Internal Energy & the First Law

Internal energy is the energy of a system's internal degrees of freedom, measured in the rest frame of the system as a whole. A 2.0 kg block sliding at 5.0 m s⁻¹ carries 25 J of bulk kinetic energy, and lifting it 3.0 m gives it 59 J of gravitational potential energy; neither is part of U. What counts is the translational kinetic energy of the molecules, their rotational and vibrational energy, and the potential energy of the intermolecular forces. The payoff is that U belongs to the state. For a closed system of n moles, name any two of p, V and T and U is determined, so ΔU = U_f − Uᵢ and every route between the same two states gives the same number. You never need an absolute value of U, only differences, which is why no zero of internal energy is ever specified.

There is no answer to the question of how much heat is in a gas. Heat is energy transferred because of a temperature difference; work is energy transferred by a force acting through a displacement at the boundary. Both exist only while a process is running. When it stops there is no heat and no work anywhere — only a system whose U has changed. Joule made the point with a paddle wheel: a 12.0 kg mass falling 1.80 m, twenty times over, does 20 × 12.0 × 9.81 × 1.80 = 4.24 × 10³ J of work on insulated water, and with c = 4186 J kg⁻¹ K⁻¹ that warms 0.250 kg of it by 4.05 K. A flame delivering the same 4.24 kJ produces the same warming, and nothing in the final state records which route was used. That experimental fact is what licenses putting Q and W into one equation.

Two decisions come before any arithmetic: what is inside the system, and which direction counts as positive. The physics convention makes Q positive when heat enters and W positive when the system does work on its surroundings, giving ΔU = Q − W. The chemistry convention tracks the work done on the system, Wₒₙ = −W, and writes ΔU = Q + Wₒₙ. Both are correct and both are in print; mixing them inside one problem is the commonest way to lose a sign. Take a gas that absorbs 850 J of heat while the surroundings compress it, doing 340 J of work on it. In the first convention W = −340 J and ΔU = 850 − (−340) = +1190 J. In the second, Wₒₙ = +340 J and ΔU = 850 + 340 = +1190 J. The bookkeeping changes; the physics does not.

Carry a gas from state A to state B twice. On route 1, 500 J of heat enters and the gas does 180 J of work, so ΔU = 500 − 180 = 320 J. On route 2, 420 J enters and the gas does 100 J, so ΔU = 420 − 100 = 320 J. The heats differ by 80 J and the works differ by 80 J, and the difference is identical because the end states are identical. Read it backwards and it becomes a tool: know ΔU from the end states and one of Q or W, and the other is forced. The extreme case is free expansion — an insulated rigid vessel split in two, gas on one side and vacuum on the other, partition removed. No heat crosses the insulation and no boundary moves against anything, so Q = 0, W = 0 and ΔU = 0 exactly, even though the volume has doubled.

Kinetic theory assigns ½kT per molecule to each active degree of freedom, which sums to U = (f/2)nRT with f = 3 for a monatomic gas and f = 5 for a diatomic one near room temperature. Volume and pressure do not appear. So heating 2.00 mol of helium from 295 K to 345 K raises U by (3/2)(2.00)(8.314)(50.0) = 1.25 × 10³ J, and raises it by exactly that whether the gas was held rigid, allowed to expand, or compressed and then expanded. Writing C_V = (f/2)R lets you state it as ΔU = n C_V ΔT on any path. The subscript V is historical: constant volume is where the quantity is easiest to measure, because W = 0 there makes Q equal ΔU. Real gases drift from this, since attraction between molecules makes U depend weakly on volume too.

Build a table with one row per stage and columns for Q, W and ΔU, every entry signed. Fill in what is given, use ΔU = Q − W to finish each row, and use the end states to finish the column. If the system returns to where it started, U returns with it, so the ΔU column must sum to zero — a free check, and often the quickest route to a missing entry. Suppose one stage has Q = +1.24 kJ and W = +0.42 kJ, so ΔU = +0.82 kJ, and a second has Q = −0.95 kJ and W = −0.78 kJ, so ΔU = −0.17 kJ. If a third stage closes the cycle, its ΔU must be −0.65 kJ. Summing the columns then gives Qₙₑₜ = Wₙₑₜ automatically; if those two totals disagree, either an entry is wrong or the process never closed.