Compressibility
- Compressibility is the measure of compression of a substance.
- Reciprocal of bulk modulus is termed as ‘Compressibility’.
- Mathematically:
- k=1/B = – (1/p) (ΔV/V)
- It is denoted by ‘k’.
- k(solids)<k(liquids)<k(gases)
Problem:-The average depth of Indian Ocean is about 3000 m.
Calculate the fractional compression, ΔV/V, of water at the bottom of the ocean, given that the bulk modulus of water is 2.2 × 109 N m–2. (Take g = 10 m s–2)
Answer: – The pressure exerted by a 3000 m column of water on the bottom layer
p = hρ g = 3000 m × 1000 kg m–3 × 10 m s–2
= 3 × 107 kg m–1 s-2
= 3 × 107 N m–2
Fractional compression ΔV/V, is
ΔV/V = stress/B = (3 × 107 N m-2)/ (2.2 × 109 N m–2)
= 1.36 × 10-2 or 1.36 %
Problem: The bulk modulus for water is 2.1GPa.Calculate the contraction in volume of 200ml of water is subjected to a pressure of 2MPa.
Answer:-
B=2.1GPa = 2.1 x109Pa.
V=200ml = 2×10-6ml.
P=2MPa = 200×106 Pa.
B=- (1/p) (ΔV/V) = ΔV = pV/B
= (2×106x200x10-6)/ = 2.1 x109
B=0.19ml
Problem: – 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.
Answer:
Initial volume, V1 = 100.0l = 100.0 × 10–3 m3
Final 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
= (100×1.013 × 105x100x10-3)/0.5×10-3
=2.026×106 Pa
Bulk modulus of air= 1.0×105Pa
Therefore,
Bulk modulus of water/ Bulk modulus of air
=2.026×106/1.0×105 =2.026×104
This ratio is very high because air is more compressible than water.
Problem:-
What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 103 kg m–3?
Answer:-
Let the given depth be h.
Pressure at the given depth, p = 80.0 atm = 80 × 1.01 × 105 Pa
Density of water at the surface, ρ1 = 1.03 × 103 kg m–3
Let ρ2 be the density of water at the depth h.
Let V1 be the volume of water of mass m at the surface.
Let V2 be the volume of water of mass m at the depth h.
Let ΔV be the change in volume.
ΔV = V1 – V2
=m (1/ ρ1 – 1/ ρ2)
Therefore, Volumetric strain = ΔV/ V1
=m (1/ ρ1 – 1/ ρ2) x ρ1/m
Therefore, ΔV/ V1 = (1- ρ1/ ρ2) … (i)
Bulk modulus, B = p V1/ ΔV
ΔV/ V1 = p/B
Compressibility of water =1/B =45.8×10-11 Pa-1
Therefore, ΔV/ V1= 80×1.013×105x45.8×10-11 = 3.71×10-3 … (ii)
For equations (i) and (ii), we get:
1- ρ1 / ρ2 = 3.71 x10-3
ρ2 = 1.03×103/(1-(3.71×10-3)
=1.034×103kgm-3
Therefore, the density of water at the given depth (h) is 1.034 × 103 kg m–3.
Problem: – Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.
Answer:-
Hydraulic pressure exerted on the glass slab, p = 10 atm = 10 × 1.013 × 105 Pa
Bulk modulus of glass, B = 37 × 109 Nm–2
Bulk modulus, B= p/ (ΔV/V)
Where,
ΔV/V = Fractional change in volume
Therefore,
ΔV/ V = p/B
=10×1.013×105
=2.73×10-5
Hence, the fractional change in the volume of the glass slab is 2.73 × 10–5.
Problem: – How much should the pressure on a litre of water is changed to compress it by 0.10%?
Answer:
Volume of water, V = 1 L
It is given that water is to be compressed by 0.10%.
Fractional change = ΔV/V =0.1/100×1 = 10-3
Bulk modulus, B= ρ/ ΔV/V
p=B x ΔV/V
Bulk Modulus of water, B = 2.2×109Nm-2
p=2.2×109x10-3
=2.2x106Nm-2
Therefore, the pressure on water should be 2.2 ×106 Nm–2.
Problem: – The average depth of Indian Ocean is about 3000 m. Calculate the fractional compression, ΔV/V, of water at the bottom of the ocean, given that the bulk modulus of water is 2.2 × 109 N m–2. (Take g = 10 m s–2)
Answer:- The pressure exerted by a 3000 m column of water on the bottom layer
p = hρ g = 3000 m × 1000 kg m–3 × 10 m s–2
= 3 × 107 kg m–1 s-2
= 3 × 107 N m–2
Fractional compression ΔV/V, is
ΔV/V = stress/B = (3 × 107 N m-2)/ (2.2 × 109 N m–2)
= 1.36 × 10-2 or 1.36 %