First law of thermodynamics

A hot cup of tea cools, a compressed gas in a bicycle pump warms your hand, a chemical reaction in a flask fizzes and heats the beaker — in every case, energy is moving as heat or as work, and the total never appears from nowhere or vanishes into nothing. The first law of thermodynamics is exactly this bookkeeping rule, the law of conservation of energy written for chemical and physical processes: whatever energy a closed system gains or loses as heat and work, its internal energy changes by precisely that amount, no more, no less. This article builds the single equation that captures this idea, $\Delta U = q + w$ , fixes the IUPAC sign convention ISC expects, and shows how the equation collapses to simple special cases for gas expansion, constant-volume reactions, and insulated (adiabatic) processes — the exact toolkit used throughout the Chemical Thermodynamics unit.

The core insight

For a closed system, the change in internal energy equals the heat absorbed plus the work done on the system, $\Delta U = q + w$ , with $q$ and $w$ each following the IUPAC sign convention — a direct statement that energy is conserved.

The core idea

Energy is conserved, full stop. The first law says energy can neither be created nor destroyed, only converted between heat, work, and the internal energy stored in a system's molecules. For a closed system exchanging only heat and work with its surroundings, the change in internal energy $U$ is:

\Delta U = q + w

where: $\Delta U$ is the change in internal energy of the system, $q$ is the heat exchanged with the surroundings, and $w$ is the work exchanged with the surroundings.

Sign convention matters more than the formula. ISC/CISCE follows the IUPAC convention: $q > 0$ when heat flows into the system (endothermic), $q < 0$ when heat flows out (exothermic); $w > 0$ when work is done on the system (e.g., compression), $w < 0$ when the system does work on the surroundings (e.g., expansion). Every sign is judged from the system's point of view — energy coming in is positive, energy leaving is negative.

Work of expansion against a constant pressure. When a gas pushes back a piston against a steady external pressure $P_{ext}$ , the work exchanged is:

w = -P_{ext}\,\Delta V

where: $P_{ext}$ is the constant external pressure and $\Delta V = V_f - V_i$ is the change in volume. Expansion ( $\Delta V > 0$ ) makes $w$ negative, matching the convention that the system loses energy by doing work; compression ( $\Delta V < 0$ ) makes $w$ positive.

Three special cases students must recognise instantly:

Isochoric (rigid container, $\Delta V = 0$ ): $w = 0$ , so $\Delta U = q_v$ — all heat exchanged at constant volume shows up entirely as internal-energy change.
Adiabatic (insulated, $q = 0$ ): $\Delta U = w$ — any internal-energy change comes entirely from work, with no heat crossing the boundary.
Reversible isothermal expansion of an ideal gas: since $U$ depends only on temperature for an ideal gas, $\Delta U = 0$ when $T$ is constant, so $q = -w$ ; the qualitative result worth remembering is that reversible work exceeds irreversible work for the same $\Delta V$ (the exact logarithmic formula is beyond this syllabus).

The figure below shows the energy balance as a simple accounting diagram: heat and work flow across the system boundary, and their sum is exactly the change stored inside as internal energy — nothing is lost in transit.

Figure

A box labelled system with heat q entering from the left and work w entering from the right, both arrows pointing inward, and a third arrow inside pointing to a stored quantity labelled change in internal energy equal to q plus w

The core idea (continued): why this is a genuine conservation law

Internal energy $U$ is a state function — it depends only on the current condition of the system (temperature, pressure, amount), not on the path taken to reach it. So $\Delta U$ is fixed once you know the initial and final states. Heat $q$ and work $w$ , individually, are path functions: their split can differ depending on how a process is carried out (fast or slow, against high or low external pressure), but their sum $\Delta U$ always comes out the same for the same start and end points. This is precisely why the first law is powerful: even though $q$ and $w$ are each path-dependent bookkeeping entries, the total energy stored is not — it is conserved, exactly as the law demands.

Derivation AFinding work, heat, and internal energy change from an irreversible gas expansion

The goal is to compute the work done during an irreversible expansion of an ideal gas against constant external pressure, then combine it with a given heat exchange to find the resulting change in internal energy, using nothing beyond the defining relations $w = -P_{ext}\Delta V$ and $\Delta U = q + w$ , so that every intermediate quantity carries the correct IUPAC sign throughout.

Step 1. Identify the volume change. For one mole of ideal gas expanding from $V_i = 10\ \text{L}$ to $V_f = 20\ \text{L}$ against a constant external pressure $P_{ext} = 1.5\ \text{atm}$ , the volume change is $\Delta V = V_f - V_i = 20 - 10 = 10\ \text{L}$ .

Step 2. Apply the work formula. Since the gas expands against a constant external pressure, $w = -P_{ext}\,\Delta V = -(1.5\ \text{atm})(10\ \text{L}) = -15\ \text{L}\,\cdot\,\text{atm}$ . The negative sign is expected: the system (gas) is doing work on the surroundings as it expands, so energy leaves the system through work.

Step 3. Convert to SI energy units. Using $1\ \text{L}\,\cdot\,\text{atm} = 101.3\ \text{J}$ , $w = -15 \times 101.3 = -1519.5\ \text{J} \approx -1.52\ \text{kJ}$ .

Step 4. Combine with the given heat using the first law. If the gas absorbs $q = +400\ \text{J}$ of heat while expanding, then $\Delta U = q + w = 400 + (-1519.5) = -1119.5\ \text{J} \approx -1.12\ \text{kJ}$ .

Step 5. Interpret the sign of the result. Even though heat flowed into the gas, its internal energy fell overall, because the work the gas performed while expanding (1519.5 J) exceeded the heat it absorbed (400 J). The first law does not say heat and internal-energy change move together — it says their combination with work is exactly conserved.

Worked example

One mole of an ideal gas expands irreversibly against a constant external pressure of 1.5 atm, its volume increasing from 10 L to 20 L, while absorbing 400 J of heat from the surroundings. Calculate $w$ and $\Delta U$ . (1 L·atm = 101.3 J)

Step 1. Read off $\Delta V = 20 - 10 = 10\ \text{L}$ and $P_{ext} = 1.5\ \text{atm}$ , exactly as set up in the derivation above.

Step 2. Compute the work: $w = -P_{ext}\Delta V = -(1.5)(10) = -15\ \text{L}\,\cdot\,\text{atm} = -15 \times 101.3 = -1519.5\ \text{J} \approx -1.52\ \text{kJ}$ .

Step 3. Apply the first law with $q = +400\ \text{J}$ : $\Delta U = q + w = 400 - 1519.5 = -1119.5\ \text{J} \approx -1.12\ \text{kJ}$ .

Competitive stretch

Extend the same gas sample: suppose instead it is compressed adiabatically and 250 J of work is done on it by the surroundings, and separately, in an unrelated process, the gas releases 60 J of heat while its internal energy falls by 300 J — find $q$ , $\Delta U$ for the first case and $w$ for the second.

Step 1. Adiabatic means $q = 0$ by definition, and compression means work is done on the system, so $w = +250\ \text{J}$ .

Step 2. First law gives $\Delta U = q + w = 0 + 250 = +250\ \text{J}$ : internal energy rises because compression work converts entirely into stored energy with no heat escaping.

Step 3. For the separate process, $q = -60\ \text{J}$ (heat released) and $\Delta U = -300\ \text{J}$ are given; rearranging $\Delta U = q + w$ gives $w = \Delta U - q = -300 - (-60) = -240\ \text{J}$ .

Final answer: adiabatic compression gives $q = 0$ , $\Delta U = +250\ \text{J}$ ; the second process has $w = -240\ \text{J}$ (the gas expanded, doing work on the surroundings).

Common traps

Mixing sign conventions — some older textbooks write $\Delta U = q - w$ with work done by the system positive; ISC/CISCE uses $\Delta U = q + w$ with $w = -P_{ext}\Delta V$ . Never switch conventions mid-problem.
Forgetting to convert L·atm to joules (or vice versa) before adding a work term to a heat value already given in joules.
Assuming $q = 0$ automatically forces $\Delta U = 0$ — true only if $w$ is also zero. In an adiabatic process, $\Delta U = w$ , which need not vanish.
Believing work is always negative — it is negative for expansion but positive for compression; the sign follows the direction of energy transfer, not a fixed rule of thumb.

Ask the tutor: If sign conventions or L·atm-to-joule conversions keep tripping you up, ask the tutor to work through a few more $\Delta U = q + w$ problems step by step.

Why does expansion make work negative if the gas is clearly doing something, using energy?

The sign is defined from the system's own point of view, not from whether energy is "used." When the gas expands, it pushes the surroundings and transfers energy out through work, so from the system's perspective that energy is leaving — hence $w < 0$ . If instead the surroundings compress the gas, energy is being pushed into the system through work, so $w > 0$ . Think of the sign as tracking the direction energy crosses the boundary, into the system is positive, out of the system is negative, applied consistently to both q and w.

If Delta U is a state function, why do q and w individually change with the path?

Internal energy only depends on where the system starts and ends, like altitude on a mountain depending only on your current position. Heat and work are more like the distance you walked and the height you climbed on different trails to reach that altitude, they can differ trail to trail even though your net altitude gain is the same. A gas can go from state A to state B doing a lot of work and absorbing a lot of heat, or doing little work and absorbing little heat, but the combination q + w always equals the same Delta U because energy bookkeeping must balance regardless of route.

How do I know whether to use w equals minus P external Delta V or the reversible isothermal case?

Use $w = -P_{ext}\Delta V$ whenever the problem states or implies expansion or compression against a fixed external pressure, this is the standard irreversible case and covers almost every numerical problem at this level. The reversible isothermal case only applies when the process is described as infinitely slow, with external pressure adjusted to match the gas pressure at every instant, and ISC syllabus expects only the qualitative comparison (reversible work magnitude exceeds irreversible work for the same volume change) rather than the logarithmic formula, so you will rarely need to compute a number for it.