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Thermal Physics

The first law of thermodynamics

  • 3 things to know
  • Component 1

Overview

Recall questions on ΔU = Q − W with the sign convention for each term, applying it to a gas being compressed and expanding, the work done as pΔV, adiabatic and isothermal changes, and worked problems in which two of the three quantities are given.

What you need to know

Answer each one before you reveal it.

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Key term

State the first law of thermodynamics in terms of the heat supplied and the work done by the system, defining every symbol

Key term

What underlying physical principle does the first law of thermodynamics express?

Key term

Interpret a negative value of ΔU.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    State the first law of thermodynamics in the form used in this course, defining every symbol.

    Model answer

    The first law is ΔU = Q − W, where ΔU is the increase in internal energy in J, Q is the heat supplied to the system in J and W is the work done by the system in J.

  2. 2

    What underlying physical principle does the first law of thermodynamics express?

    Model answer

    The first law is a statement of the conservation of energy applied to a thermodynamic system.

  3. 3

    Interpret a negative value of ΔU.

    Model answer

    A negative ΔU means the internal energy of the system has decreased, so for an ideal gas the temperature has fallen.

  4. 4

    Interpret a negative value of Q.

    Model answer

    A negative Q means heat has flowed out of the system to its surroundings rather than into it.

  5. 5

    Interpret a negative value of W.

    Model answer

    A negative W means work has been done on the system by its surroundings, which for a gas corresponds to compression.

  6. 6

    A gas absorbs 500 J of heat and does 200 J of work on its surroundings. Calculate the change in its internal energy.

    Model answer

    Applying ΔU = Q − W gives 500 − 200 = 300 J, so the internal energy rises by 300 J. Adding the 300 J stored to the 200 J of work done returns the 500 J supplied.

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