Compute the bulk modulus of water from the following data: Initial volume $= 100.0 \; L$,Pressure increase $= 100.0 \; atm$ $(1 \; atm = 1.013 \times 10^{5} \; Pa)$,Final volume $= 100.5 \; L$. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

(N/A) Initial volume,$V_{1} = 100.0 \; L = 100.0 \times 10^{-3} \; m^{3}$.
Final volume,$V_{2} = 100.5 \; L = 100.5 \times 10^{-3} \; m^{3}$.
Change in volume,$\Delta V = V_{2} - V_{1} = 0.5 \times 10^{-3} \; m^{3}$.
Increase in pressure,$\Delta p = 100.0 \; atm = 100 \times 1.013 \times 10^{5} \; Pa = 1.013 \times 10^{7} \; Pa$.
Bulk modulus $B = -\frac{\Delta p}{\Delta V / V_{1}} = \frac{\Delta p \cdot V_{1}}{\Delta V}$.
$B = \frac{1.013 \times 10^{7} \times 100.0 \times 10^{-3}}{0.5 \times 10^{-3}} = \frac{1.013 \times 10^{6}}{0.5 \times 10^{-3}} = 2.026 \times 10^{9} \; Pa$.
The bulk modulus of air is approximately $1.0 \times 10^{5} \; Pa$.
The ratio of the bulk modulus of water to that of air is $\frac{2.026 \times 10^{9}}{1.0 \times 10^{5}} \approx 2.026 \times 10^{4}$.
This ratio is very large because water is a liquid and is nearly incompressible,whereas air is a gas and is highly compressible due to the large intermolecular spaces.