Comprehensive Guide to Kinetic Molecular Theory, Gas Laws, Manometers, and Phase Diagrams

Kinetic Molecular Theory & Properties of Gases

• Kinetic Molecular Theory (KMT):

  • Kinetic means motion.

  • All gases consist of tiny particles that are in continuous, rapid, random, and straight-line motion.

  • Gas particles move independently until they collide with each other or the container walls.

• Four Fundamental Assumptions of Ideal Gases:

  • Constant Random Motion: Gas particles move continuously in rapid, random, and straight-line paths until colliding with another object.

  • Negligible Particle Volume: The actual physical volume occupied by individual gas particles is so insignificant compared to the total space of the container that it is treated as effectively zero.

  • Perfectly Elastic Collisions: When gas particles collide with each other or container walls, no net kinetic energy is lost; total kinetic energy remains conserved.

  • No Intermolecular Interactions: Gas particles experience no attractive or repulsive forces toward one another. They interact exclusively during physical collisions.

• Compressibility of Gases:

  • Gases are easily compressed (pushed closer together) because individual gas particles are separated by large amounts of empty space.

• Gas Pressure ($P$):

  • Gas pressure results directly from collisions of gas particles with the walls of their container.

  • Definition: Average force exerted per unit area ($P = \frac{F}{A}$).

  • Atmospheric/Air pressure results from air particles colliding with objects from all directions.

• Temperature ($T$) and Absolute Zero:

  • Temperature in gas calculations must always be expressed in Kelvin ($\text{K}$).

  • Kelvin temperature is directly proportional to the average kinetic energy ($KE$) of the gas particles.

  • Absolute zero ($0\,\text{K}$) represents the absolute absence of particle motion ($KE = 0$).

• Two Factors That Cause an Increase in Pressure:

  • Increasing the total frequency of particle collisions.

  • Increasing the force of individual collisions.

• Variables Affecting Gas Pressure:

  • Amount of Gas ($n$): Measured in moles ($\text{mol}$). Increasing the number of particles increases collision frequency, directly increasing pressure ($\uparrow n \rightarrow \uparrow P$).

  • Volume ($V$): Measured in Liters ($\text{L}$) or cubic centimeters ($\text{cm}^3$). Decreasing the available volume forces particles into a smaller space, increasing collision frequency and pressure ($\downarrow V \rightarrow \uparrow P$).

  • Temperature ($T$): Measured in Kelvin ($\text{K}$). Increasing temperature increases particle kinetic energy and speed, increasing both collision frequency and force, directly increasing pressure ($\uparrow T \rightarrow \uparrow P$).

• Standard Pressure Equivalences:

  • $1\,\text{atm} = 101.3\,\text{kPa} = 760\,\text{mm\,Hg} = 14.7\,\text{psi} \text{ (lbs/in}^2\text{)}$

• Pressure Unit Conversion Worked Example:

  • Convert $120\,\text{mm\,Hg}$ to $\text{kPa}$:

  • $\text{Pressure} = 120\,\text{mm\,Hg} \times \left( \frac{101.3\,\text{kPa}}{760\,\text{mm\,Hg}} \right) = 16\,\text{kPa}$

• Ideal vs. Real Gases:

  • Ideal Gases: Follow all gas laws perfectly under all conditions of pressure and temperature. Assume zero particle volume and zero intermolecular attractions.

  • Real Gases: Behave most ideally under conditions of high temperature and low pressure.

  • Real gases deviate most from ideal behavior at low temperature and high pressure, because close particle proximity makes intermolecular attractive forces and particle volumes significant.

Gas Pressure Measurement: Barometers and Manometers

• Barometer:

  • Used to measure atmospheric pressure.

  • Consists of a glass tube filled with mercury submerged inverted in an open reservoir of mercury.

  • Mercury inside the tube exerts downward pressure due to its weight.

  • Atmospheric pressure pushes down on the open surface of the mercury reservoir, supporting the mercury column.

  • Under standard atmospheric conditions, air pressure supports a mercury column exactly $760\,\text{mm}$ high.

  • The space above the mercury column inside the sealed glass tube is a vacuum (contains no particles).

• Manometer Overview:

  • Instrument used to measure the pressure of an enclosed gas.

  • Consists of a bulb containing a gas attached to a U-shaped tube partially filled with mercury ($\text{Hg}$).

• Closed-Ended Manometer:

  • The arm opposite the gas bulb is sealed, containing a vacuum above the mercury.

  • The gas pressure balances directly against the height of the mercury column.

  • Formula: $P_{\text{gas}} = \Delta h$

• Open-Ended Manometer:

  • The arm opposite the gas bulb is open to atmospheric pressure ($P_{\text{air}}$).

  • Case 1: Gas pressure is greater than atmospheric pressure ($P_{\text{gas}} > P_{\text{air}}$):

  • Mercury is pushed toward the open end.

  • Formula: $P_{\text{gas}} = P_{\text{air}} + \Delta h$

  • Case 2: Gas pressure is less than atmospheric pressure ($P_{\text{gas}} < P_{\text{air}}$):

  • Mercury is pushed toward the gas bulb.

  • Formula: $P_{\text{gas}} = P_{\text{air}} - \Delta h$

• Universal 3-Step Method for Open Manometers:

  • Step 1: Determine which pressure is bigger ($P_{\text{gas}}$ or $P_{\text{air}}$).

  • Step 2: Convert all pressure values to the unit required for the unknown variable.

  • Step 3: Apply the formula: $\text{BIG} = \text{small} + \text{height}$ or $\text{height} = \text{BIG} - \text{small}$.

• Sample Manometer Worked Problems:

  • Problem 1: An open manometer has $P_{\text{air}} = 790\,\text{mm\,Hg}$ and the mercury level is $20\,\text{mm}$ higher on the gas bulb side ($P_{\text{air}} > P_{\text{gas}}$). Find $P_{\text{gas}}$.

  • $\text{Big} = P_{\text{air}}$, $\text{Small} = P_{\text{gas}}$

  • $P_{\text{air}} = P_{\text{gas}} + \Delta h$

  • $790\,\text{mm\,Hg} = P_{\text{gas}} + 20\,\text{mm\,Hg}$

  • $P_{\text{gas}} = 770\,\text{mm\,Hg}$

  • Problem 2: An open manometer has $P_{\text{air}} = 753\,\text{mm\,Hg}$ and the mercury level is $13\,\text{mm}$ higher on the open side ($P_{\text{gas}} > P_{\text{air}}$). Find $P_{\text{gas}}$.

  • $\text{Big} = P_{\text{gas}}$, $\text{Small} = P_{\text{air}}$

  • $P_{\text{gas}} = P_{\text{air}} + \Delta h$

  • $P_{\text{gas}} = 753\,\text{mm\,Hg} + 13\,\text{mm\,Hg} = 766\,\text{mm\,Hg}$

  • Problem 3: An open manometer has $P_{\text{air}} = 765\,\text{mm\,Hg}$ and $\Delta h = 27\,\text{mm}$ with the open arm mercury level higher ($P_{\text{gas}} > P_{\text{air}}$). Find $P_{\text{gas}}$.

  • $P_{\text{gas}} = P_{\text{air}} + \Delta h$

  • $P_{\text{gas}} = 765\,\text{mm\,Hg} + 27\,\text{mm\,Hg} = 792\,\text{mm\,Hg}$

  • Problem 4: Atmospheric pressure is $112.8\,\text{kPa}$ and enclosed gas pressure is $0.78\,\text{atm}$. Find the mercury height difference $\Delta h$ in $\text{mm\,Hg}$.

  • Step 1: Determine bigger pressure.

    • $112.8\,\text{kPa} \times \left( \frac{760\,\text{mm\,Hg}}{101.3\,\text{kPa}} \right) = 846\,\text{mm\,Hg}$ ($P_{\text{air}}$, BIG)

    • $0.78\,\text{atm} \times \left( \frac{760\,\text{mm\,Hg}}{1\,\text{atm}} \right) = 593\,\text{mm\,Hg}$ ($P_{\text{gas}}$, small)

  • Step 2: Use formula $\text{height} = \text{BIG} - \text{small}$.

    • $\Delta h = 846\,\text{mm\,Hg} - 593\,\text{mm\,Hg} = 253\,\text{mm\,Hg}$

Fundamental Gas Laws & The Combined Gas Law

• Boyle's Law (Pressure & Volume):

  • At constant temperature ($T$) and particle number ($n$), pressure and volume are inversely proportional.

  • Mathematical expression: $P \times V = k$

  • Two-state equation: $P_1 V_1 = P_2 V_2$

  • As volume decreases, particle collisions per unit area increase, increasing pressure ($\downarrow V \rightarrow \uparrow P$).

• Charles's Law (Volume & Temperature):

  • At constant pressure ($P$) and particle number ($n$), volume and temperature are directly proportional.

  • Mathematical expression: $\frac{V}{T} = k$

  • Two-state equation: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$

  • Extrapolation of $\text{Volume}$ vs. $\text{Temperature}$ curves for all gases intersects zero volume at $-273.15^\circ\text{C}$ or $0\,\text{K}$.

• Gay-Lussac's Law (Pressure & Temperature):

  • At constant volume ($V$) and particle number ($n$), pressure and temperature are directly proportional.

  • Mathematical expression: $\frac{P}{T} = k$

  • Two-state equation: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$

• Avogadro's Law (Volume & Amount):

  • At constant pressure ($P$) and temperature ($T$), volume is directly proportional to the number of moles of gas ($n$).

  • Mathematical expression: $\frac{V}{n} = k$

  • Two-state equation: $\frac{V_1}{n_1} = \frac{V_2}{n_2}$

• Amount of Gas & Pressure Relationship:

  • At constant volume ($V$) and temperature ($T$), pressure is directly proportional to the number of moles of gas ($n$).

  • Mathematical expression: $\frac{P}{n} = k$

• Combined Gas Law:

  • Combines Boyle's Law, Charles's Law, and Gay-Lussac's Law into a single relation.

  • Equation: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$ or $P_1 V_1 T_2 = P_2 V_2 T_1$

  • Extended form incorporating particle numbers ($n$): $\frac{P_1 V_1}{n_1 T_1} = \frac{P_2 V_2}{n_2 T_2}$

• Combined Gas Law Worked Examples:

  • Example 1: A gas occupies $12.3\,\text{L}$ at a pressure of $40.0\,\text{mm\,Hg}$. What is the volume when the pressure is increased to $60.00\,\text{mm\,Hg}$ at constant temperature and particle count?

  • Given: $P_1 = 40.0\,\text{mm\,Hg}$, $V_1 = 12.3\,\text{L}$, $P_2 = 60.00\,\text{mm\,Hg}$, $V_2 = ?$

  • Formula: $P_1 V_1 = P_2 V_2 \rightarrow V_2 = \frac{P_1 V_1}{P_2}$

  • Calculation: $V_2 = \frac{(40.0\,\text{mm\,Hg})(12.3\,\text{L})}{60.00\,\text{mm\,Hg}} = 8.20\,\text{L}$

  • Example 2: A gas occupies $473\,\text{cm}^3$ at $36^\circ\text{C}$. Find its volume at $94^\circ\text{C}$ under constant pressure.

  • Temperature conversions: $T_1 = 36^\circ\text{C} + 273 = 309\,\text{K}$, $T_2 = 94^\circ\text{C} + 273 = 367\,\text{K}$

  • Given: $V_1 = 473\,\text{cm}^3$, $V_2 = ?$

  • Formula: $\frac{V_1}{T_1} = \frac{V_2}{T_2} \rightarrow V_2 = \frac{V_1 T_2}{T_1}$

  • Calculation: $V_2 = \frac{(473\,\text{cm}^3)(367\,\text{K})}{309\,\text{K}} = 562\,\text{cm}^3$

  • Example 3: A gas occupies $7.84\,\text{mL}$ at $71.8\,\text{kPa}$ and $25^\circ\text{C}$. Find its volume at STP.

  • Initial state: $P_1 = 71.8\,\text{kPa}$, $V_1 = 7.84\,\text{mL}$, $T_1 = 25^\circ\text{C} + 273 = 298\,\text{K}$

  • STP state: $P_2 = 101.325\,\text{kPa}$, $T_2 = 273\,\text{K}$, $V_2 = ?$

  • Formula: $V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}$

  • Calculation: $V_2 = \frac{(71.8\,\text{kPa})(7.84\,\text{mL})(273\,\text{K})}{(101.325\,\text{kPa})(298\,\text{K})} = 5.09\,\text{mL}$

Ideal Gas Law & Standard Conditions

• Standard Temperature and Pressure (STP):

  • Standard Temperature: $273\,\text{K}$ or $0^\circ\text{C}$

  • Standard Pressure: $1\,\text{atm} = 101.3\,\text{kPa} = 760\,\text{mm\,Hg}$

• Avogadro's Hypothesis & Molar Volume:

  • At STP, $1\,\text{mole}$ of any ideal gas occupies a standard volume of $22.4\,\text{L}$.

  • $1\,\text{mole} = 22.4\,\text{L} = 6.02 \times 10^{23}\,\text{particles}$

• The Ideal Gas Law Equation:

  • Used to solve for one variable ($P, V, n,\text{ or } T$) under a single set of conditions.

  • Derived from $\frac{P V}{n T} = R$

  • Formula: $P V = n R T$

• Universal Gas Constant ($R$) Values:

  • The value of $R$ depends explicitly on the pressure units used (Volume must be in $\text{L}$, Temperature in $\text{K}$, Amount in $\text{mol}$):

  • $R = 0.08205\,\text{atm\,L\,mol}^{-1}\text{K}^{-1}$ (or $0.08206\,\text{atm\,L\,mol}^{-1}\text{K}^{-1}$) for pressure in $\text{atm}$

  • $R = 8.314\,\text{kPa\,L\,mol}^{-1}\text{K}^{-1}$ for pressure in $\text{kPa}$

  • $R = 62.36\,\text{mm\,Hg\,L\,mol}^{-1}\text{K}^{-1}$ for pressure in $\text{mm\,Hg}$

• Ideal Gas Law Worked Example:

  • Problem: What pressure ($\text{atm}$) is exerted by $0.622\,\text{mol}$ of gas contained in a $9.22\,\text{L}$ vessel at $16^\circ\text{C}$?

  • Given: $n = 0.622\,\text{mol}$, $V = 9.22\,\text{L}$, $T = 16^\circ\text{C} + 273 = 289\,\text{K}$, $R = 0.08206\,\text{atm\,L\,mol}^{-1}\text{K}^{-1}$

  • Formula: $P = \frac{n R T}{V}$

  • Calculation: $P = \frac{(0.622\,\text{mol})(0.08206\,\text{atm\,L\,mol}^{-1}\text{K}^{-1})(289\,\text{K})}{9.22\,\text{L}} = 1.60\,\text{atm}$

Gas Mixtures & Dalton's Law of Partial Pressures

• Law of Conservation of Mass in Gas Mixtures:

  • Matter cannot be created or destroyed.

  • The total moles of gas in a mixture equals the sum of the moles of individual gases:

  • $n_{\text{total}} = n_a + n_b$

  • Example: Combining $1\,\text{mole}$ of $\text{H}_2$ and $2\,\text{moles}$ of $\text{He}$ yields $3\,\text{moles}$ of total gas mixture.

  • Substituting the Ideal Gas Law ($n = \frac{P V}{R T}$) into mass conservation yields:

  • $\frac{P_a V_a}{R T_a} + \frac{P_b V_b}{R T_b} = \frac{P_t V_t}{R T_t}$

• Dalton's Law of Partial Pressures:

  • In a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of each individual gas.

  • Formula: $P_{\text{total}} = P_1 + P_2 + P_3 + \dots$

• Partial Pressures of Atmospheric Air Calculation:

  • Standard atmospheric barometric pressure ($P_B$) = $760\,\text{mm\,Hg}$

  • Partial pressure of Nitrogen ($\text{N}_2$): $0.7904 \times 760\,\text{mm\,Hg} = 600.7\,\text{mm\,Hg}$

  • Partial pressure of Oxygen ($\text{O}_2$): $0.2093 \times 760\,\text{mm\,Hg} = 159.1\,\text{mm\,Hg}$

  • Partial pressure of Carbon Dioxide ($\text{CO}_2$): $0.0003 \times 760\,\text{mm\,Hg} = 0.2\,\text{mm\,Hg}$

  • Total Atmospheric Pressure = $600.7 + 159.1 + 0.2 = 760.0\,\text{mm\,Hg}$

• Collection of Gas Over Water:

  • Gases collected in laboratory experiments by water displacement become saturated with water vapor.

  • The total measured pressure includes both the produced dry gas pressure and the vapor pressure of water at that specific temperature.

  • Formula: $P_{\text{total}} = P_{\text{atm}} = P_{\text{gas}} + P_{\text{water}}$

  • Re-arranged: $P_{\text{gas}} = P_{\text{total}} - P_{\text{water}}$

  • $P_{\text{water}}$ is obtained from standard vapor pressure of water tables based on temperature.

  • Worked Example:

  • Hydrogen gas is collected over water at $20^\circ\text{C}$. Total barometric pressure $P_{\text{total}} = 739.6\,\text{mm\,Hg}$. Water vapor pressure at $20^\circ\text{C} = 17.3\,\text{mm\,Hg}$.

  • Calculate dry gas pressure $P_{\text{gas}}$:

  • $P_{\text{gas}} = P_{\text{total}} - P_{\text{water}} = 739.6\,\text{mm\,Hg} - 17.3\,\text{mm\,Hg} = 722.3\,\text{mm\,Hg}$

Gas Movements: Diffusion, Effusion, and Molar Mass

• Diffusion vs. Effusion:

  • Diffusion: The movement of gas particles from an area of higher concentration to an area of lower concentration until uniformly distributed.

  • Effusion: The process by which gas particles pass through a tiny opening or pore into a vacuum or region of lower pressure.

• Graham's Law of Effusion:

  • The rate of effusion or diffusion of a gas is inversely proportional to the square root of its molar mass (or formula mass).

  • Equation: $\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}}$ or $\frac{(\text{Rate}_1)^2}{(\text{Rate}_2)^2} = \frac{M_2}{M_1}$

  • Key Concept: Lighter gas molecules move faster than heavier gas molecules at the same temperature.

• The Mole & Molar Mass Definitions:

  • $1\,\text{mole} = 6.02 \times 10^{23}\text{ particles}$ (Avogadro's number).

  • Atomic Mass: Mass of a single atom measured in atomic mass units ($\text{amu}$).

  • Molar Mass: Mass in grams of $1\,\text{mole}$ ($6.02 \times 10^{23}\text{ particles}$) of a substance ($\text{g\,mol}^{-1}$).

  • Conversion equivalences:

  • $12.01\,\text{amu} = 1\text{ C atom}$

  • $12.01\,\text{g C} = 1\,\text{mole C} = 6.02 \times 10^{23}\text{ atoms C}$

  • Compound Molar Mass Calculation Example: Magnesium Carbonate ($\text{MgCO}_3$):

  • $\text{Mg} = 24.3050\,\text{g\,mol}^{-1}$

  • $\text{C} = 12.0107\,\text{g\,mol}^{-1}$

  • $3 \times \text{O} = 3 \times 15.9994\,\text{g\,mol}^{-1} = 47.9982\,\text{g\,mol}^{-1}$

  • Total Molar Mass of $\text{MgCO}_3 = 24.3050 + 12.0107 + 47.9982 = 84.3\,\text{g\,mol}^{-1}$

• Graham's Law Worked Examples:

  • Example 1: Balloon A contains gas particles weighing $6.50\,\text{g}$; Balloon B contains gas particles weighing $17.5\,\text{g}$. Which effuses faster?

  • Balloon A effuses faster because its particles have lower mass.

  • Example 2: Compare effusion rates of Helium ($\text{He}$, Molar Mass = $4.0\,\text{g\,mol}^{-1}$) and Nitrogen ($\text{N}_2$, Molar Mass = $28.0\,\text{g\,mol}^{-1}$):

  • $\frac{\text{Rate}_{\text{He}}}{\text{Rate}_{\text{N}_2}} = \sqrt{\frac{28.0\,\text{g\,mol}^{-1}}{4.0\,\text{g\,mol}^{-1}}} = \sqrt{7.0} = 2.7$

  • $\text{He}$ effuses 2.7 times faster than $\text{N}_2$

  • Example 3: Compare effusion rates of Helium ($\text{He}$, $4.0\,\text{g\,mol}^{-1}$) and Methane ($\text{CH}_4$, Molar Mass = $12 + 4(1) = 16.0\,\text{g\,mol}^{-1}$):

  • $\frac{\text{Rate}_{\text{He}}}{\text{Rate}_{\text{CH}_4}} = \sqrt{\frac{16.0\,\text{g\,mol}^{-1}}{4.0\,\text{g\,mol}^{-1}}} = \sqrt{4.0} = 2.0$

  • $\text{He}$ effuses 2.0 times faster than $\text{CH}_4$

• Gas Diffusion Reaction Demonstration:

  • Reaction: $\text{NH}_{3(g)} + \text{HCl}_{(g)} \rightarrow \text{NH}_4\text{Cl}_{(s)}$

  • Ammonia ($\text{NH}_3$, Molar Mass = $17\,\text{g\,mol}^{-1}$) diffuses faster than Hydrochloric Acid gas ($\text{HCl}$, Molar Mass = $36.5\,\text{g\,mol}^{-1}$).

  • The white ring of solid ammonium chloride ($\text{NH}_4\text{Cl}$) forms closer to the $\text{HCl}$ end of the tube.

Vapor Pressure, Phase Changes, and Phase Diagrams

• Vapor Pressure Concepts:

  • Pressure exerted by vaporized particles above a liquid or solid state in a sealed container.

  • Dynamic Equilibrium: Occurs when the rate of evaporation equals the rate of condensation.

  • Increasing temperature increases particle kinetic energy, raising the evaporation rate and directly increasing vapor pressure.

• Boiling Point:

  • Boiling occurs when the vapor pressure of a liquid equals the surrounding atmospheric pressure.

  • At $25^\circ\text{C}$, water vapor pressure is far below atmospheric pressure; bubbles cannot form.

  • At $100^\circ\text{C}$, water vapor pressure equals $1\,\text{atm}$ ($101.3\,\text{kPa}$); bubbles form and rise freely.

• Phase Changes Definitions:

  • Melting: Solid $\rightarrow$ Liquid

  • Freezing: Liquid $\rightarrow$ Solid

  • Vaporization: Liquid $\rightarrow$ Gas

  • Condensation: Gas $\rightarrow$ Liquid

  • Sublimation: Conversion of a solid directly into a gas/vapor without passing through the liquid state.

  • Deposition: Conversion of a gas/vapor directly into a solid.

• Examples and Applications of Sublimation:

  • Examples: Dry Ice (solid $\text{CO}_2$ at $-78^\circ\text{C}$), Iodine, Camphor, Naphthalene mothballs, Snow, Icicles, frozen $\text{CO}_2$ in space, $\text{CO}_2$ in fire extinguishers, Perfume.

  • Applications:

  • Freeze-drying coffee: Coffee is frozen and water vapor is removed using a vacuum pump.

  • Cold shipping: Dry ice keeps food cold without melting into liquid that damages packaging.

  • Chemical purification: Organic chemists use sublimation to separate and purify mixtures.

• Phase Diagram Fundamentals:

  • Graph representing the relationships among solid, liquid, and gas phases of a substance as a function of temperature and pressure.

  • Regions: Indicate conditions where a single pure phase exists.

  • Curves/Lines: Represent equilibrium boundaries between two coexisting phases.

  • Triple Point: The specific single point of temperature and pressure where solid, liquid, and gas phases coexist in thermodynamic equilibrium.

  • Critical Point: End point of the liquid-gas equilibrium curve beyond which liquid and gas phases are indistinguishable.

• Phase Diagram of Water ($\text{H}_2\text{O}$):

  • Triple Point Values: Temperature = $0.016^\circ\text{C}$, Pressure = $0.61\,\text{kPa}$ ($4.58\,\text{torr}$).

  • All 3 states of matter (ice, liquid water, water vapor) coexist at the triple point.

  • Normal Melting Point: $0^\circ\text{C}$ at standard pressure ($101.3\,\text{kPa}$ / $1\,\text{atm}$).

  • Normal Boiling Point: $100^\circ\text{C}$ at standard pressure ($101.3\,\text{kPa}$ / $1\,\text{atm}$).

  • The solid-liquid equilibrium line has a negative slope (slopes left) because liquid water is denser than solid ice.

• Phase Diagram of Carbon Dioxide ($\text{CO}_2$):

  • Triple Point Values: Temperature = $-56.4^\circ\text{C}$, Pressure = $5.11\,\text{atm}$.

  • Sublimation Point at $1\,\text{atm}$ = $-78.5^\circ\text{C}$.

  • Liquid $\text{CO}_2$ cannot exist at standard atmospheric pressure ($1\,\text{atm}$) because the triple point pressure ($5.11\,\text{atm}$) is well above standard pressure.

  • Solid $\text{CO}_2$ (dry ice) sublimes directly to gas at room pressure.