For the same bulk modulus and volume, pressure change follows volume change. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Keep bulk modulus and starting volume fixed

This contrast changes only the volume-change marker. The smaller marker should produce a smaller pressure change.

K=12 PaΔV=12 m3K=12\ \text{Pa}\qquad \Delta V=\tfrac{1}{2}\ \text{m}^{3}
Smaller volume-change inputOnly the volume-change marker is smaller.pressurevolume changevolume=6 m^3volumeChange=1/2 m^3pressureChange=1 Pamodulus=12 Pa

Smaller volume changes make smaller pressure changes

The table keeps the same starting volume and modulus. Each row is an exact positive-magnitude compression ledger.

VΔVKΔP6 m314 m312 Pa12 Pa6 m312 m312 Pa1 Pa6 m31 m312 Pa2 Pa\begin{array}{c|c|c|c}V&\Delta V&K&\Delta P\\6\ \text{m}^{3}&\tfrac{1}{4}\ \text{m}^{3}&12\ \text{Pa}&\tfrac{1}{2}\ \text{Pa}\\6\ \text{m}^{3}&\tfrac{1}{2}\ \text{m}^{3}&12\ \text{Pa}&1\ \text{Pa}\\6\ \text{m}^{3}&1\ \text{m}^{3}&12\ \text{Pa}&2\ \text{Pa}\\\end{array}

A smaller volume change lowers the pressure change

The modulus and starting volume stay fixed; only the checked volume change is smaller.

ΔP=12 Pa12 m36 m3=1 Pa\Delta P=12\ \text{Pa}\cdot\frac{\tfrac{1}{2}\ \text{m}^{3}}{6\ \text{m}^{3}}=1\ \text{Pa}
Smaller bulk-compression ledgerThe pressure marker shrinks with the volume-change marker.pressurevolume changevolume=6 m^3volumeChange=1/2 m^3pressureChange=1 Pamodulus=12 Pa