Question

The compound 1-iodododecane is a nonvolatile liquid with a density of 1.20g/mL. The density of mercury is 13.6g/mL. What do you predict for the height of a barometer column based on 1-iodododecane, when the atmospheric pressure is 749 torr? What is the pressure, in atmospheres, on the body of a diver if he is 21 ft below the surface of the water when the atmospheric pressure is 742 torr?

Answer 1

Answer:

a. $h=0.823m$

b. $P_{diver}=161653.04Pa$

Explanation:

Hello,

a.

One could consider the following equation for this exercise's first part:

$P_{liq}=P_{h}+P_{atm}\\d_{liq}hg=d_{Hg}hg+P_{atm}$

Whereas $d$ is the density, $g$ the acceleration of gravity, $h$ the height and $P_{atm}$ the atmospheric pressure, thus:

$h=\frac{P_{atm}}{g*(d_{liq}-d_{Hg})} =\frac{749torr*\frac{101325Pa}{760torr} }{-9.8\frac{m}{s^2}*(1.2-13.6)\frac{g}{mL}*\frac{1kg}{1000g}*\frac{1x10^6mL}{1m^3} } \\h=0.823m$

b.

Now, to know the pressure on the diver's body, one uses the following equation considering that the pressure is exerted downwards:

$P_{diver}=P_{atm}+d_{water}gh\\\\P_{diver}=742torr*\frac{101325Pa}{760torr}+1000\frac{kg}{m^3} *9.8\frac{m}{s^2} *21ft*\frac{0.3048m}{1 ft} \\P_{diver}=161653.04Pa$

Best regards.