Thermal Physics and Ideal Gases

pV = nRT; kinetic theory; internal energy; specific heat capacity; latent heat

# Thermal Physics and Ideal Gases — A-Level Physics

Thermal physics connects the macroscopic properties of gases (pressure, volume, temperature) with the microscopic behaviour of molecules. The ideal gas model and kinetic theory are powerful tools.


1. Temperature and Thermal Equilibrium

  • Temperature measures the average kinetic energy of particles
  • Thermal equilibrium: no net heat flow between objects at the same temperature
  • Absolute zero (0 K = −273.15°C): particles have minimum possible energy
  • T(K)=T(°C)+273.15T(K) = T(°C) + 273.15

2. Internal Energy

Internal energy = total kinetic energy + total potential energy of all particles.

  • For an ideal gas: internal energy = kinetic energy only (no intermolecular forces)
  • Increasing temperature → increases internal energy
  • During a change of state: temperature constant but internal energy changes (PE changes)

3. Specific Heat Capacity

Q=mcΔθQ = mc\Delta\theta

Where:

  • QQ = heat energy (J)
  • mm = mass (kg)
  • cc = specific heat capacity (J kg⁻¹ K⁻¹)
  • Δθ\Delta\theta = temperature change (K or °C)

Example: cwater=4200c_{\text{water}} = 4200 J kg⁻¹ K⁻¹

4. Specific Latent Heat

Q=mLQ = mL

  • LfL_f = specific latent heat of fusion (solid ↔ liquid)
  • LvL_v = specific latent heat of vaporisation (liquid ↔ gas)
  • During phase change: temperature remains constant

5. Gas Laws

Boyle's Law: pV=constantpV = \text{constant} (at constant TT) → p1/Vp \propto 1/V

Charles's Law: V/T=constantV/T = \text{constant} (at constant pp) → VTV \propto T

Pressure Law: p/T=constantp/T = \text{constant} (at constant VV) → pTp \propto T

Combined Gas Law

p1V1T1=p2V2T2\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}

Ideal Gas Equation

pV=nRT=NkT\boxed{pV = nRT = NkT}

Where:

  • nn = number of moles
  • R=8.314R = 8.314 J mol⁻¹ K⁻¹
  • NN = number of molecules
  • k=1.38×1023k = 1.38 \times 10^{-23} J K⁻¹ (Boltzmann constant)
  • k=R/NAk = R/N_A where NA=6.02×1023N_A = 6.02 \times 10^{23} mol⁻¹

6. Kinetic Theory of Gases

Pressure from Molecular Motion

pV=13Nmc2\boxed{pV = \frac{1}{3}Nm\overline{c^2}}

Where c2\overline{c^2} = mean square speed. crms=c2c_{\text{rms}} = \sqrt{\overline{c^2}}.

Mean Kinetic Energy

12mc2=32kT\boxed{\frac{1}{2}m\overline{c^2} = \frac{3}{2}kT}

Average KE per molecule depends only on temperature.

Assumptions of Ideal Gas

  1. Large number of identical molecules
  2. Volume of molecules << volume of container
  3. Collisions are elastic
  4. No intermolecular forces (except during collisions)
  5. Time of collision << time between collisions
  6. Random motion

Worked Example: Ideal Gas

Problem

A gas at 20°C occupies 0.05 m³ at 100 kPa. How many moles?

n=pV/(RT)=100000×0.05/(8.314×293)=5000/2436=2.05n = pV/(RT) = 100000 \times 0.05/(8.314 \times 293) = 5000/2436 = 2.05 mol

Solution

Worked Example: RMS Speed

Problem

Find the rms speed of nitrogen molecules (M=28M = 28 g/mol) at 300 K.

12mc2=32kT\frac{1}{2}m\overline{c^2} = \frac{3}{2}kT crms=3kT/m=3RT/M=3×8.314×300/0.028=267000=517c_{\text{rms}} = \sqrt{3kT/m} = \sqrt{3RT/M} = \sqrt{3 \times 8.314 \times 300/0.028} = \sqrt{267000} = 517 m/s

Solution

Worked Example: Heating Water

Problem

2 kg of water heated from 20°C to 100°C (c=4200c = 4200 J/kg/K). Then 0.5 kg boils (Lv=2.26×106L_v = 2.26 \times 10^6 J/kg).

Heating: Q1=mcΔθ=2×4200×80=672,000Q_1 = mc\Delta\theta = 2 \times 4200 \times 80 = 672{,}000 J Boiling: Q2=mL=0.5×2.26×106=1,130,000Q_2 = mL = 0.5 \times 2.26 \times 10^6 = 1{,}130{,}000 J Total: Q=1,802,000Q = 1{,}802{,}000 J = 1.8 MJ

Solution

8. Practice Questions

    1. A gas has volume 2 L at 300 K and 1 atm. Find the volume at 400 K and 2 atm. (3 marks)
    1. Calculate the rms speed of helium atoms (M=4M = 4 g/mol) at 273 K. (3 marks)
    1. State the assumptions of an ideal gas. (3 marks)
    1. 500 g of ice at 0°C is heated until it becomes water at 50°C. Lf=334,000L_f = 334{,}000 J/kg, c=4200c = 4200 J/kg/K. Find total energy needed. (3 marks)

    Answers

    1. V2=V1×T2/T1×p1/p2=2×400/300×1/2=1.33V_2 = V_1 \times T_2/T_1 \times p_1/p_2 = 2 \times 400/300 \times 1/2 = 1.33 L.

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Summary

  • pV=nRT=NkTpV = nRT = NkT
  • Kinetic theory: pV=13Nmc2pV = \frac{1}{3}Nm\overline{c^2}; 12mc2=32kT\frac{1}{2}m\overline{c^2} = \frac{3}{2}kT
  • Q=mcΔθQ = mc\Delta\theta (heating); Q=mLQ = mL (phase change)
  • rms speed: crms=3kT/mc_{\text{rms}} = \sqrt{3kT/m}
  • Internal energy of ideal gas = KE only

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