Pancyprian Physics study materials
Pancyprian Physics study resources

Pancyprian Physics: Thermodynamics and Kinetic Theory: Heat, Ideal Gases, the Laws

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Thermodynamics and Kinetic Theory: Heat, Ideal Gases, the Laws

The thermodynamics question tests whether you can distinguish heat from temperature, apply calorimetry without sign errors, and use the ideal gas law with absolute temperature. Kinetic theory connects the macroscopic gas laws to molecular motion, and the first law appears as an energy-accounting statement that must be applied with a consistent sign convention.

Heat and temperature

Temperature (K) = θ(°C) + 273.15. Heating without a change of state: Q = mcΔT. At a change of state the temperature stays constant and Q = mL, where L is the specific latent heat. In a calorimeter, heat lost by the hot body equals heat gained by the cold body plus the calorimeter, assuming no losses to the surroundings. Specific heat capacity is per kilogram; heat capacity is per object.

Ideal gases

pV = nRT = NkT, with R = 8.31 J/(mol·K) and k = 1.38 × 10⁻²³ J/K. Temperature must be in kelvin. Kinetic theory gives the mean translational kinetic energy (3/2)kT per molecule independent of gas identity, and the rms speed √(3RT/M), where M is the molar mass in kg/mol. For a monatomic ideal gas the internal energy is U = (3/2)nRT, so U depends only on T.

The laws of thermodynamics

First law: ΔU = Q − W, with W the work done by the gas. Isothermal: ΔU = 0, so Q = W. Adiabatic: Q = 0, so ΔU = −W and the gas cools on expansion. Isochoric: W = 0, so ΔU = Q. Isobaric: W = pΔV. The second law forbids a device that converts heat entirely into work with no other change; efficiency η = W/Q_H, and the Carnot limit is 1 − T_c/T_h with both temperatures in kelvin.

Worked example

0.20 kg of water at 90 °C is poured into 0.30 kg of water at 20 °C in a calorimeter of negligible heat capacity. Find the final temperature. Heat lost = heat gained: 0.20c(90 − θ) = 0.30c(θ − 20). The c cancels: 18 − 0.20θ = 0.30θ − 6, so 0.50θ = 24 and θ = 48 °C. The result lies between the two starting temperatures, as it must.

Traps that cost marks

  • Substituting Celsius temperatures into pV = nRT or into an efficiency formula.
  • Reversing the sign of W in the first law, or forgetting that W is work done by the gas.
  • Adding Q = mcΔT across a phase change where Q = mL should be used.
  • Confusing specific heat capacity (J kg⁻¹ K⁻¹) with molar heat capacity.
  • Quoting efficiency as a number greater than 1, or leaving it as a decimal when a percentage is asked.