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Thermal Physics

Work done by a gas at constant pressure

  • 3 things to know
  • A2 Unit 3

Overview

Recall questions on Work done by a gas at constant pressure.

What you need to know

Answer each one before you reveal it.

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Key term

Derive W = pΔV from the definition of work as force multiplied by distance.

Key term

Explain why work, like heat, is described as energy in transit.

Key term

State the equation for the work done by a gas expanding at constant pressure, and define each symbol with its unit.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Explain why work, like heat, is described as energy in transit.

    Model answer

    Work names energy only while it is being transferred across the boundary of a system. Once transferred it becomes part of the system's internal energy rather than remaining as work.

  2. 2

    State the equation for the work done by a gas expanding at constant pressure, and define each symbol with its unit.

    Model answer

    W = pΔV, where W is the work done by the gas in J, p is the constant pressure in Pa and ΔV is the increase in volume in m³.

  3. 3

    Derive W = pΔV from the definition of work as force multiplied by distance.

    Model answer

    The gas pushes a piston of area A with a force pA. If the piston moves out a distance d, the work done is pA × d, and Ad is the increase in volume ΔV, so W = pΔV.

  4. 4

    What is the sign of the work when a gas expands, and what does that sign mean?

    Model answer

    The work is positive. The gas does work on its surroundings and loses that energy through its boundary.

  5. 5

    What is the sign of the work when a gas is compressed?

    Model answer

    The change in volume is negative, so W is negative. Work is done on the gas by the surroundings.

  6. 6

    A gas at a constant pressure of 1.5 × 10⁵ Pa expands from 2.0 × 10⁻³ m³ to 5.0 × 10⁻³ m³. Calculate the work it does.

    Model answer

    W = pΔV = 1.5 × 10⁵ × 3.0 × 10⁻³ = 450 J. Dividing back, 450 ÷ 1.5 × 10⁵ gives a volume change of 3.0 × 10⁻³ m³.

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