Question

Compute the bulk modulus of water from the following data:
Initial volume=100.0 litres
Pressure increase = 100.0 atm
Final volume = 100.5 litre.
Compare the bulk modulus of water with that of air. Explain in simple terms why the ratio is so large .
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Answer 1

Explanation:

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

5

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 2

$\huge\rm\underline\purple{Question :-}$

Compute the bulk modulus of water from the following data: Initial volume = 100.0 litres. Pressure increase = 100.0 atm. Final volume = 100.5 litre. Compare the bulk modulus of water with that of air. Explain in simple terms why the ratio is so large .

$\large{\underbrace{\sf{\red{ Required\:Solution:}}}}$

${\large{\bold{\sf{\bf {\underline{Given,}}}}}}$

P = 100 atmosphere

= 100 × 1.013 ×$10^5\: Pa$

( ∵ 1 atm = 1.013 × $10^5\: Pa$)

Initial volume, V1 = 100 litre

= 100 × $10^-^3$ $m^3$

Final volume, V2 = 100.5 litre

= 100.5 × $10^-^3$ $m^3$

${\large{\bold{\sf{\bf {\underline{Now,}}}}}}$

${\large{\bold{\sf{\bf {\underline{Change\: in\: volume,}}}}}}$

ΔV = V2 - V1

= (100.5 - 100) × $10^-^3$ $m^3$

= 0.5 × $10^-^3$ $m^3$

${\large{\bold{\sf{\bf {\underline{Using\: formula\: of \:bulk \:modulus,}}}}}}$

B = $\frac{P}{\Delta\:V/V}$ = $\frac{PV}{\Delta\:V}$

= $\frac{100×1.013×10^5×100×10^-3}{0.5×10^-3}$

= 2.026 × $10^9\: Pa$

${\large{\bold{\sf{\bf {\underline{We\: know,}}}}}}$

Bulk modulus of air = 1.0 × $10^5\:Pa$

${\large{\bold{\sf{\bf {\underline{Now,}}}}}}$

$\frac{Bulk \:modulus\: of\: water}{Bulk\: modulus\: of \:air}$ = $\frac{2.026×10^9}{1.0×10^5}$

= 2.026 × $10^4$

= 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.

${\large{\bold{\sf{\bf {\underline{In\: other\:words,}}}}}}$

The intermolecular distances in case of liquids are very small as compared to that in gases.

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