Lesson 9.5 · Unit 9 · CED topic 9.4

Internal energy and the first law of thermodynamics

Pump a bicycle tire quickly and the pump barrel gets warm. Nobody lit a fire in it; you did work on the air by compressing it, and the energy went into the motion of its atoms. Rub your hands together and the same thing happens. Energy can enter a system as work or as heating, and what the system does with it is raise its internal energy. The first law of thermodynamics is the statement that none of this energy is lost: it is conservation of energy written for systems of many particles. This lesson defines internal energy, gives the first law with its sign convention, and shows how to use it for a gas.

Internal energy of an ideal gas

In Physics 1 you kept track of the kinetic energy of a whole object moving through space and its potential energy in a field. A gas in a closed box sitting on a table has none of that: its center of mass is at rest. But its atoms are all moving, and the energy of that hidden motion is the gas’s internal energy \(U\). Internal energy is the sum of the kinetic energies of the particles and the potential energy of their interactions.

An ideal gas has no interactions except during collisions, so it has no potential energy at all, and its internal energy is just the total kinetic energy of its atoms. From lesson 9.1 each monatomic atom has average kinetic energy \(\tfrac{3}{2}k_BT\). With \(N\) atoms, \[ U = N\cdot\tfrac{3}{2}k_BT = \tfrac{3}{2}nRT. \] Two consequences are used constantly. First, \(U\) depends only on the temperature (and the amount of gas): if the temperature does not change, \(\Delta U = 0\). Second, \(\Delta U = \tfrac{3}{2}nR\,\Delta T\), so the sign of \(\Delta T\) tells you the sign of \(\Delta U\). Because \(PV = nRT\), we can also write \(U = \tfrac{3}{2}PV\), which is helpful when you are given a PV diagram instead of a temperature.

Internal energy is a state function: it depends only on the current state of the gas (its \(P\), \(V\), \(T\)), not on how the gas got there. Work and heating are not state functions; they depend on the path.

The first law

There are two ways to change the internal energy of a closed system: transfer energy by heating (or cooling), \(Q\), or transfer energy by work, \(W\). The first law says \[ \Delta U = Q + W. \] Sign convention. This course defines both \(Q\) and \(W\) as energy transferred into the system. So \(Q\) is positive when the system is heated and negative when it is cooled, and \(W\) is positive when work is done on the system and negative when the system does work on its surroundings. Whenever you see a sign problem on an exam, check it against this table.

For a gas in a cylinder with a movable piston, work is done by changing the volume. If the pressure is constant (or if we use an average pressure), pushing the piston inward so the volume decreases does positive work on the gas, and letting the gas expand does negative work on it: \[ W = -P\Delta V. \] With \(\Delta V \lt 0\) (compression) \(W \gt 0\); with \(\Delta V \gt 0\) (expansion) \(W \lt 0\). A force that does not change the volume (a rigid container) does no work, so \(W = 0\) if \(\Delta V = 0\).

Applying the first law in three steps avoids most errors: (1) decide whether the temperature increases, decreases or stays the same, which gives the sign of \(\Delta U\); (2) find \(W\) from the volume change; (3) solve for \(Q\) or whatever is unknown. For an isolated system, \(Q = 0\) and \(W = 0\), so \(\Delta U = 0\) and the energy is constant.