Subjective Type

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

Solution

Initial volume, $$V_1 = 100.0l = 100.0 \times 10^{-3}$$ $$ m^3$$
Final volume, $$V_2 = 100.5 l = 100.5 \times 10^{-3} $$ $$m^3$$
Increase in volume, $$V = V_2 - V_1 = 0.5 \times 10^{-3}$$ $$m^3$$
Increase in pressure, $$p = 100.0 atm = 100 \times 1.013 \times 10^{5}$$ Pa
Bulk modulus = $$p / (\Delta V/V_1) = p V_1 / \Delta V$$
= $$ 100 \times 1.013 \times 10^{5} \times 100 \times 10^{-3} / (0.5 \times 10^{-3}$$ )
= $$2.026 \times 10^{9}$$ 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.


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