Internal Energy and the First Law of Thermodynamics
The first law of thermodynamics states that the change in the internal energy of a system equals the heat added to the system plus the work done on it: ΔU = q + w. In brief: if a system absorbs 500 J of heat and does 200 J of work on its surroundings, its internal energy rises by 500 − 200 = 300 J.
The first law is the statement of energy conservation applied to heat and work. It tells you how much the energy of a system changes, but not in which direction a process will go; that is the role of the second law. This article defines internal energy, explains the law with sign conventions, relates ΔU to enthalpy and works through examples of the kind found in chemistry courses.
What is internal energy?
The internal energy U is the total energy of all the particles in a system: the kinetic energy of their translation, rotation and vibration, plus the potential energy of the attractions between them, the energy in chemical bonds and the energy in the nuclei and electrons. The absolute value of U cannot be measured, but changes in U can be. Internal energy is a state function. It does not include the kinetic energy or potential energy of the system as a whole, for example when the whole container is moving.
What does the first law say?
The internal energy of an isolated system is constant. When a closed system exchanges energy with the surroundings, the change in internal energy is:
ΔU = q + w
where q is the heat absorbed by the system and w is the work done on it. The energy lost by the system is gained by the surroundings: ΔUsystem = −ΔUsurroundings. For a cyclic process in which the system returns to its starting state, ΔU = 0, so the net heat absorbed equals the net work done by the system.
Four combinations of heat and work
| q | w | ΔU = q + w | Example |
|---|---|---|---|
| +500 J | −200 J | +300 J | A heated gas expands and does work |
| −400 J | +150 J | −250 J | A gas is compressed and loses heat to the surroundings |
| +300 J | +100 J | +400 J | Stirring a liquid that is also being heated |
| 0 | −120 J | −120 J | Adiabatic expansion: the gas cools |
What happens at constant volume or constant pressure?
Constant volume
If the volume does not change, no pressure–volume work is done (w = 0), so ΔU = qv. The heat measured in a sealed bomb calorimeter equals the change in internal energy.
Constant pressure and enthalpy
Most reactions take place in vessels open to the atmosphere, at constant pressure. The system then does pressure–volume work w = −PΔV, so ΔU = qp − PΔV. Rearranging gives qp = ΔU + PΔV. This quantity is defined as the change in enthalpy:
ΔH = ΔU + PΔV = qp
For reactions with gases, PΔV = ΔngasRT, where Δngas is the change in the number of moles of gas. The difference between ΔH and ΔU is therefore small if there are no gases involved or if the number of moles of gas does not change.
Worked examples
Example 1: heat and work on a gas
A gas absorbs 500 J of heat and does 200 J of work. Here q = +500 J and w = −200 J (the work is done by the system), so ΔU = 500 + (−200) = +300 J.
Example 2: compression
A gas is compressed by a piston that does 150 J of work on the gas, while the gas releases 400 J of heat. Then q = −400 J and w = +150 J, so ΔU = −400 + 150 = −250 J.
Example 3: converting between ΔH and ΔU
The combustion of hydrogen is 2H2(g) + O2(g) → 2H2O(l), with ΔH = −571.6 kJ at 298.15 K. Find ΔU.
- Δngas = 0 − 3 = −3 mol.
- PΔV = ΔngasRT = (−3)(8.314 × 10−3 kJ/mol·K)(298.15 K) = −7.44 kJ.
- ΔU = ΔH − PΔV = −571.6 − (−7.44) = −564.2 kJ.
The internal energy decreases by less than the enthalpy, because the surroundings do 7.44 kJ of work on the system as the gas volume shrinks.
Example 4: work from a reaction
When 1.00 mol of zinc dissolves in acid at 25 °C and 1.00 atm, 1.00 mol of H2 is produced. The work done by the system is |w| = nRT = 1.00 × 8.314 × 298 = 2.48 kJ. The gas pushes the atmosphere back, so w = −2.48 kJ.
Why is a perpetual motion machine impossible?
A machine that produces work without any source of energy would produce energy from nothing and violate the first law. This was the basis for the refusal of patent offices to consider such “machines of the first kind”. The first law also implies that a heat engine cannot produce more work than the heat it takes in, and a second law limit makes the maximum fraction even smaller.
Who formulated the first law?
The law was developed by several scientists independently in the 1840s. Julius Robert Mayer, a German physician, proposed the equivalence of heat and work in 1842. James Prescott Joule determined the mechanical equivalent of heat experimentally in 1843 and the following years, finding that about 4.2 J of work gives 1 cal of heat; the currently defined value is 4.184 J per calorie. Hermann von Helmholtz gave the law a general mathematical form in 1847, and Rudolf Clausius stated it in the form used today, with internal energy as a state function, in 1850.
Which mistakes are common?
- Sign errors on work: if the system expands, w is negative in the convention ΔU = q + w.
- Using ΔH when ΔU is needed: the two differ by ΔngasRT.
- Forgetting that the law applies to closed systems: in open systems, matter also carries energy.
- Using Celsius in nRT: use kelvins.
- Treating internal energy as heat content: heat is not stored; ΔU tracks the energy change.