Question 9.12:
Compute the bulk modulus of water from the following data: Initial volume = 100.0 litres,
Pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), Final volume = 100.5 litres.
Compare the bulk modulus of water with that of air (at constant temperature). Explain in
simple terms why the ratio is so large.
Initial volume, V1=100.01 litres=100.0×10-3 m3
Final volume, V2=100.51 litres=100.5×10-3 m3
Change in volume, ∆V=V2-V1=0.5×10-3 m3
Change in pressure, ΔP = 100.0 atm = 100 × 1.013 × 105 Pa
Bulk modulus is given by,
B=∆P∆VV1=∆P×V1∆V=100×1.013×105×100×10-30.5×10-3=2.026×109Pa
The bulk modulus of air =1.0×105Pa
∴Bulk modulus of water Bulk modulus of air =2.026×1091.0×105=2.026×104
This ratio is very high because air is more compressible than water.
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