Q. 124.4( 22 Votes )

# Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 10^{5} 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 :

Given data,

Initial volume, V_{1} = 100.0 L = 100.0 × 10^{-3} m ^{3}.

Final volume, V_{2} = 100.5 L = 100.5 × 10^{-3} m ^{3}.

Increase in volume, Δv = v_{2}-v_{1} = 0.5 × 10^{-3} m ^{3}.

Increase in pressure, Δp = 100.0 atm = 100× 1.013× 10^{5} 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/v_{1}) = Δp×v_{1}/ Δv

⇒ Bulk modulus = 100× 1.013× 10^{5}× 100× 10^{-3}/⇒ (0.5× 10^{-3})

⇒ Bulk modulus = 2.026× 10^{9} Pa

⇒ Bulk modulus of air = 1 × 10^{5} Pa.

∴ Bulk modulus of water/ Bulk modulus of air = 2.026× 10^{9}/(1× 10^{5}) = 2.026× 10^{4}

This ratio is very high because air is more compressible than water.

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