MCAT topics → Thermodynamics and Kinetics → First Law of Thermodynamics and Internal Energy

First Law of Thermodynamics and Internal Energy

MCAT trap: Confuses physics sign convention (w > 0 for work by system) with chemistry convention (w > 0 for work on system). In the MCAT/chemistry convention ΔU = q + w, work done ON the system is positive and work done BY the system is negative.

The First Law of Thermodynamics is a heavily tested MCAT concept: conservation of energy applied to a thermodynamic system, expressed as ΔU = q + w. Internal energy (U) is a state function — it depends only on where you start and end, not the path. Heat (q) and work (w) are path-dependent, but their sum always equals the same ΔU for a given change of state. across all three modes: pure recall of sign conventions and process definitions, application to specific scenarios (what happens to temperature in an adiabatic expansion?), and data interpretation from P-V diagrams in passages.

The trickiest part for most students is the sign convention. The MCAT uses the chemistry convention — ΔU = q + w — where work done ON the system is positive. Physics courses often flip this, using ΔU = q − w where work done BY the system is positive. These conventions give the same answer if applied correctly, but mixing them up mid-problem is a very common error. Memorize which convention your equation assumes and stick with it.

The four standard process types (isothermal, adiabatic, isobaric, isochoric) are fair game on the MCAT, and students routinely confuse adiabatic with isothermal. Adiabatic means no heat exchange (q = 0), which forces temperature to change when work is done — it does NOT mean constant temperature. Isothermal means constant temperature, which requires heat flow to compensate for any work. Getting these definitions locked in prevents a cascade of wrong answers on passage-based questions.

Common misconceptions

Common mistake
Wrong: Work done BY the system is positive (chemistry convention is sometimes confused with physics convention).
Right: In the MCAT/chemistry convention ΔU = q + w, work done ON the system is positive and work done BY the system is negative.
The physics convention defines positive work as work done BY the system, leading to ΔU = q − w. The chemistry convention — which the MCAT uses — defines positive work as work done ON the system, giving ΔU = q + w. These are two valid frameworks, not two different answers, but you cannot mix them. If a gas expands and does work on the surroundings, w is negative in the chemistry convention, which correctly reduces ΔU. Always check which sign convention your equation assumes before plugging in numbers.
Common mistake
Wrong: In an adiabatic process, temperature is constant because no heat flows.
Right: In an adiabatic process q = 0, so ΔU = w and temperature changes; it is the isothermal process where temperature is constant.
Adiabatic means thermally isolated — no heat is exchanged with the surroundings (q = 0). It does NOT mean temperature stays constant. In fact, when q = 0, the first law gives ΔU = w, so any work done on or by the system directly changes internal energy and therefore temperature. Isothermal is the process where temperature is constant (ΔT = 0), which for an ideal gas also means ΔU = 0, requiring q = −w. These two concepts are essentially opposites in terms of heat flow versus temperature change.
Common mistake
Wrong: Work can be done even when volume is constant because pressure changes.
Right: PV work requires a volume change (w = -PΔV), so no PV work is done in an isochoric (constant volume) process regardless of pressure change.
PV work is defined as w = −PΔV. If volume does not change (isochoric process, ΔV = 0), then w = 0 — full stop. It doesn't matter how dramatically the pressure changes; without a volume change, there is no boundary work done. Pressure change alone cannot push a piston or expand a gas against surroundings if the container is rigid. In an isochoric process, the first law simplifies to ΔU = q, so all energy exchange is in the form of heat.
Common mistake
Wrong: The area under a P-V curve represents the change in internal energy.
Right: The area under a P-V curve represents the work done by or on the system, not ΔU.
The area under a P-V curve gives work (w = −∫P dV), not internal energy. Internal energy is a state function determined by the endpoints, not the path — you cannot read ΔU directly off a curve's area. Work, by contrast, is path-dependent and equals the area swept out as the system moves along the curve. To find ΔU from a P-V diagram, you still need to know q (or use additional information about the process type), then apply ΔU = q + w.
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What the exam tests

  1. Know the formula ΔU = q + w and apply the correct sign convention: work done ON the system is positive, work done BY the system is negative (chemistry/MCAT convention).
  2. Identify what is zero or constant in each process type: isothermal (ΔT = 0, ΔU = 0 for ideal gas), adiabatic (q = 0), isobaric (ΔP = 0), isochoric (ΔV = 0, w = 0).
  3. Calculate PV work using w = −PΔV at constant pressure, or determine work from the area under a P-V curve when pressure is not constant.
  4. Read a P-V diagram and correctly identify which region represents work done, determine the sign of work based on expansion versus compression, and use the first law to find q or ΔU.

Can you avoid these mistakes?

A gas expands against a constant external pressure of 2 atm, increasing its volume by 3 L. Using the chemistry convention, calculate the work done and state its sign. What does this tell you about the internal energy change if the process is adiabatic?
A sealed rigid container of gas is heated. Which process type is this, and what can you conclude about the work done? Write the simplified first law equation that applies here.
On a P-V diagram, a gas undergoes a cycle: it expands at high pressure, then is compressed at low pressure back to its starting volume. Is the net work done by the gas positive, negative, or zero? How do you determine this from the diagram?
A student reads that 'in an adiabatic expansion, temperature stays constant because no heat flows in.' Identify the two errors in this statement and write the corrected version.

Related topics

Enthalpy and Hess's Law Calorimetry and Heat Capacity Entropy and the Second Law Gibbs Free Energy and Spontaneity

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