Internal Energy and the First Law

Introduction

The first law states that energy is conserved. For a closed system using the chemistry sign convention, its internal-energy change equals heat received plus work done on it: ΔU = q + w. This simple equation is powerful only when the system, transfer directions and work modes are defined.

Core explanation

Internal energy includes microscopic kinetic and potential contributions associated with the particles of a system. Its absolute value is usually not measured; a change ΔU between two states is. Heat and work are ways that energy crosses a system boundary. If 400 J enters as heat and 100 J leaves as work, the system's internal energy rises by 300 J. No energy is created or destroyed; the transfers account for the change.

For a closed system, matter does not cross the boundary, so the simple ΔU = q + w balance is appropriate when q and w include all relevant energy-transfer modes. If the only work is pressure-volume work, w can be calculated from pressure and volume change under specified path conditions. If electrical work also occurs, include it in total w. If matter enters or leaves, energy carried by that matter requires a more complete open-system balance.

Internal energy is a state function. If the system returns to its initial state, ΔU cycle = 0, and the net heat and work must sum to zero. Neither q cycle nor w cycle must separately vanish. A heat engine may receive net heat and produce net work during each cycle while its working fluid returns to its starting internal energy.

The equation does not decide whether a process is spontaneous. Both a spontaneous expansion and a forced compression obey energy conservation. The second law and Gibbs energy address direction under specified constraints. Likewise, ΔU alone does not determine a reaction rate. Keeping conservation, direction and rate as separate questions prevents one formula from being asked to do too much.

For an ideal gas, internal energy depends only on temperature, so an isothermal change has ΔU = 0. The first law then requires q = −w. During a reversible isothermal expansion the gas can absorb heat and do equal-magnitude work, leaving its temperature and internal energy unchanged. During free expansion into a vacuum in an insulated vessel, both q and simple P–V work may be zero, also giving ΔU = 0. Different paths can share the same state-function change while transfers differ.

In practical calorimetry, one often defines a larger insulated system containing both reaction and calorimeter. Energy lost by the reaction appears as energy gained by the solution and apparatus, apart from leakage. The first law explains the heat-balance equation used to infer reaction energy from a thermometer reading.

Step-by-step reasoning

1. Confirm a closed system and list heat and every relevant work transfer. 2. Choose the chemistry sign convention explicitly. 3. Convert words such as “released” and “compressed” into signed q and w. 4. Compute ΔU = q + w with consistent units. 5. Check whether the sign and magnitude fit energy conservation.

Visual explanation

Draw an energy reservoir labelled U as a box. Put arrows for +q and +w entering, and −q and −w leaving. Show that the change in the level inside equals the signed sum of arrows. Below, draw a closed loop returning to the original level with net q and w opposite.

Real-world analogy

A water tank level changes according to water entering minus water leaving through different pipes. The level is analogous to a state quantity, while flow through each pipe is analogous to transfer. Energy is not material water, but the accounting structure is similar.

Real-world example

An insulated piston-cylinder is compressed from outside. With q ≈ 0 and positive work done on the gas, ΔU > 0. For an ideal gas this raises temperature. The first law connects a mechanical action to a thermal outcome without claiming that heat entered.

Why?

Why can temperature rise in an insulated compression? Mechanical work enters the gas system and increases its internal energy. Heat flow is not the only route to a temperature change.

Common misconception

“Conservation of energy means ΔU is always zero.” The energy of the universe is conserved, but a selected system can gain or lose energy through heat and work. Only an isolated system or a complete cycle has zero ΔU under the stated conditions.

Worked example

A closed system gives 250 J of heat to the room and receives 600 J of electrical work. For the system, q = −250 J and w = +600 J. Therefore ΔU = −250 + 600 = +350 J. The gain is possible because work input exceeds heat loss. If the system returns to its starting state later, subsequent transfers must total −350 J for the return leg.

Quick check

1. In a complete cycle, what is ΔU? Answer: Zero, because the system returns to the same state and U is a state function.

Exam focus

Write the defined system and ΔU = q + w before substituting values. Include all stated work modes. Do not infer spontaneity or rate from the first law alone.

Advanced insight

The first law is a conservation principle, whereas the internal-energy function depends on the material and state. Thermodynamic equations of state and heat capacities specify how U changes with temperature, volume and composition. The balance alone does not provide those material relationships.

Summary

Internal energy is a state function whose change equals signed heat plus work for a closed system. Energy conservation permits the system to gain or lose energy. A cycle has zero net ΔU, but heat and work transfers may be nonzero and offset.

Practice questions

1. A system absorbs 90 J of heat and does 40 J of work. Find ΔU. Answer: q = +90 J, w = −40 J, so ΔU = +50 J. 2. An insulated gas receives 120 J of compression work. What is ΔU if no other work occurs? Answer: +120 J, because q = 0 and w = +120 J. 3. Does ΔU = 0 require q = 0 and w = 0 separately? Answer: No. They may be nonzero but equal and opposite, so q + w = 0.