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.

#### Solution

Initial volume, V_{1 }= 100.0l = 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

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.