AeroLab Reading room

Physics / A2 level

Thermodynamics

Apply the first law with an explicit work sign convention.

Learning objectives

The big idea

Internal energy is a property of a system. Heating and work describe ways energy crosses its boundary; a system does not contain a separate store called heat. Internal energy includes microscopic kinetic and potential contributions.

Choose the system and state whether work is done by it or on it. Here ΔU = Q − Wby: energy supplied by heating increases internal energy, while energy transferred out through work decreases it. Area under a quasi-static p–V path represents work done by the gas.

1. Choose the system and sign convention first

Internal energy belongs to a system’s state; heating and work describe transfers across its boundary. Here use ΔU = Q − Wby, with Q positive into the system and Wby positive for work done by it. A system receiving 300 J by heating while doing 100 J work gains 200 J internal energy. A source using work done on the system will write an equivalent equation with a different work sign.

2. Expansion work depends on the path

For a quasi-static constant-pressure expansion, work done by a gas is pΔV. Expanding by 0.002 m³ at 100000 Pa gives 200 J. More generally, area under a pressure–volume path gives work by the gas, using the appropriate boundary pressure. Two paths between the same endpoints can enclose different areas, so work is not determined only by the initial and final states.

3. Compare isothermal and adiabatic changes

For an ideal gas, internal energy depends only on temperature. Isothermal expansion therefore has ΔU = 0, so energy entering by heating balances work done. Adiabatic means Q = 0, not constant temperature. If an adiabatically expanding gas does positive work, its internal energy falls and its temperature decreases. Free expansion into a vacuum is a different limiting case because external work can be zero.

4. Cycles test your energy accounting

After a complete cycle, a system returns to its initial state and has zero net internal-energy change. Net energy supplied by heating then equals net work done over the cycle under this convention. On a suitable p–V diagram, the enclosed area represents net work, with direction setting its sign. A working device can exchange energy along several paths; observing one hot component is not enough to calculate efficiency.

A method you can reuse

  1. Draw a boundary around the chosen system.
  2. Write the meaning and signs of Q and work.
  3. Identify the process and any constant quantities.
  4. Calculate work from pressure and volume or graph area.
  5. Apply the first law and check the temperature consequence for an ideal gas.
Can a gas absorb energy by heating and still have no temperature change?

Yes. During ideal isothermal expansion, incoming energy can leave as work, leaving internal energy and temperature unchanged. Heating does not necessarily produce a temperature rise.

Equations and relationships

ΔU=Q−Wby\Delta U=Q-W_{\text{by}}

Q enters by heating; Wby is work done by the system.

Wby=pΔVW_{\text{by}}=p\Delta V

Work at constant pressure for a quasi-static change.

Worked example

A gas receives 100 J by heating and does 40 J of work. What is its internal-energy change?

Follow the solution

ΔU = Q − Wby = 100 − 40 = +60 J.

Investigate and explain

Use a teacher-provided p–V data set to compare expansion along two paths between the same volumes; estimate the different work transfers.

Use gas and energy-transfer models as related illustrations. They do not constitute a full reversible heat-engine simulation; distinguish imposed temperature controls from physically adiabatic or isolated conditions.

Open the interactive investigation

Check your understanding

1. Which is a property of a system rather than a transfer process?

  1. Heating
  2. Work
  3. Internal energy
Check the answer and reasoning

Internal energy

Internal energy describes the state; heating and work describe energy crossing a boundary.

2. A gas receives 100 J by heating and does 40 J of work. What is its internal-energy change?

  1. +60 J
  2. +140 J
  3. −60 J
Check the answer and reasoning

+60 J

ΔU = Q − Wby = 100 − 40 = +60 J.

3. An ideal gas expands isothermally. What is its internal-energy change?

  1. Always positive because volume increases
  2. Zero
  3. Equal to its final pressure
Check the answer and reasoning

Zero

Ideal-gas internal energy depends only on temperature.