Question

A. Calculate the fractional change in volume of a fixed mass of seawater as it moves from the surface of the ocean to a depth of 1500 m.

b. Calculate the fractional change in density of a fixed mass of air as it moves isothermally from the bottom to the top of the Empire State Building (a height of 350 m, equivalent to a change in pressure of about 4100 P a).
c. Compare the results from part a and part b and discuss the nature of the fluid medium. (2 pts)

Answer 1

a) 0.645%

b) 4.046%

c) Seawater is highly incompressible and the air is much more compressible

Step-by-step explanation:

a)

First we take a look at BULK MODULUS;

change in pressure is expressed as;

dp = -K(dV/V)

bulk modulus K = 2.28*10^9 N/m^2 for seawater

now one atm is given as;

1 atm = 101325 N/m^2

by conversion factor,

1 atm = 10.3m equivalent in height of water

now writing this pressure in terms of equivalent height of water at 1500m

we say;

dp = 1500/10.3 * 101325 N/m^2

now substituting into dp = -K(dV/V)

dV/V = - dp/K

= - ( 1500/10.3 * 101325 N/m^2 ) / (2.28*10^9 N/m^2)

= - (14756067.96 / 2.28*10^9)

= - (0.00647)

= - 0.645%

the fractional change in volume of fixed mass of seawater is 0.645%, the negative sign indicates the decrease in the change in volume.

b)

Lets look at the ideal gas equation

pV = mRT

P = pRT

now we know that T and T are constant

dP = RTdp

so

dP/P = dp/p

given that the equivalent change in the pressure is 4100 Pa;

dp/p = 4100 / 101325

= 0.04046

= 4.046%

the fractional change in density of a fixed mass of air is 4.046%

c)

Taking a close look of the fractional change in volume of seawater in partA and the fractional change in density of a fixed mass of air in partB, we can say that Seawater is highly incompressible and the air is much more compressible.