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.
Given data,
Initial volume, V1 = 100.0 L = 100.0 × 10-3 m 3.
Final volume, V2 = 100.5 L = 100.5 × 10-3 m 3.
Increase in volume, Δv = v2-v1 = 0.5 × 10-3 m 3.
Increase in pressure, Δp = 100.0 atm = 100× 1.013× 105 Pa.
The bulk modulus (or) of a substance is a measure of how incompressible/resistant to compressibility that substance is. It is defined as the ratio of the infinitesimal pressure increase to the resulting relative decrease of the volume.
Bulk modulus of water = Δp/(Δv/v1) = Δp×v1/ Δv
⇒ Bulk modulus = 100× 1.013× 105× 100× 10-3/⇒ (0.5× 10-3)
⇒ Bulk modulus = 2.026× 109 Pa
⇒ Bulk modulus of air = 1 × 105 Pa.
∴ Bulk modulus of water/ Bulk modulus of air = 2.026× 109/(1× 105) = 2.026× 104
This ratio is very high because air is more compressible than water.