# 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.

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