The gravitational force exerted by an object is given by F

Chapter 6, Problem 10P

(choose chapter or problem)

QUESTION: Problem 10P

The gravitational force exerted by an object is given by F = mg, where F is the force in newtons, m is the mass in kilograms, and g is the acceleration due to gravity (9.81 m/s2).

(a) Use the definition of the pascal to calculate the mass (in kg) of the atmosphere above 1 m2 of ocean.

(b) Osmium (Z = 76) is a transition metal in Group 8B(8) and has the highest density of any element (22.6 g/mL). If an osmium column is 1 m2 in area, how high must it be for its pressure to equal atmospheric pressure? [Use the answer from part (a) in your calculation.]

Questions & Answers

QUESTION: Problem 10P

The gravitational force exerted by an object is given by F = mg, where F is the force in newtons, m is the mass in kilograms, and g is the acceleration due to gravity (9.81 m/s2).

(a) Use the definition of the pascal to calculate the mass (in kg) of the atmosphere above 1 m2 of ocean.

ANSWER:

Solution 10P:

Step 1: (a)

Here, we have to calculate the mass of the atmosphere above 1 m2 of the ocean using pascal.

We know that,

$Pressure\ =\frac{Force}{Area}$

1.0 Pascal = $\frac{1\ Newton\ (N)}{1\ {m}^{2}}$ ----(1)

Given that,

The gravitational force exerted by an object is given by F = mg,

Where, m is the mass in kg

g is the acceleration due to gravity (9.81 m/s2)

F is the force in N.

We know that

1.0 atm = 1.01325 $\times{}{10}^{5}Pa$

Putting the values in equation (1) we get

1.01325 $\times{}{10}^{5}Pa$ = $\frac{mg}{1\ {m}^{2}}$

$\Rightarrow{}mg=1.01325\times{}{10}^{5}Pa\ \times{}{1.0\ m}^{2}$

$\Rightarrow{}m=\frac{1.01325\times{}{10}^{5}Pa\ \times{}{1.0\ m}^{2}}{9.81\ m/{s}^{2}}$

$\Rightarrow{}m=\frac{1.01325\times{}{10}^{5}Pa\ \times{}{1.0\ m}^{2}}{9.81\ m/{s}^{2}}$ $\times{}\frac{1\ N{m}^{-2}}{1\ Pa\ }$ $\times{}\frac{1.0\ kg\ m\ {s}^{-2}}{1.0\ N}$