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

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 × 10 ⁵ Pa

Bulk modulus = \(\frac{Δp }{ ΔV / V_1} =\frac{ Δp × V_1 }{ ΔV} \)

\(= \frac{100 × 1.013 × 10 ^5 × 100 × 10 ^{- 3 }}{ 0.5 × 106^{ - 3 }}\)

= 2.026 × 10 ⁹ Pa
Bulk modulus of air = 1.0 × 10 ⁵ Pa

\(∴\frac{ \text{Bulk modulus of water }}{ \text{Bulk modulus of air} }\)

\(= \frac{2.026 × 10 ^9 }{ 1.0 × 10^ 5 }\)

= 2.026 × 10 ⁴

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