AP Chemistry Chapter 5 Practice Test: Gas Laws and Kinetic Molecular Theory Notes

Gas Density and Molar Mass Calculations

• The Ideal Gas Law relates pressure ($P$), volume ($V$), moles ($n$), the ideal gas constant ($R$), and absolute temperature ($T$):

    $P \cdot V = n \cdot R \cdot T$

• Rearranging the ideal gas equation to solve for the number of moles ($n$):

    $n = \frac{P \cdot V}{R \cdot T}$

• For a $2.0\,L$ container at standard temperature and pressure ($0^{\circ}C = 273.15\,K$ and $1.0\,atm$):

    $n = \frac{(1.0\,atm)(2.0\,L)}{(0.08206\,L\,atm\,mol^{-1}\,K^{-1})(273.15\,K)} = 0.0892\,mol$

• The molar mass ($M$) of a gas sample weighing $4.0\,g$ is calculated as:

    $M = \frac{\text{mass}}{n} = \frac{4.0\,g}{0.0892\,mol} \approx 44.8\,g\,mol^{-1}$

• Comparing molar masses of common gases:

  • $\text{SO}_2$: $32.06 + 2(16.00) = 64.06\,g\,mol^{-1}$

  • $\text{N}_2$: $2(14.01) = 28.02\,g\,mol^{-1}$

  • $\text{CO}_2$: $12.01 + 2(16.00) = 44.01\,g\,mol^{-1}$

  • $\text{NH}_3$: $14.01 + 3(1.008) = 17.03\,g\,mol^{-1}$

• A $2.0\,L$ container holding $4.0\,g$ of gas under standard conditions corresponds to $\text{CO}_2$.

Kinetic Molecular Theory and Temperature Effects

• Five identical rigid $2.0\,L$ vessels are filled with $\text{N}_2(g)$ at $27^{\circ}C$ ($300.15\,K$) to an initial pressure of $720\,mm\,Hg$. Four contain $0.050\,mol$ of solid samples ($\text{NaHCO}_3(s)$, $\text{NaBr}(s)$, $\text{Cu}(s)$, or $\text{I}_2(s)$).

• According to Kinetic Molecular Theory (KMT), the average kinetic energy ($KE_{\text{avg}}$) of gas particles is directly proportional to the absolute temperature ($T$):

    $KE_{\text{avg}} = \frac{3}{2} R \cdot T = \frac{1}{2} m \cdot v_{\text{rms}}^2$

• Heating a gas sample from $27^{\circ}C$ ($300\,K$) to $127^{\circ}C$ ($400\,K$) increases the root-mean-square speed ($v_{\text{rms}}$) of the particles:

    $v_{\text{rms}} = \sqrt{\frac{3 R \cdot T}{M}}$

• Graphical particle representations of gas mixtures at elevated temperatures reflect this increase in average kinetic energy by depicting longer velocity vectors (arrows) for the gas particles.

Conservation of Mass in Sealed Systems

• When a sealed vessel containing $0.050\,mol$ of $\text{I}_2(s)$ and $\text{N}_2(g)$ is heated to $127^{\circ}C$, the iodine completely vaporizes into $\text{I}_2(g)$.

• The total mass of the sealed system remains strictly constant throughout the phase change.

• The law of conservation of mass dictates that the number and identity of atoms ($\text{N}$ and $\text{I}$) inside a closed container cannot change, regardless of transitions between solid and gas phases or changes in temperature, pressure, and average kinetic energy.

Stoichiometric Gas Generation and Pressure

• When heated to $127^{\circ}C$, different solid additives affect system pressure differently depending on their thermal stability:

  • $\text{NaBr}(s)$ remains thermally stable as a solid and contributes zero additional moles of gas.

  • $\text{NaHCO}_3(s)$ undergoes thermal decomposition according to the chemical reaction:

        $2\,\text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{CO}_2(g) + \text{H}_2\text{O}(g)$

• The thermal decomposition yields two moles of gaseous products ($1\,mol\,\text{CO}_2$ and $1\,mol\,\text{H}_2\text{O}$) for every two moles of $\text{NaHCO}_3(s)$ reacted.

• The generation of additional gas molecules inside a rigid vessel increases total gas moles ($n_{\text{total}}$), resulting in a higher total pressure:

    $P_{\text{total}} = \frac{n_{\text{total}} R \cdot T}{V}$

Deviations from Ideal Gas Behavior

• The Ideal Gas Law assumes:

  • Gas particles occupy zero volume (point masses).

  • No attractive or repulsive intermolecular forces exist between gas particles.

• When actual gas volume exceeds ideal predicted volume, the deviation occurs because gas particles possess finite real volume (molecular volume).

• At high particle densities, the physical volume occupied by molecules ($n \cdot b$) reduces the free space available for movement, making actual available volume larger than ideal predictions:

    $V_{\text{real}} = V_{\text{container}} - n \cdot b$

• Gases behave most ideally under conditions of high temperature and low pressure:

  • High Temperature: Provides high thermal kinetic energy, overcoming weak intermolecular attractive forces ($a$).

  • Low Pressure: Ensures low particle density and large intermolecular distances, rendering molecular volume ($b$) negligible.

• For example, $1.0\,mol$ of $\text{CO}_2(g)$ exhibits maximum ideal behavior at $800\,K$ and $0.10\,atm$, compared to lower temperatures ($50\,K$ or $100\,K$) or higher pressures ($5.0\,atm$ or $100\,atm$).

Maxwell-Boltzmann Speed Distribution

• The distribution of molecular speeds for a gas depends on molar mass ($M$) and temperature ($T$), defined by the Maxwell-Boltzmann distribution:

    $f(v) = 4\pi \left(\frac{M}{2\pi R \cdot T}\right)^{3/2} v^2 \exp\left(-\frac{M \cdot v^2}{2 R \cdot T}\right)$

• Comparing $\text{H}_2(g)$ ($M = 2.016\,g\,mol^{-1}$) and $\text{He}(g)$ ($M = 4.003\,g\,mol^{-1}$) at identical temperature ($273\,K$) and pressure ($1.0\,atm$):

  • Because $\text{He}$ has a larger molar mass than $\text{H}_2$, its most probable speed ($v_{\text{mp}} = \sqrt{\frac{2 R \cdot T}{M}}$) and average speed are lower.

  • The Maxwell-Boltzmann distribution curve for $\text{He}$ shifts to the left (toward lower speeds) and exhibits a taller, narrower peak than $\text{H}_2$ to maintain a constant total area of $1.0$ under the curve.

Ideal Gas Law Calculations and Variable Relationships

• During an isothermal expansion of gas (e.g., air expanding from $2.0\,L$ at $P_1$ to $5.0\,L$ at $P_2$ at constant temperature):

  • Because temperature is constant ($T_1 = T_2$), the average kinetic energy ($KE_{\text{avg}} = \frac{3}{2} R \cdot T$) of the molecules remains completely unchanged.

  • By Boyle's Law ($P_1 \cdot V_1 = P_2 \cdot V_2$), final pressure $P_2$ decreases proportionally:

        $P_2 = P_1 \left(\frac{V_1}{V_2}\right) = P_1 \left(\frac{2.0\,L}{5.0\,L}\right) = 0.40 \cdot P_1$

• To calculate the pressure ($P$) exerted by $1.85\,mol$ of ideal gas in a $3.00\,L$ vessel at $35.0^{\circ}C$ ($308.15\,K$):

    $P = \frac{n \cdot R \cdot T}{V} = \frac{(1.85)(0.0821)(308)}{3.00}$

Gas Compression and Kinetic Molecular Theory

• Gases are compressible because individual gas molecules are separated by large distances of empty space relative to molecular dimensions.

• Compressing a sample of butane ($\text{C}_4\text{H}_{10}$) at constant temperature from $0.50\,L$ down to $0.20\,L$ increases gas pressure:

  • Reducing volume increases particle number density per unit space.

  • Gas molecules collide with container walls with greater frequency.

  • Because temperature remains constant, the force of individual collisions stays the same, but higher collision frequency per unit area yields higher overall pressure ($P = \frac{F_{\text{total}}}{A}$).

Reaction Stoichiometric Gas Pressure Changes

• Consider the synthesis reaction in a rigid container at constant temperature:

    $2\,\text{NO}(g) + \text{O}_2(g) \rightarrow 2\,\text{NO}_2(g)$

• Molar balance analysis:

  • Reactants: $2\,mol\,\text{NO} + 1\,mol\,\text{O}_2 = 3\,mol$ of gas.

  • Products: $2\,mol\,\text{NO}_2 = 2\,mol$ of gas.

• As the reaction goes to completion, total moles of gas decrease from $3$ to $2$ ($\Delta n_{\text{gas}} = -1\,mol$).

• Because pressure in a rigid container at constant temperature is directly proportional to moles of gas ($P \propto n$), total system pressure decreases.

Laboratory Gas Collection and Wet vs. Dry Gas Calculations

• Reaction of magnesium with hydrochloric acid:

    $\text{Mg}(s) + 2\,\text{H}^+(aq) \rightarrow \text{Mg}^{2+}(aq) + \text{H}_2(g)$

• Stoichiometric analysis for a $0.0360\,g$ sample of pure $\text{Mg}$ reacting with excess $2.0\,\text{M}\,\text{HCl}(aq)$:

  • Moles of $\text{Mg}$:

        $n_{\text{Mg}} = \frac{0.0360\,g}{24.305\,g\,mol^{-1}} = 1.48 \times 10^{-3}\,mol$

*   Moles of $\text{HCl}(aq)$ required for complete reaction:

        $n_{\text{HCl}} = 2 \times n_{\text{Mg}} = 2 \times (1.481 \times 10^{-3}\,mol) = 2.96 \times 10^{-3}\,mol$

• When $\text{H}_2(g)$ is collected over water at $23.0^{\circ}C$ ($296.15\,K$) under a total measured pressure of $749\,torr$ with equilibrium water vapor pressure $P_{\text{H}_2\text{O}} = 21\,torr$:

  • Applying Dalton's Law of Partial Pressures:

        $P_{\text{total}} = P_{\text{H}_2} + P_{\text{H}_2\text{O}}$

        $P_{\text{H}_2} = P_{\text{total}} - P_{\text{H}_2\text{O}} = 749\,torr - 21\,torr = 728\,torr$

*   Converting partial pressure to atmospheres:

        $P_{\text{H}_2} = \frac{728\,torr}{760\,torr\,atm^{-1}} = 0.9579\,atm$

*   Calculating the volume ($V$) of dry $\text{H}_2(g)$ produced:

        $V = \frac{n \cdot R \cdot T}{P_{\text{H}_2}} = \frac{(1.481 \times 10^{-3}\,mol)(0.08206\,L\,atm\,mol^{-1}\,K^{-1})(296.15\,K)}{0.9579\,atm} = 0.0376\,L$

Solution Preparation and Dilution Safety

• To prepare $50.0\,mL$ ($0.0500\,L$) of $2.0\,\text{M}\,\text{HCl}(aq)$ using a stock solution of $12.3\,\text{M}\,\text{HCl}$:

• Dilution calculation:

    $M_1 \cdot V_1 = M_2 \cdot V_2$

    $(12.3\,\text{M}) \cdot V_1 = (2.0\,\text{M}) \cdot (50.0\,mL)$

    $V_1 = \frac{(2.0\,\text{M})(50.0\,mL)}{12.3\,\text{M}} = 8.13\,mL$

• Laboratory procedure and safety steps:

  1. Measure approximately $20\text{--}30\,mL$ of distilled water into a $50.00\,mL$ volumetric flask.

  2. Using a $10.0\,mL$ graduated cylinder and a dropper, measure $8.13\,mL$ of the concentrated $12.3\,\text{M}\,\text{HCl}$ solution.

  3. Slowly pour the concentrated acid into the volumetric flask containing water. Always add acid to water to prevent extreme heat generation and violent acid splattering caused by the exothermic dissolution process.

  4. Add distilled water to the flask until the volume approaches the calibration mark, then use a dropper to add final drops until the bottom of the meniscus rests on the $50.00\,mL$ mark.

  5. Stopper the volumetric flask and invert it repeatedly to ensure uniform concentration.

Gas Behavior, Phase Changes, and Condensation

• For a sample of $\text{O}_2(g)$ in a cylinder with a movable piston at initial conditions ($V = 2.0\,L$, $T = 298\,K$, $P = 1.0\,atm$):

• Gas density ($d$) under initial conditions:

    $d = \frac{P \cdot M}{R \cdot T} = \frac{(1.0\,atm)(32.00\,g\,mol^{-1})}{(0.08206\,L\,atm\,mol^{-1}\,K^{-1})(298\,K)} = 1.31\,g\,L^{-1}$

• Releasing gas through a side valve while keeping temperature and pressure constant:

  • Density remains defined as $d = \frac{P \cdot M}{R \cdot T}$.

  • Because temperature ($T$), pressure ($P$), and molar mass ($M$) do not change, the density of the remaining gas stays constant ($1.31\,g\,L^{-1}$) regardless of the mass of gas vented.

• Cooling the cylinder to $-55^{\circ}C$ ($218.15\,K$):

  • Kinetic Molecular Theory Explanation: Lowering temperature reduces average kinetic energy and velocity of $\text{O}_2$ particles.

  • Particles strike the piston walls less frequently and with smaller impact force.

  • To maintain equilibrium against constant external atmospheric pressure, the movable piston moves downward, reducing volume until internal pressure matches external pressure.

• Cooling the cylinder further to $-183^{\circ}C$ ($90.15\,K$):

  • The boiling point of $\text{O}_2$ is $-183^{\circ}C$ ($90.18\,K$).

  • At $-183^{\circ}C$, $\text{O}_2$ condenses from gas to liquid.

  • Liquids occupy dramatically smaller molar volumes than gases, causing the measured volume to drop significantly below the volume predicted by the Ideal Gas Law.