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The first law of thermodynamics connects heat transfer, work, and changes in internal energy. This cheat sheet helps students organize the sign conventions and equations needed to solve worked examples correctly. It is especially useful for problems involving gases, pistons, pressure-volume graphs, and heating or cooling processes.

Clear formulas and example patterns make it easier to decide what is known, what is unknown, and which form of the law to use.

The core idea is conservation of energy applied to a thermal system: energy can enter as heat, leave as work, or change the system's internal energy. A common physics convention is ΔU=QW\Delta U = Q - W, where WW is work done by the system. For gases at constant pressure, work is often found with W=PΔVW = P\Delta V, and for pressure-volume graphs, work is the area under the curve.

Special processes such as isochoric, isobaric, isothermal, and adiabatic changes simplify the first law in different ways.

Key Facts

  • The first law of thermodynamics is ΔU=QW\Delta U = Q - W, where QQ is heat added to the system and WW is work done by the system.
  • If heat enters the system, then Q>0Q > 0, and if heat leaves the system, then Q<0Q < 0.
  • If the system expands and does work on the surroundings, then W>0W > 0, and if the surroundings compress the system, then W<0W < 0.
  • For a constant-pressure process, the work done by a gas is W=PΔV=P(VfVi)W = P\Delta V = P(V_f - V_i).
  • On a pressure-volume graph, the work done by the gas is the area under the curve, so W=ViVfPdVW = \int_{V_i}^{V_f} P\,dV.
  • For an isochoric process, ΔV=0\Delta V = 0, so W=0W = 0 and the first law becomes ΔU=Q\Delta U = Q.
  • For an adiabatic process, Q=0Q = 0, so the first law becomes ΔU=W\Delta U = -W.
  • For an ideal gas isothermal process, ΔT=0\Delta T = 0 and ΔU=0\Delta U = 0, so Q=WQ = W.

Vocabulary

System
The system is the matter or region being studied, such as a gas inside a piston.
Internal Energy
Internal energy UU is the total microscopic kinetic and potential energy of the particles in a system.
Heat
Heat QQ is energy transferred because of a temperature difference between the system and its surroundings.
Work
Work WW is energy transferred when a force moves a boundary, such as a gas expanding a piston.
Isochoric Process
An isochoric process occurs at constant volume, so ΔV=0\Delta V = 0 and W=0W = 0.
Adiabatic Process
An adiabatic process has no heat transfer, so Q=0Q = 0.

Common Mistakes to Avoid

  • Mixing up the sign of work, which is wrong because ΔU=QW\Delta U = Q - W assumes WW is work done by the system, not work done on the system.
  • Using W=PΔVW = P\Delta V when pressure is not constant, which is wrong because variable pressure requires the graph area or W=ViVfPdVW = \int_{V_i}^{V_f} P\,dV.
  • Forgetting that compression gives ΔV<0\Delta V < 0, which is wrong because compression usually makes W<0W < 0 under the work-done-by-system convention.
  • Assuming temperature always changes when heat is added, which is wrong because an isothermal ideal gas can have Q>0Q > 0 while ΔU=0\Delta U = 0.
  • Ignoring units for pressure and volume, which is wrong because 1Pam3=1J1\,\text{Pa}\cdot\text{m}^3 = 1\,\text{J} and mismatched units give incorrect energy values.

Practice Questions

  1. 1 A gas absorbs 500J500\,\text{J} of heat and does 180J180\,\text{J} of work on its surroundings. Find ΔU\Delta U using ΔU=QW\Delta U = Q - W.
  2. 2 A gas expands at a constant pressure of 2.0×105Pa2.0 \times 10^5\,\text{Pa} from 0.010m30.010\,\text{m}^3 to 0.016m30.016\,\text{m}^3. Find the work done by the gas using W=PΔVW = P\Delta V.
  3. 3 During an adiabatic compression, 350J350\,\text{J} of work is done on a gas. Find QQ, WW, and ΔU\Delta U using the convention ΔU=QW\Delta U = Q - W.
  4. 4 Explain why the area under a pressure-volume graph represents work, and describe how the sign of the work changes for expansion versus compression.

Understanding First Law of Thermodynamics Worked Examples

A thermodynamics problem becomes much clearer when you first draw a boundary around the system. The system may be the gas inside a cylinder, not the piston, container, or room. Everything outside that boundary is the surroundings.

Heat is energy transferred because of a temperature difference. Work is energy transferred when a force causes motion, such as a moving piston. Internal energy is different.

It is energy stored in the random motion and interactions of particles within the system. For an ideal gas, internal energy depends only on temperature. This means a rise in gas temperature signals an increase in internal energy, even if the gas volume changes in a complicated way.

Sign conventions cause many wrong answers because textbooks sometimes use opposite definitions for work. Before calculating, write one sentence stating which convention the question uses. Under the convention where work means work done by the gas, an expanding gas transfers energy to its surroundings.

Its work is positive. A compressed gas receives energy through work, so its work is negative. Think about a bicycle pump.

Pushing the handle compresses air. Your hand does work on the air, and the air often warms up. This is why the direction of energy transfer matters more than simply seeing the word work in a question.

Pressure and volume graphs provide a visual way to check calculations. During expansion, the path moves to the right. During compression, it moves to the left.

The work depends on the whole path, not just the starting and ending points. Two paths can connect the same states yet enclose different areas below their lines. They therefore involve different amounts of work.

Internal energy, however, is a state property. Its change is fixed once the initial and final states are known.

Heat is not a state property either. A gas may reach the same final temperature by receiving heat slowly, being compressed, or following a mixture of both processes.

For worked examples, make a short energy record before inserting numbers. Identify whether the gas expands or compresses. Decide whether heat enters or leaves.

Then predict whether internal energy should increase, decrease, or possibly remain unchanged. This prediction catches sign errors early. Keep pressure in pascals and volume in cubic metres when calculating work, since their product gives joules.

A pressure given in kilopascals or a volume given in litres must usually be converted first. In real engines, refrigerators, aerosols, and pumps, processes are rarely perfectly constant pressure, constant volume, or adiabatic.

The ideal process names are models. They help isolate one energy transfer at a time and build reliable reasoning for more realistic situations.