Solution Found!
?(a) How high in meters must a column of glycerol be to exert a pressure equal to that of a 760-mm column of mercury? The density of glycerol is 1.26 g
Chapter 10, Problem 10.17(choose chapter or problem)
(a) How high in meters must a column of glycerol be to exert a pressure equal to that of a 760-mm column of mercury? The density of glycerol is 1.26 g/mL, whereas that of mercury is 13.6 g/mL.
(b) What pressure, in atmospheres, is exerted on the body of a diver if she is 15 ft below the surface of the water when the atmospheric pressure is 750 torr? Assume that the density of the water is \(1.00 \mathrm{~g} / \mathrm{cm}^{3}=1.00 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). The gravitational constant is \(9.81 \mathrm{~m} / \mathrm{s}^{2}, \text { and } 1 \mathrm{~Pa}=1 \mathrm{~kg} / \mathrm{m}-\mathrm{s}^{2}\).
Questions & Answers
QUESTION:
(a) How high in meters must a column of glycerol be to exert a pressure equal to that of a 760-mm column of mercury? The density of glycerol is 1.26 g/mL, whereas that of mercury is 13.6 g/mL.
(b) What pressure, in atmospheres, is exerted on the body of a diver if she is 15 ft below the surface of the water when the atmospheric pressure is 750 torr? Assume that the density of the water is \(1.00 \mathrm{~g} / \mathrm{cm}^{3}=1.00 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). The gravitational constant is \(9.81 \mathrm{~m} / \mathrm{s}^{2}, \text { and } 1 \mathrm{~Pa}=1 \mathrm{~kg} / \mathrm{m}-\mathrm{s}^{2}\).
ANSWER:Step 1 of 5
(a)
We need to identify the height of glycerol, in meters, to exert a pressure equal to that of a column of mercury.
The equation for pressure is
\(P=d h g\)
Where,
P is the pressure
d is the density
h is the height
g is the acceleration because of the gravity.
Since the density of glycerol is \(1.26 \mathrm{~g} / \mathrm{mL}\) and the density of mercury is \(13.6 \mathrm{~g} / \mathrm{mL}\).
Since the pressure needed to exert by the column of glycerol is 760 mm, that means that
\(\mathrm{P}_{\text {glycerol }}=\mathrm{P}_{\text {mercury }}\)
Using the relation, we can derive it to get the height of the column of glycerol. The expression will be
\(\mathrm{h}_{\text {glycerol }}=\frac{(\mathrm{dh})_{\text {mercury }}}{\mathrm{g}} \mathrm{d}_{\text {glycerol }}\)
The given data
\(\begin{array}{l}760 \mathrm{~mm}: \mathrm{h}_{\text {mercury }}\\
13.6 \mathrm{~g} / \mathrm{mL}: \mathrm{d}_{\text {mercury }}\\
1.26 \mathrm{~g} / \mathrm{mL}: d_{\text {glycerol }}\end{array}\)