9478 / 2027
Thermodynamic Systems overview

Lesson 2 of 4

Heating, work and the first law

  • I can distinguish work done on and by a gas and calculate displacement work against constant external pressure.Syllabus

    Syllabus 9478, 13(d). Distinguish work done on and by a gas and calculate displacement work against constant external pressure.

  • I can recall and apply the first law as Delta U = Q + W, with W the work done on the system.Syllabus

    Syllabus 9478, 13(f). Recall and apply the first law as Delta U = Q + W, with W the work done on the system.

Internal energy can change through heating, work or both. Use one work convention throughout the calculation and identify which transfers enter the chosen system.

Set the signs before substituting

For the systems considered here, with no change in the whole system's bulk kinetic or gravitational potential energy, the first law is:

ΔU = Q + Won

Q is positive for net energy transferred into the system by heating. Won is positive for net work done on it. A positive ΔU means its internal energy increases.

Signs for the work-on form of the first law
TransferInto the systemOut of the system
Heating QPositiveNegative
Work WonPositive work on itNegative when it does work

If work done by the gas is given instead, Wby = -Won, so the same law becomes ΔU = Q - Wby. Do not add positive work by the gas to Q in the work-on convention.

Derive constant-pressure displacement work

Choose the gas as the system. It expands against a constant absolute external pressure pext. For piston area A, the external resisting force has magnitude pextA. Moving the boundary outward through Δx increases the gas volume by ΔV = AΔx.

Wby = force × outward displacement
= pextAΔx = pextΔV
Won = -pextΔV

For expansion, ΔV is positive: the gas does positive work and Won is negative. For compression, ΔV is negative and work on the gas is positive. Use the total stated external pressure, including the surroundings and load, rather than only a gauge-pressure difference.

In a slow, frictionless piston process, the gas pressure is effectively equal to the external pressure. The area under that constant-pressure p-V path then gives the work magnitude. A general gas-pressure curve cannot automatically give work for an irreversible process in which gas and external pressures differ.

Estimate the work scale

Pressure of order 105 Pa acting through a volume increase of order 10-3 m3 gives work of order 102 J. If heating supplies a few hundred joules, an internal-energy increase of a few hundred joules is plausible, with expansion work taking some energy out.

Worked constant-pressure expansion

Separate the input from the energy retained

A fixed amount of monatomic ideal gas expands slowly against pext = 1.20 × 105 Pa, from 2.00 × 10-3 to 3.50 × 10-3 m3. It receives 450 J by heating. The piston is frictionless, and the equilibrium gas pressure is the stated external pressure.

Expansion is outward motion against an inward load

The gas expands slowly with a frictionless piston against constant external pressure pext = 1.20 × 105 Pa. In this quasistatic model, gas pressure is effectively equal to the external pressure.

Use piston area and its outward displacement

A 0.150-metre piston displacement increases gas volume by 0.00150 cubic metresA horizontal cylinder has its fixed gas-facing end at drawing coordinate sixty. The initial piston gas-facing position is 180, shown dashed; the final position is 270, shown solid. Axial distances therefore use six hundred drawing units per metre: initial gas length 0.200 metres, final 0.350 metres and outward displacement 0.150 metres. The cross-section is schematic; its given constant area is 0.0100 square metres. The blue arrow points right for piston displacement. The separate purple external-load force on the piston points left and has magnitude p external times area, 1200 newtons. It is the external load force, not a claimed net force on the slowly moving piston. The two positions are alternative states of one piston. Work by the gas is positive 180 joules; work on the gas is negative 180 joules.GasΔx = +0.150 mFext = 1200 N0.200 m0.350 mArea A = 0.0100 m2

Dashed and solid piston faces mark the initial and final states on one length scale. The given area makes ΔV = 0.0100 × 0.150 = 1.50 × 10-3 m3. Blue is displacement; purple is the external load force.

The constant-pressure rectangle gives the work magnitude

The expansion from two to 3.5 thousandths of a cubic metre does 180 joules of workThe volume axis is in units of ten to the minus three cubic metres, and the pressure axis is in units of ten to the five pascals. A horizontal path from volume two to 3.50 has pressure 1.20. The expansion arrow points right. The shaded rectangle has width 1.50 and height 1.20 in these axis units, so its area represents 1.50 times 1.20 times one hundred joules, or 180 joules. With the stated slow frictionless condition, plotted gas pressure equals the constant external pressure. This gives positive work by the gas and negative work on it during expansion. Reversing the path changes the work signs while leaving the positive geometric rectangle unchanged.0.01.02.03.54.00.00.40.81.21.6180 Jp / 105 PaV / 10-3 m3

For expansion, Wby = +180 J and Won = -180 J. With Q = +450 J, the gas gains ΔU = 450 - 180 = 270 J. Reversing the same path changes the work signs, not the geometric area.

The piston area is 0.0100 m2, with initial and final gas lengths 0.200 and 0.350 m. External force and outward displacement are opposite. The constant-pressure graph uses the slow-process pressure equality; its 180 J area is work by the expanding gas.
ΔV = (3.50 - 2.00) × 10-3
= 1.50 × 10-3 m3
Wby = (1.20 × 105)(1.50 × 10-3)
= +180 J
Won = -180 J
ΔU = 450 + (-180) = +270 J

The p-V graph's scaled rectangle gives the same result: 1.20 × 1.50 × 105 × 10-3 = 180 J. Its geometric area is positive; the direction of the path determines the signed work.

The 450 J input supplies both the 180 J work output and the 270 J internal-energy increase. They are separate destinations for the same input.

There is an independent endpoint check. For this monatomic ideal gas, U = 3nRT/2 = 3pV/2:

U1 = (3/2)(1.20 × 105)(2.00 × 10-3) = 360 J
U2 = (3/2)(1.20 × 105)(3.50 × 10-3) = 630 J
U2 - U1 = 270 J

If its initial temperature is 300 K, fixed amount and constant pressure give T2 = 300(3.50/2.00) = 525 K. The expansion is not isothermal.

Try it step by step

Work it out in steps: Work done on the gas W = -p x (change in V) in J, then Increase in internal energy = Q + W in J.

  1. Problem 1. A gas absorbs 500 J of heat and expands by 2.0 x 10-3 m3 against a constant pressure of 1.0 x 105 Pa. Find the work done on the gas and the change in its internal energy.
  2. Problem 2. A gas absorbs 800 J of heat and expands by 5.0 x 10-3 m3 against a constant pressure of 1.0 x 105 Pa. Find the work done on the gas and the change in its internal energy.
  3. Problem 3. A gas loses 150 J of heat while it is compressed by 1.5 x 10-3 m3 at a constant pressure of 2.0 x 105 Pa. Find the work done on the gas and the change in its internal energy.
Show the answer

Problem 1.

Work done on the gas W = -p x (change in V) in J: W = -1.0 x 105 x 2.0 x 10-3 = -200 J (the gas does 200 J of work on the surroundings)

Increase in internal energy = Q + W in J: change in U = Q + W = 500 + (-200) = 300 J

Problem 2.

Work done on the gas W = -p x (change in V) in J: W = -1.0 x 105 x 5.0 x 10-3 = -500 J

Increase in internal energy = Q + W in J: change in U = 800 + (-500) = 300 J

Problem 3.

Work done on the gas W = -p x (change in V) in J: W = -2.0 x 105 x (-1.5 x 10-3) = +300 J (work is done on the gas)

Increase in internal energy = Q + W in J: change in U = (-150) + 300 = 150 J

Check your understanding

A gas receives 450 J by heating and does 180 J of work during expansion. Using Delta U = Q + W_on, what is its internal-energy change?

  1. +270 J, since W_on = -180 J.
  2. +450 J, because heating equals the internal-energy increase.
  3. +630 J, because both quoted energy values are positive.
Show the answer

+270 J, since W_on = -180 J.

Heating enters the system, while expansion work transfers energy out. With the work-on convention, Delta U = +450 + (-180) = +270 J.

Apply the same convention to different processes

Separate thermal processes, all using work done on the system
ProcessQ / JWon / JΔU / J
Reverse along the same constant-pressure path-450+180-270
Insulated compression with 90 J work input0+90+90
Rigid closed system, 250 J heating and no other work+2500+250

In the reverse path, compression work enters the gas but more energy leaves by heating. The latter two rows are independent examples; they are not stated to reach the same endpoints as the expansion.

An insulated system can warm when work is done on it. Insulation makes Q zero under the stated model; it does not make ΔU zero or ensure constant temperature. A rigid boundary removes volume-displacement work, but electrical or other work could still enter unless excluded.

For expansion into a vacuum, pext = 0, so the displacement work is zero even though the volume increases. A volume increase alone therefore does not establish a positive work output.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

State the first law of thermodynamics, defining each term. A gas receives 500 J of energy by heating and expands, doing 200 J of work on its surroundings. Calculate the change in its internal energy. [3]

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