Unit 2 Practice Quiz: Diffusion, Temperature, and Pressure

Kinetic Molecular Theory and Particle Diffusion

• Concentration Gradients and Diffusion Origins:

  • Initial Particle Localization: When a sealed container of volatile compounds (such as a freshly popped bag of microwave popcorn) is opened, the highest concentration of scent particles is initially located inside the container.

  • Direction of Net Diffusion: Diffusion is the spontaneous net movement of particles from an area of higher concentration to an area of lower concentration down a concentration gradient. Gas particles never spontaneously diffuse from an area of least concentration to an area of greatest concentration, as that would violate the second law of thermodynamics.

  • Microscopic Mechanism of Diffusion: Diffusion results entirely from random, constant thermal motion and successive collisions between moving particles (Brownian motion), causing particles to spread out uniformly over time until dynamic equilibrium is reached.

• Temperature Dependence of Diffusion Rates:

  • Kinetic Energy and Velocity: The average kinetic energy of particles is directly proportional to the absolute temperature in Kelvin ($T$), defined by the relation:         $\text{KE}_{\text{avg}} = \frac{3}{2} k_B T$

  • Diffusion in Hot versus Cold Media: Solute particles (such as food coloring) diffuse significantly faster in hot water than in cold water. In a hot liquid, the solvent molecules possess greater average kinetic energy, move at higher velocities, and collide with solute particles more frequently and forcefully, dispersing them at an accelerated rate.

• Thermal Expansion of Thermometric Liquids:

  • Volume Changes with Temperature: Liquid particles (such as alcohol or mercury) inside a thermometer expand to occupy a larger volume when placed in a hot environment, causing the liquid column to rise.

  • Mechanism of Expansion: As temperature increases, the liquid molecules gain kinetic energy and vibrate or move more vigorously, increasing their average intermolecular separation distance. Conversely, particles contract to a smaller volume only when placed in a cold environment.

Atmospheric Pressure, Suction Mechanics, and Barometry

• The Physics of Suction Through a Straw:

  • The Vacuum Misconception: Suction does not exert an active tensile or pulling force on a liquid. A vacuum itself has no physical power to pull liquid upward into the mouth.

  • Mechanics of Liquid Ingestion: Inhaling or lowering the diaphragm and expanding the oral cavity increases volume, thereby reducing internal pressure ($P_{\text{mouth}} < P_{\text{atm}}$).

  • Role of External Pressure: The external atmospheric pressure ($P_{\text{atm}}$) pushing down on the surface of the liquid outside the straw is the force that pushes the liquid up the straw into the lower-pressure region. If external atmospheric pressure were removed (such as in the vacuum of space), drinking through an open straw would be physically impossible.

• Construction and Principles of a Mercury Barometer:

  • Apparatus Components:

    • Glass Tube: A long vertical glass tube (typically $\sim 1\,\text{m}$ in length) sealed at the top end and open at the bottom.

    • Liquid Mercury Reservoir: An open dish or basin filled with dense liquid mercury ($\text{Hg}$).

    • Torricellian Vacuum: The sealed space at the top of the tube above the liquid mercury column, which contains virtually no air particles (only an imperceptible trace of mercury vapor pressure, $\sim 0.0016\,\text{mmHg}$ at room temperature).

    • Air Particles: Gaseous molecules in the ambient atmosphere that collide with and exert downward force upon the exposed surface of the mercury pool.

  • Standard Atmospheric Pressure: At sea level and $0^\circ\text{C}$, standard atmospheric pressure supports a vertical mercury column of exactly $760\,\text{mmHg}$.

  • Hydrostatic Equilibrium in the Barometer Column:

    • The mercury remains suspended inside the column rather than emptying into the dish because the downward gravitational weight of the liquid column is exactly balanced by the upward force exerted by atmospheric pressure pushing on the surface of the open mercury reservoir.

    • Hydrostatic balance equation:             $P_{\text{atm}} = \rho g h$             where $\rho$ is the density of mercury ($13.6\,\text{g/cm}^3$ or $13,600\,\text{kg/m}^3$), $g$ is the acceleration due to gravity ($9.8\,\text{m/s}^2$), and $h$ is the height of the mercury column ($0.760\,\text{m}$ at standard pressure).

Gas Pressure Measurement with Barometers and Manometers

• Comparative Functions of Pressure Measurement Devices:

  • Barometer (Left Device): A closed-end device containing a vacuum above a mercury column; it measures the absolute atmospheric pressure of the surrounding classroom environment.

  • Open-Ended Manometer (Right Device): A U-shaped tube partially filled with mercury, with one end connected to a contained gas sample (a circular bulb holding helium gas, $\text{He}_{(g)}$) and the other end open to the ambient atmosphere; it measures the pressure of the gas relative to atmospheric pressure.

• Atmospheric Pressure Determination:

  • Instrument Used: The closed barometer on the left.

  • Reading: The height of the mercury column indicates an ambient classroom atmospheric pressure of $746\,\text{mmHg}$.

• Helium Gas Pressure Calculation in an Open-Ended Manometer:

  • Identifying Levels:

    • Gas-side arm height ($h_{\text{gas}}$, left arm): $114\,\text{mm}$

    • Open-atmosphere arm height ($h_{\text{open}}$, right arm): $350\,\text{mm}$

  • Height Differential ($\Delta h$):         $\Delta h = h_{\text{open}} - h_{\text{gas}}$         $\Delta h = 350\,\text{mm} - 114\,\text{mm} = 236\,\text{mm} = 236\,\text{mmHg}$

  • Pressure Relationship: Because the mercury level in the open arm is higher than the mercury level in the gas arm, the trapped helium gas is pushing harder than the atmosphere, meaning $P_{\text{gas}} > P_{\text{atm}}$.

  • Total Gas Pressure Formulation:         $P_{\text{He}} = P_{\text{atm}} + \Delta h$         $P_{\text{He}} = 746\,\text{mmHg} + 236\,\text{mmHg} = 982\,\text{mmHg}$

Pressure Units and Dimensional Analysis Conversions

• Standard Pressure Equivalences:     $1.00\,\text{atm} = 760\,\text{mmHg} = 760\,\text{torr} = 101.3\,\text{kPa} = 101,325\,\text{Pa} = 14.7\,\text{psi}$

• Step-by-Step Conversion Problems Using Conversion Tables:

  • Conversion 1: Kilopascals ($\text{kPa}$) to Millimeters of Mercury ($\text{mmHg}$)

    • Problem Statement: Convert $88.4\,\text{kPa}$ to $\text{mmHg}$.

    • Conversion Factor: $\frac{760\,\text{mmHg}}{101.3\,\text{kPa}}$

    • Calculation Setup:             $\text{Pressure} = 88.4\,\text{kPa} \times \left(\frac{760\,\text{mmHg}}{101.3\,\text{kPa}}\right)$             $\text{Pressure} = \frac{67,184}{101.3}\,\text{mmHg} = 663.22\,\text{mmHg}$

    • Final Answer: $663\,\text{mmHg}$ (or $663.2\,\text{mmHg}$)

  • Conversion 2: Torr ($\text{torr}$) to Pounds per Square Inch ($\text{psi}$)

    • Problem Statement: Convert $515\,\text{torr}$ to $\text{psi}$.

    • Conversion Factor: $\frac{14.7\,\text{psi}}{760\,\text{torr}}$

    • Calculation Setup:             $\text{Pressure} = 515\,\text{torr} \times \left(\frac{14.7\,\text{psi}}{760\,\text{torr}}\right)$             $\text{Pressure} = \frac{7,570.5}{760}\,\text{psi} = 9.961\,\text{psi}$

    • Final Answer: $9.96\,\text{psi}$ (or $10.0\,\text{psi}$)

  • Conversion 3: Atmospheres ($\text{atm}$) to Pascals ($\text{Pa}$)

    • Problem Statement: Convert $2.75\,\text{atm}$ to $\text{Pa}$.

    • Conversion Factor: $\frac{101,325\,\text{Pa}}{1.00\,\text{atm}}$

    • Calculation Setup:             $\text{Pressure} = 2.75\,\text{atm} \times \left(\frac{101,325\,\text{Pa}}{1.00\,\text{atm}}\right)$             $\text{Pressure} = 278,643.75\,\text{Pa}$

    • Final Answer: $279,000\,\text{Pa}$ (expressed in scientific notation as $2.79 \times 10^5\,\text{Pa}$)