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.
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Answer 1

Answer:

Initial volume, V1 = 100.0l = 100.0 × 10 –3 m3

Final volume, V2 = 100.5 l = 100.5 ×10 –3 m3

Increase in volume, ΔV = V2 – V1 = 0.5 × 10–3 m3

Increase in pressure, Δp = 100.0 atm = 100 × 1.013 × 105 Pa

Bulk modulus = Δp / (ΔV/V1) = Δp × V1 / ΔV

= 100 × 1.013 × 105 × 100 × 10-3 / (0.5 × 10-3)

= 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.

Answer 2

Answer:

Initial volume, V1 = 100.0l = 100.0 × 10 –3 m3

Final volume, V2 = 100.5 l = 100.5 ×10 –3 m3

Increase in volume, ΔV = V2 – V1 = 0.5 × 10–3 m3

Increase in pressure, Δp = 100.0 atm = 100 × 1.013 × 105 Pa

Bulk modulus = Δp / (ΔV/V1) = Δp × V1 / ΔV

= 100 × 1.013 × 105 × 100 × 10-3 / (0.5 × 10-3)

= 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.