic; volume of molecules is negligible.

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Summary

  • pV=nRT=NkTpV = nRT = NkT
  • Kinetic: pV=13Nmc2pV = \frac{1}{3}Nm\overline{c^2}; avg KE = 32kT\frac{3}{2}kT
  • crms=3kT/mc_{\text{rms}} = \sqrt{3kT/m}
  • Ideal gas: high TT, low pp (molecules far apart, forces negligible)

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