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 × 10⁵ Pa

Bulk modulus = ${\displaystyle \frac{\frac{△p}{△V}}{V_1}=\frac{△p\times V_1}{△V}}$

${\displaystyle =\frac{100\times 1.013\times {10}^5\times 100\times {10}^{-3}}{0.5\times {10}^{-3}}}$

${\displaystyle =2.026\times {10}^{9}Pa}$

Bulk modulus of air =${\displaystyle 1.0\times {10}^{5}Pa}$

${\displaystyle \therefore \frac{\text{Bulk modulus of water}}{\text{Bulk modulus of air}}=\frac{2.026\times {10}^9}{1.0\times {10}^5}=2.026\times {10}^4}$

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