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4 years ago

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(a) show that the pressure exerted by a fluid P (in pascals) is given by P= hdg, where h is the column of the fluid in metres, d

is density in kg/m3, and g is the acceleration due to gravity (9.81 m/s2). (Hint: see appendix 2.). (b) The volume of an air bubble that starts at the bottom of a lake at 5.24 degree celsius increases by a factor of 6 as it rises to the surface of water where the temperature is 18.73 degree celsius and the air pressure is 0.973 atm. The density of the lake water is 1.02 g/cm3. Use the equation in (a) to determine the depth of the lake in metres.

1 answer:

Murrr4er [49]4 years ago

3 0

Answer:

<em>56.4 m</em>

<em></em>

Explanation:

volume increases by factor of 6, i.e $\frac{V2}{V1}$ = 6

Initial temperature T1 at bottom of lake = 5.24°C = 278.24 K

Final temperature T2 at top of lake = 18.73°C = 291.73 K

NB to change temperature from °C to K we add 273

Final pressure P2 at the top of the lake = 0.973 atm

Initial pressure P1 at bottom of lake = ?

Using the equation of an ideal gas

$\frac{P1V1}{T1}$ = $\frac{P2V2}{T2}$

P1 = $\frac{P2V2T1}{V1T2}$ = $\frac{0.973*6*278.24}{291.73}$

P1 = 5.57 atm

5.57 atm = 5.57 x 101325 = 564380.25 Pa

Density Ρ of lake = 1.02 g/ $cm^{3}$ = 1020 kg/ $m^{3}$

acceleration due to gravity g = 9.81 $m/s^{2}$

Pressure at lake bottom = pgd

where d is the depth of the lake

564380.25 = 1020 x 9.81 x d

d = $\frac{564380.25}{10006.2}$ = <em>56.4 m</em>

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