Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 10⁵ 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.
The initial and final volume of the water is 100.0 litre and 100.5 litre respectively. The increase in the pressure is 100.0 atm .
The bulk modulus is given by the equation,
B = Δ P Δ V V
Substituting the values in the above equation, we get:
B = 100 × 1.013 × 10 5 Pa ( 100.5 × 10 − 3 m 3 − 100.0 × 10 − 3 m 3 ) 100.0 × 10 − 3 m 3 = 2.026 × 10 9 Pa
The bulk modulus of air is 1 × 10 5 Pa .
The ratio of the bulk modulus of water to that of air is,
B water B air = 2.026 × 10 9 Pa 1 × 10 5 Pa = 2.026 × 10 4
From the above calculation, it is observed that the ratio is very large. It is because water is less compressible as compared to air.
The initial and final volume of the water is
The bulk modulus is given by the equation,
Substituting the values in the above equation, we get:
The bulk modulus of air is
The ratio of the bulk modulus of water to that of air is,
From the above calculation, it is observed that the ratio is very large. It is because water is less compressible as compared to air.
