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First Law of Thermodynamics Worked Examples cheat sheet - grade 11-12

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Physics Grade 11-12

First Law of Thermodynamics Worked Examples Cheat Sheet

A printable reference covering the first law, internal energy, heat, work, sign conventions, and thermodynamic process examples for grades 11-12.

TopicsThermodynamicsHeatEnergyPhysicsStudy Guide
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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=Q−W\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=Q−W\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(Vf−Vi)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=∫ViVfP dVW = \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=Q−W\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=∫ViVfP dVW = \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 1 Pa⋅m3=1 J1\,\text{Pa}\cdot\text{m}^3 = 1\,\text{J} and mismatched units give incorrect energy values.

Practice Questions

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