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Physics (AS & A Level) 9702 · Chapter 16

Thermodynamics

Internal energy as the sum of random kinetic and potential energies, why a temperature rise means more internal energy, work done when a gas changes volume, and the first law of thermodynamics.

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16.1 Internal energy

The internal energy of a system is the sum of the random distribution of kinetic and potential energies of its molecules. It is determined by the state of the system: its temperature, volume and phase.

  • The kinetic part depends on temperature.
  • The potential part depends on the separation of the molecules, so it changes when a substance melts or boils.

A rise in temperature means a rise in the mean kinetic energy of the molecules, which means a rise in internal energy.

During melting or boiling, the temperature is constant, so the kinetic energy is unchanged, but the potential energy rises, so the internal energy still rises.

16.2 The first law of thermodynamics

Work done by a gas

When a gas expands at constant pressure pp, pushing a piston of area AA through a distance dd, the work it does is Fd=pAdFd = pAd:

W=pΔVW = p\Delta V

  • When a gas expands, it does work on the surroundings. Work done on the gas is negative.
  • When a gas is compressed, work is done on it. Work done on the gas is positive.

On a pressure–volume graph, the work done is the area under the line, which still applies when the pressure varies.

A pressure-volume graph with a shaded rectangle under a constant-pressure expansion showing the work done

A pressure-volume graph with a shaded rectangle under a constant-pressure expansion showing the work done

The first law

ΔU=q+W\Delta U = q + W

SymbolMeaningPositive when
ΔU\Delta Uincrease in internal energyinternal energy rises
qqenergy transferred to the system by heatingthe system is heated
WWwork done on the systemthe gas is compressed

Common processes

ProcessConditionConsequence for an ideal gas
Constant volumeΔV=0\Delta V = 0, so W=0W = 0ΔU=q\Delta U = q
Constant temperatureΔT=0\Delta T = 0ΔU=0\Delta U = 0, so q=−Wq = -W
Thermally insulated (or very fast)q=0q = 0ΔU=W\Delta U = W: compressing warms the gas
Complete cyclereturns to its startΔU=0\Delta U = 0, so net q=−q = - net WW

A bicycle pump warms up because you do work on the gas quickly, leaving little time for heat to escape, so q≈0q \approx 0 and ΔU≈W\Delta U \approx W.

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