Mechanical Properties of Solids

Question : 12

Total: 21

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

p = 100 atmosphere = 100×1.013×105Pa
Initial volume,V1=100‌litre=100×10–3m3
Final volume, V2=100.5‌litre=100.5×10–3m3
∴ΔV= change in volume =V2−V1
=(100.5−100)×10−3m3
=0.5×10−3m3
Bw= bulk modulus of water =?
∴ From formula, Bw=

ΔV

, we get

Btw=

ΔV

100×1.013×105×100×10−3

0.5×10−3

(∵V=V1)

or Bw=2.026×109Pa
Also we know that the bulk modulus of air at S.T.P is given by
Bair=105Pa
∴

Bair

2.026×109

105

=2.026×104=20260
The ratio is too large. This is due to the fact that the strain for air is much larger than for water at the same temperature. In other words the intermolecular distances in case of liquids are very small as compared to the corresponding distances in the case of gases. Hence there are larger interatomic forces in liquids than in gases.

Go to Question:

Prev Question

Next Question