Atmospheric Pressure Dynamics and Molar Chemistry Calculations

Definition of Atmospheric Pressure and the Barometer

• The concept of a barometer can be visualized using a straw with a finger over one end. As long as the end is closed and the weather remains constant, the liquid (mercury) will stay inside the tube.
• In a mercury-filled barometer, the average height of the liquid column is found to be $76\,\text{cm}$.
• One atmosphere ($1\,\text{atm}$) of pressure is defined by the average height of a mercury barometer at sea level.
• Equivalent measures for one atmosphere of pressure include:
  • $76\,\text{cm}$ of mercury ($\text{Hg}$).
  • $760\,\text{mm}$ of mercury.
  • $760\,\text{torr}$.
• The unit "torr" is named after Torr Kelly, identified as the inventor of the barometer.
• Atmospheric pressure is fundamentally defined as a mercury barometer reading exactly $76\,\text{cm}$. This definition is based on the average of readings taken at sea level over the course of a full year.

Determining Pressure through Fluid Dynamics and Mathematical Derivations

• A barometer is designed to measure atmospheric pressure. It works based on a pressure equilibrium: the downward pressure exerted by the mercury in the tube is equal to the upward pressure exerted by the air in the room.
• Pressure ($P$) is defined generally as force ($F$) divided by area ($A$):
  • $P = \frac{F}{A}$
• To find the pressure of the mercury in the tube, the weight of the mercury and the cross-sectional area of the tube must be known.
• The cross-sectional area involves the internal diameter ($d$). The radius is defined as:
  • $r = \frac{d}{2}$
• The area is calculated using the formula for the area of a circle:
  • $A = \pi \times r^2$
• Weight ($W$) is equivalent to force ($F$), which is calculated as mass ($m$) multiplied by the acceleration of gravity ($g$):
  • $F = W = m \times g$
• Therefore, pressure can be expressed as:
  • $P = \frac{m \times g}{A}$
• Because the mass of mercury in the tube changes as the pressure changes, mass can be substituted using the density formula ($\text{Density} = \frac{\text{Mass}}{\text{Volume}}$). Mass is therefore:
  • $m = V \times \rho$
  • Where $\rho$ is the density of mercury, which is a constant given as $13.58\,\text{g\,cm}^{-3}$.
• Substituting mass into the pressure equation:
  • $P = \frac{\rho \times V \times g}{A}$
• Since the barometer tube is a cylinder, its volume is equal to the cross-sectional area multiplied by the height ($h$):
  • $V = A \times h$
• Substituting the cylindrical volume into the pressure equation results in the final general formula for pressure exerted by a liquid:
  • $P = \frac{\rho \times (A \times h) \times g}{A}$
• The area terms cancel out, leaving:
  • $P = \rho \times g \times h$
• This equation ($P = \rho gh$) shows that the pressure exerted by any liquid is proportional to its density, the gravitational constant, and the depth (or height) of the liquid column. This explains why pressure increases as depth increases in a lake or ocean.

Dimensional Analysis and Calculating Pressure in CGS Units

• To determine the true units of pressure in the CGS (centimeter-gram-second) system, units are analyzed using the density of mercury ($13.58\,\text{g\,cm}^{-3}$), gravity ($980\,\text{cm\,s}^{-2}$), and the standard height ($76\,\text{cm}$).
• The unit product is:
  • $(\text{g\,cm}^{-3}) \times (\text{cm\,s}^{-2}) \times (\text{cm})$
• Breaking down the units:
  • $(\text{g}) \times (\text{cm}^{-3}) \times (\text{cm}^2) \times (\text{s}^{-2})$
  • $(\text{g}) \times (\text{cm}^{-1}) \times (\text{s}^{-2})$
• Pressure is force per area. In CGS, the unit of force is the "dime" (calculated as $\text{g\,cm\,s}^{-2}$). The unit of area is $\text{cm}^2$.
• By rearranging the terms ($\frac{\text{g\,cm\,s}^{-2}}{\text{cm}^2}$), it is proven that $\rho gh$ results in the unit "dimes per square centimeter."
• Example Problem: A liquid has a density of $2\,\text{g\,cm}^{-3}$, gravity is $980\,\text{cm\,s}^{-2}$, and height is $4.8 \times 10^3\,\text{cm}$.
  • $P = (2) \times (4.8 \times 10^3) \times (980)$
  • Rounded calculation: $2 \times 5 \times 10^3 \times 1000 = 1 \times 10^7\,\text{dimes/cm}^2$.
  • Exact calculation: $9,408,000\,\text{dimes/cm}^2$ or $9.4 \times 10^6\,\text{dimes/cm}^2$.
• Comparison to standard atmosphere:
  • Calculation for standard mercury barometer: $13.58 \times 76 \times 980 \approx 1 \times 10^6\,\text{dimes/cm}^2$.
  • Therefore, a pressure of $10,000,000\,\text{dimes/cm}^2$ is approximately equivalent to $10\,\text{atmospheres}$.

Foundations of Atomic Chemistry and the Periodic Table

• Definitions of periodic table notation:
  • $Z$: The atomic number (located in the lower-left corner), which identifies the element. Examples include Hydrogen ($Z = 1$) and Uranium ($Z = 92$), the latter being the last naturally occurring "nonmandane" element.
  • $A$: The atomic weight (located in the upper-left corner), measured in AMUs per atom.
• Specific atomic weights provided:
  • Lithium ($Li$): $7\,\text{AMUs/atom}$.
  • Nitrogen ($N$): $14\,\text{AMUs/atom}$.
  • Magnesium ($Mg$): $24\,\text{AMUs/atom}$.
  • Sulfur ($S$): $32\,\text{AMUs/atom}$.
• The atomic standard is Carbon-12 ($Z=6, A=12$), which is defined as exactly $12$ to infinity AMUs per atom. It is the only isotope with a perfect integer value for atomic weight.
• Gram Atomic Weight (GAW) is the atomic weight expressed in grams per mole ($\text{g/mol}$):
  • Lithium: $7\,\text{g/mol}$.
  • Nitrogen: $14\,\text{g/mol}$.
  • Magnesium: $24\,\text{g/mol}$.
  • Sulfur: $32\,\text{g/mol}$.

Avogadro's Number and Molar Conversions

• One mole ($n$) of any element always contains exactly $6.02 \times 10^{23}$ atoms.
• The Gram Molecular Weight or Gram Atomic Weight of an element is the mass of one mole of that element.
• Relationship Examples:
  • $7\,\text{g}$ of Lithium = $1\,\text{mole}$ = $6.02 \times 10^{23}$ atoms.
  • $14\,\text{g}$ of Lithium = $2\,\text{moles}$.
  • $70\,\text{g}$ of Lithium = $10\,\text{moles}$.
  • $28\,\text{g}$ of Nitrogen = $2\,\text{moles}$ = $2 \times (6.02 \times 10^{23})$ atoms.
  • $64\,\text{g}$ of Sulfur = $2\,\text{moles}$.
  • $16\,\text{g}$ of Sulfur = $0.5\,\text{moles}$ = $3.01 \times 10^{23}$ atoms.
• Comprehensive Calculation Problem: Given $8 \times 10^{36}$ atoms of Lithium, convert to moles, grams, and AMUs.
  • Converting to Moles: Divide atoms by Avogadro's number.
    • $\frac{8 \times 10^{36}}{6.02 \times 10^{23}\text{ atoms/mole}} \approx 1.32 \times 10^{13}\,\text{moles}$.
  • Converting to Grams: Multiply moles by gram atomic weight ($7\,\text{g/mol}$).
    • $(1.32 \times 10^{13}\,\text{moles}) \times (7\,\text{g/mole}) = 9.24 \times 10^{13}\,\text{grams}$.
  • Converting to AMUs: Multiply atoms by atomic weight per atom ($7\,\text{AMU/atom}$).
    • $(8 \times 10^{36}\,\text{atoms}) \times (7\,\text{AMU/atom}) = 56 \times 10^{36}\,\text{AMUs} = 5.6 \times 10^{37}\,\text{AMUs}$.

Questions & Discussion

• Question: If the barometer reads $77\,\text{cm}$, is the pressure greater or less than one atmosphere?

• Answer: Greater.

• Question: If it reads $72\,\text{cm}$, is the pressure greater or less?

• Answer: Less.

• Question: What happens to the mercury if the atmospheric pressure drops from $76\,\text{cm}$ to $74\,\text{cm}$?

• Answer: Mercury will drip out of the tube until the level reaches $74\,\text{cm}$.

• Question: How does altitude affect a barometer?

• Answer: As you walk up a mountain, pressure decreases, causing the mercury level to go down. Thus, a barometer can function as an altimeter under specific conditions.

• Question: Why doesn't the mercury drip out when we take the dish away (in the straw demonstration)?

• Answer: Mercury stays in because there is a pressure equilibrium: the downward pressure of mercury equals the upward pressure of the air in the room.

• Question: Can you provide a written equation to find all these answers?

• Answer: No. One must understand the concept rather than just remembering formulas. Like knowing there are $16$ ounces in a pound, the units themselves should guide the calculation through understanding. If the concept is understood, confusion should be eliminated.