Question 9.12:
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.
Solution:
Initial volume, V1 = 100.0l = 100.0 × 10 –3 m3Final volume, V2 = 100.5 l = 100.5 ×10 –3 m3
Increase in volume, ΔV = V2 – V1 = 0.5 × 10–3 m3
Increase in pressure, Δp = 100.0 atm = 100 × 1.013 × 105 Pa
Bulk modulus = Δp / (ΔV/V1) = Δp X V1 / ΔV
= 100 X 1.013 X 105 X 100 X 10-3 / (0.5 X 10-3)
= 2.026 X 109 Pa
Bulk modulus of air = 1 X 105 Pa
∴ Bulk modulus of water / Bulk modulus of air = 2.026 X 109 / (1 X 105) = 2.026 X 104
This ratio is very high because air is more compressible than water.
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