91Ó°ÊÓ

The gravitational force exerted by an object is given by \(F=m g\) where \(F\) is the force in newtons, \(m\) is the mass in kilograms, and \(g\) is the acceleration due to gravity \(\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\) (a) Use the definition of the pascal to calculate the mass (in \(\mathrm{kg}\) ) of the atmosphere above \(1 \mathrm{~m}^{2}\) of ocean. (b) Osmium \((Z=76)\) is a transition metal in Group \(8 \mathrm{~B}(8)\) and has the highest density of any element ( \(22.6 \mathrm{~g} / \mathrm{mL}\) ). If an osmium column is \(1 \mathrm{~m}^{2}\) in area, how high must it be for its pressure to equal atmospheric pressure? [Use the answer from part (a) in your calculation.]

Step by step solution

01

Understand the Problem

Given an object with mass (m), the gravitational force (F) it exerts is calculated using the formula: \[ F = m \times g \] where \( g \) is the acceleration due to gravity (\(9.81 \, \mathrm{m/s^2}\)).

02

Identify the Given Information for Part (a)

The problem states that we need to calculate the mass of the atmosphere above \(1 \, \mathrm{m^2}\) of ocean. The atmospheric pressure at sea level is approximately \(101325 \, \mathrm{Pa} \) (Pascals).

03

Define the Pascals

Pressure in Pascals is defined as force per unit area: \[ 1 \, \mathrm{Pa} = 1 \, \mathrm{N/m^2} \]. Given that the atmospheric pressure is \(101325 \, \mathrm{Pa} \), the force exerted on \(1 \, \mathrm{m^2}\) is: \[ F = 101325 \, \mathrm{N} \].

04

Calculate the Mass in Part (a)

Rearrange the gravitational force formula to solve for mass (m): \[ m = \frac{F}{g} \] Substituting the values, we get: \[ m = \frac{101325 \, \mathrm{N}}{9.81 \, \mathrm{m/s^2}} \approx 10329.67 \, \mathrm{kg} \]

05

Identify the Given Information for Part (b)

For an osmium column with a density of \(22.6 \, \mathrm{g/cm^3}\) (or \(22600 \, \mathrm{kg/m^3}\), remember to convert the units), we need to find its height such that the pressure it exerts equals atmospheric pressure (\(101325 \, \mathrm{Pa} \)).

06

Calculate the Volume of Osmium Required

Given that density (\(\rho\)) is mass per unit volume: \[ \rho = \frac{m}{V} \to V = \frac{m}{\rho} \]. We know the mass of the atmosphere is \(10329.67 \, \mathrm{kg}\). Therefore: \[ V = \frac{10329.67}{22600} \approx 0.457 \, \mathrm{m^3} \]

07

Calculate the Height of the Osmium Column

The volume of a column is given by the area of the base times the height (\(A \times h \)). For a \(1 \, \mathrm{m^2}\) base area: \[ h = \frac{V}{A} \approx \frac{0.457}{1} = 0.457 \mathrm{m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Force

Gravitational force is the attractive force that bodies exert on one another due to their masses. It is governed by Newton's law of gravitation. In this context, the force an object exerts due to gravity can be calculated using:
\( F = m \times g \) where:

• \( F \) is the gravitational force in newtons (N)
• \( m \) is the mass of the object in kilograms (kg)
• \( g \) is the acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \)

This formula is fundamental in physics and is applied in various scenarios, such as calculating the weight of an object on the Earth's surface.

Pressure

Pressure is defined as the force exerted per unit area. It is an important concept in physics and engineering and is commonly measured in pascals (Pa). The formula for pressure is:
\( P = \frac{F}{A} \) where:

• \( P \) is the pressure
• \( F \) is the force applied
• \( A \) is the area over which the force is applied

In the exercise, atmospheric pressure at sea level is given as \( 101325 \, \text{Pa} \). This pressure can be understood as the weight of the atmosphere pressing down on \( 1 \, \text{m}^2 \) of surface area. Knowing this helps in calculating forces and, as we saw in Part (a), the mass of the atmosphere above a given area.

Density

Density is a measure of how much mass is contained in a given volume. It is typically expressed in kilograms per cubic meter (kg/m³) or grams per cubic centimeter (g/cm³). The formula for density is:
\( \rho = \frac{m}{V} \) where:

• \( \rho \) is the density
• \( m \) is the mass
• \( V \) is the volume

In Part (b) of the exercise, osmium's density is given as \( 22.6 \, \text{g/cm}^3 \), which we convert to \( 22600 \, \text{kg/m}^3 \). This value is used to calculate the volume and height of the osmium column required to exert the same pressure as the atmosphere.

Transition Metals

Transition metals are elements found in groups 3-12 of the periodic table. They are characterized by having an incomplete d subshell in at least one of their oxidation states. These metals are known for their unique properties, including high melting points, high densities, and the ability to form various oxidation states.

Osmium is an example of a transition metal and has the highest density of any natural element. It belongs to Group 8B of the periodic table, with a density of \( 22.6 \, \text{g/cm}^3 \). This extraordinary density makes osmium incredibly heavy even in small volumes, which is why it is used in Part (b) of the problem to match atmospheric pressure using a compact column.