Gas Laws

Boyle's Law

pressure up, volume down

At fixed temperature, pressure times volume stays constant for this ideal gas model. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Start at fixed temperature

At the first state, pressure 2 pascals times volume 6 cubic meters gives 12 joules.

piVi=2 Pa⋅6 m3=12 Jp_iV_i = 2\ \text{Pa}\cdot 6\ \text{m}^{3} = 12\ \text{J}
Boyle comparisonBoth piston states share one volume scale.6 m^32 Pa 300 Kstate3 m^34 Pa 300 Kstate

Increase pressure and reduce volume

At the second state, pressure is 4 pascals and volume is 3 cubic meters.

pfVf=4 Pa⋅3 m3p_fV_f = 4\ \text{Pa}\cdot 3\ \text{m}^{3}

The product stays constant

The second product is also 12 joules, so higher pressure pairs with lower volume.

piVi=pfVf=12 Jp_iV_i = p_fV_f = 12\ \text{J}
Boyle comparisonBoth piston states share one volume scale.6 m^32 Pa 300 Kstate3 m^34 Pa 300 Kstate

Several rows keep the same product

The diagram draws the first and third rows. At fixed temperature, larger pressure pairs with smaller volume so the product remains the same.

pVpV2 Pa6 m312 J3 Pa4 m312 J4 Pa3 m312 J6 Pa2 m312 J\begin{array}{c|c|c}p&V&pV\\2\ \text{Pa}&6\ \text{m}^{3}&12\ \text{J}\\3\ \text{Pa}&4\ \text{m}^{3}&12\ \text{J}\\4\ \text{Pa}&3\ \text{m}^{3}&12\ \text{J}\\6\ \text{Pa}&2\ \text{m}^{3}&12\ \text{J}\\\end{array}
Boyle comparisonRows one and three are the checked piston states.6 m^32 Pa 300 Kstate3 m^34 Pa 300 Kstate