p = 100 atmosphere = $100 \times 1.013 \times 105$PaInitial volume,$V_1 = 100 \text{ litre} = 100 \times 10^{-3} \text{ m}^3$Final volume, $V_2 = 100.5 \text{ litre} = 100.5 \times 10^{-3} \text{ m}^3$ $\therefore \Delta V =$ change in volume $= V_2 - V_1$ $= (100.5 - 100) \times 10^{-3} \text{ m}^3$$= 0.5 \times 10^{-3} \text{ m}^3$$B_w =$ bulk modulus of water $= ?$$\therefore$ From formula, $B_w = \frac{P}{\frac{\Delta V}{V}},$ we getor $B_w = 2.026 \times 10^{9} \text{ Pa}$Also we know that the bulk modulus of air at S.T.P is given by$B_{\text{air}} = 10^{5} \text{ Pa}$$\therefore \frac{B_w}{B_{\text{air}}} = \frac{2.026 \times 10^{9}}{10^{5}}$$= 2.026 \times 10^{4} = 20260$The ratio is too large. This is due to the fact that the strain for air is much larger than for water at the same temperature. In other words the intermolecular distances in case of liquids are very small as compared to the corresponding distances in the case of gases. Hence there are larger interatomic forces in liquids than in gases.