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Mechanical Properties Of Fluids

Question
CBSEENPH11019979

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, V= 100.0 l = 100.0 × 10 –3 m3
Final volume, V= 100.5 l = 100.5 ×10 –3 m3 
Increase in volume, ΔV = V2 – V= 0.5 × 10–3 m

Increase in pressure, Δp = 100.0 atm = 100 × 1.013 × 10Pa
Bulk space modulus space equals space fraction numerator increment straight p over denominator left parenthesis begin display style bevelled fraction numerator increment straight V over denominator straight V subscript 1 end fraction end style right parenthesis end fraction space equals increment straight p cross times fraction numerator straight V subscript 1 over denominator increment straight V end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 100 space cross times space 1.013 space cross times space 10 to the power of 5 space end exponent cross times space 100 space cross times space 10 to the power of negative 3 end exponent over denominator left parenthesis 0.5 space cross times space 10 to the power of negative 3 end exponent right parenthesis end fraction

space space space space space space space space space space space space space space space space space space space space space space space space space equals 2.026 space cross times space 10 to the power of 9 space Pa

Bulk space modulus space of space air space equals space 1 space cross times space 10 to the power of 5 space end exponent Pa

Therefore comma space

fraction numerator Bulk space modulus space of space water over denominator space Bulk space modulus space of space air end fraction space equals space fraction numerator 2.026 space cross times space 10 to the power of 9 over denominator left parenthesis 1 space cross times space 10 to the power of 5 right parenthesis end fraction space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 2.026 space cross times space 10 to the power of 4 space

This space ratio space is space very space high space because space air space is space
more space compressible space than space water.