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 = `((trianglep)/(triangleV))/V_1 = (trianglep xx V_1)/(triangleV)`

`= (100 xx 1.013 xx 10^5 xx 100 xx 10^(-3))/(0.5xx 10^(-3))`

`= 2.026 xx 10^9 Pa`

Bulk modulus of air =`1.0 xx 10^5 Pa`

`:. "Bulk modulus of water"/"Bulk modulus of air" = (2.026 xx 10^9)/(1.0 xx 10^5) = 2.026 xx 10^4`

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