Question

Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 10⁵ Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

Answer 1

Given

• Initial volume: \( V_{1} = 100.0 \,\text{L} = 100.0 \times 10^{-3} \,\text{m}^{3} = 0.1000 \,\text{m}^{3} \)
• Final volume: \( V_{2} = 100.5 \,\text{L} = 100.5 \times 10^{-3} \,\text{m}^{3} = 0.1005 \,\text{m}^{3} \)
• Increase in pressure: \( \Delta P = 100.0 \,\text{atm} \)
• \( 1 \,\text{atm} = 1.013 \times 10^{5} \,\text{Pa} \)

1. Change in volume

\( \Delta V = V_{2} - V_{1} = 0.1005 - 0.1000 = 0.0005 \,\text{m}^{3} \)

2. Pressure increase in SI units

\( \Delta P = 100.0 \times 1.013 \times 10^{5} = 1.013 \times 10^{7} \,\text{Pa} \)

3. Bulk modulus of water

Bulk modulus \( B \) is defined as

\( B = -\,\dfrac{\Delta P}{\Delta V / V_{1}} = \dfrac{\Delta P \, V_{1}}{\Delta V} \)

\( B = \dfrac{1.013 \times 10^{7} \times 0.1000}{0.0005} = \dfrac{1.013 \times 10^{6}}{0.0005} = 2.026 \times 10^{9} \,\text{Pa} \)

Bulk modulus of water \( B_{\text{water}} \approx 2.0 \times 10^{9} \,\text{Pa} \).

4. Comparison with air

Bulk modulus of air at constant temperature (isothermal) is approximately equal to its pressure, so take \( B_{\text{air}} \approx 1.0 \times 10^{5} \,\text{Pa} \).

\( \dfrac{B_{\text{water}}}{B_{\text{air}}} = \dfrac{2.026 \times 10^{9}}{1.0 \times 10^{5}} \approx 2.0 \times 10^{4} \)

Water is about \( 2 \times 10^{4} \) times less compressible (i.e. stiffer) than air.

5. Simple explanation of large ratio

• Air is a gas: its molecules are far apart with large empty spaces between them, so it can be compressed significantly by pressure.
• Water is a liquid: its molecules are already close together with very little free space, so even large pressures produce only tiny volume changes, giving a very large bulk modulus.