Unit 02 · Lesson 05 of 09

What does a moving boundary do?

The shaded area is a process, not a property.

Explore the model. Derive the result. Check its limits.

What you’ll be able to do

Derive moving-boundary work.
Integrate a polytropic path.
The $n=1$ path needs a logarithm.

Before you touch the controls

Does doubling final volume always double expansion work?

An experiment you can change

Shade the work beneath a polytropic path

Predict the direction of change, move one control, then explain the response using the equation.

Final volume / initial volume2 Polytropic exponent n1

W_b=2\ln(2)=1.3863\,\mathrm{kJ},\quad p_2=100\,\mathrm{kPa}

Expansion traverses the shaded area left to right, giving positive work.

Model assumptions and units

Quasi-equilibrium ideal-gas path; p1 = 200 kPa, V1 = 0.01 m³. Work by the gas is positive. kPa m³ = kJ.

pV^n=C,\quad W_b=\int_{V_1}^{V_2}p\,dV=\begin{cases}\frac{p_2V_2-p_1V_1}{1-n}&n\ne1\\p_1V_1\ln(V_2/V_1)&n=1\end{cases}

Derive moving-boundary work.

A pressure force on a piston is $pA$ . A small displacement changes volume by $A\,ds$ , so the incremental work is $p\,dV$ . Integrating requires the actual pressure path at the boundary; a quasi-equilibrium model lets you use the system pressure along that path.

\delta W_b=pA\,ds=p\,dV,\quad W_b=\int_{V_1}^{V_2}p\,dV

Integrate a polytropic path.

For $pV^n=C$ , substitute $p=CV^{-n}$ . The power-law integral is valid when n differs from one. Replace C using either endpoint to obtain the expression in terms of measured pressures and volumes.

W_b=\frac{C}{1-n}(V_2^{1-n}-V_1^{1-n})=\frac{p_2V_2-p_1V_1}{1-n}

The n=1 path needs a logarithm.

At $n=1$ , pressure is C/V and the integral is logarithmic. Dividing by $1-n$ would be undefined, so this case needs its own expression. For an ideal gas it is an isothermal path.

W_b=p_1V_1\ln\!\left(\frac{V_2}{V_1}\right)\quad(n=1)

Interpret the direction.

Expansion increases volume and gives positive boundary work by the gas. Compression traverses the plot toward smaller volume and gives negative work. At constant volume the area is zero; at $n=0$ the path is constant pressure.

A worked example

An isothermal ideal-gas expansion.

Take $p_1=200\,\mathrm{kPa}$ , $V_1=0.010\,\mathrm{m^3}$ , and $V_2=0.020\,\mathrm{m^3}$ .
With $n=1$ , $p_2=p_1V_1/V_2=100\,\mathrm{kPa}$ .
Work is $200(0.010)\ln2=1.3863\,\mathrm{kJ}$ . Ideal-gas internal energy is unchanged in this isothermal process, so heat input equals work output.

Make the idea your own

Calculate. Compare. Test the assumptions.

Work out your answer, check it, then explore the visual solution. The answer checker uses a 1.5% rounding tolerance; for model-limit questions it checks the displayed model result.

1 · Calculate

Integrate the logarithmic path

A quasi-equilibrium gas starts at 200 kPa and 0.010 m³. It doubles volume with $n = 1.$ Find work by the gas.

Hint

Use the logarithmic branch rather than dividing by 1 − n.

Explore the visual solution

Final volume / initial volume2 Polytropic exponent n1

W_b=2\ln(2)=1.3863\,\mathrm{kJ},\quad p_2=100\,\mathrm{kPa}

Expansion traverses the shaded area left to right, giving positive work.

Model assumptions and units

Quasi-equilibrium ideal-gas path; p1 = 200 kPa, V1 = 0.01 m³. Work by the gas is positive. kPa m³ = kJ.

pV^n=C,\quad W_b=\int_{V_1}^{V_2}p\,dV=\begin{cases}\frac{p_2V_2-p_1V_1}{1-n}&n\ne1\\p_1V_1\ln(V_2/V_1)&n=1\end{cases}

Controls explore variations. The steps below retain the original question’s values.

Quasi-equilibrium ideal-gas path; p1 = 200 kPa, V1 = 0.01 m³. Work by the gas is positive. kPa m³ = kJ.

pV^n=C,\quad W_b=\int_{V_1}^{V_2}p\,dV=\begin{cases}\frac{p_2V_2-p_1V_1}{1-n}&n\ne1\\p_1V_1\ln(V_2/V_1)&n=1\end{cases}

2 · Compare

Compress at constant pressure

Starting from the same state, halve volume with $n = 0.$ Find signed boundary work.

Hint

At n = 0 pressure stays constant, so W = p(V2 − V1).

Explore the visual solution

Final volume / initial volume0.5 Polytropic exponent n0

W_b=\frac{2\left(0.5^{1-0}-1\right)}{1-0}=-1\,\mathrm{kJ},\quad p_2=200\,\mathrm{kPa}

Compression traverses the shaded area right to left, giving negative work.

Model assumptions and units

Quasi-equilibrium ideal-gas path; p1 = 200 kPa, V1 = 0.01 m³. Work by the gas is positive. kPa m³ = kJ.

pV^n=C,\quad W_b=\int_{V_1}^{V_2}p\,dV=\begin{cases}\frac{p_2V_2-p_1V_1}{1-n}&n\ne1\\p_1V_1\ln(V_2/V_1)&n=1\end{cases}

Controls explore variations. The steps below retain the original question’s values.

Quasi-equilibrium ideal-gas path; p1 = 200 kPa, V1 = 0.01 m³. Work by the gas is positive. kPa m³ = kJ.

pV^n=C,\quad W_b=\int_{V_1}^{V_2}p\,dV=\begin{cases}\frac{p_2V_2-p_1V_1}{1-n}&n\ne1\\p_1V_1\ln(V_2/V_1)&n=1\end{cases}

3 · Challenge the model

A steeper expansion path

Triple initial volume with $n = 1.4.$ Find work, then change n to 1 without changing endpoints in volume. Why does work differ?

Hint

Use p2 = p1/r^n, then the n ≠ 1 formula.

Explore the visual solution

Final volume / initial volume3 Polytropic exponent n1.4

W_b=\frac{2\left(3^{1-1.4}-1\right)}{1-1.4}=1.778\,\mathrm{kJ},\quad p_2=42.96\,\mathrm{kPa}

Expansion traverses the shaded area left to right, giving positive work.

Model assumptions and units

Quasi-equilibrium ideal-gas path; p1 = 200 kPa, V1 = 0.01 m³. Work by the gas is positive. kPa m³ = kJ.

pV^n=C,\quad W_b=\int_{V_1}^{V_2}p\,dV=\begin{cases}\frac{p_2V_2-p_1V_1}{1-n}&n\ne1\\p_1V_1\ln(V_2/V_1)&n=1\end{cases}

Controls explore variations. The steps below retain the original question’s values.

Quasi-equilibrium ideal-gas path; p1 = 200 kPa, V1 = 0.01 m³. Work by the gas is positive. kPa m³ = kJ.

pV^n=C,\quad W_b=\int_{V_1}^{V_2}p\,dV=\begin{cases}\frac{p_2V_2-p_1V_1}{1-n}&n\ne1\\p_1V_1\ln(V_2/V_1)&n=1\end{cases}

Your learning path

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