Question

Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm ($1atm=1.013\times {10}^{5}Pa$), 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.

Answer 1

Initial volume, $V_1=100.0l=100.0\times {10}^{-3}$ $m^3$
Final volume, $V_2=100.5l=100.5\times {10}^{-3}$ $m^3$
Increase in volume, $V=V_2-V_1=0.5\times {10}^{-3}$ $m^3$
Increase in pressure, $p=100.0atm=100\times 1.013\times {10}^5$ Pa
Bulk modulus = $p/\left(\Delta V/V_1\right)=p{V}_1/\Delta V$
= $100\times 1.013\times {10}^5\times 100\times {10}^{-3}/(0.5\times {10}^{-3}$ )
= $2.026\times {10}^9$ Pa
Bulk modulus of air = $1\times {10}^5$ Pa
Bulk modulus of water / Bulk modulus of air = $2.026\times {10}^9/(1\times {10}^5)=2.026\times {10}^4$
This ratio is very high because air is much more compressible than water.