Gases: Page-by-Page Notes

Chapter 5: Gases - Overview and Fundamentals

Description: Gases are characterized by macroscopic properties like pressure, volume, mass, and temperature, which are observable. The microscopic properties, like molecular speed and collision frequency, explain these macroscopic behaviors. Understanding the SI units for these properties is crucial for calculations.

1. Units and Pressure

Description: Pressure is caused by gas particles colliding with a surface. More frequent or forceful collisions mean higher pressure. It can be measured in atmospheres ($ext{atm}$), millimeters of mercury ($ext{mmHg}$ or $ext{Torr}$), Pascals ($ext{Pa}$), or Bar.

Practice Question 1: Unit Conversion
A gas sample has a volume of $2.5 \times 10^{3}\ \text{cm}^3$. What is this volume in liters ($ext{L}$) and cubic meters ($\text{m}^3$)?

Solution:

• To Liters: Since $1\ \text{L} = 1000\ \text{cm}^3$,
  $2.5 \times 10^3\ \text{cm}^3 \times \frac{1\ \text{L}}{1000\ \text{cm}^3} = 2.5\ \text{L}$

• To Cubic Meters: Since $1\ \text{m}^3 = 1000\ \text{L}$,
  $2.5\ \text{L} \times \frac{1\ \text{m}^3}{1000\ \text{L}} = 2.5 \times 10^{-3}\ \text{m}^3$

Practice Question 2: Pressure Unit Conversion
An aircraft tire operates at $13.8\ \text{bar}$. Convert this pressure to kilopascals ($\text{kPa}$) and atmospheres ($\text{atm}$).

Solution:

• To kPa: Use the conversion $1\ \text{bar} = 100\ \text{kPa}$.
  $13.8\ \text{bar} \times \frac{100\ \text{kPa}}{1\ \text{bar}} = 1380\ \text{kPa}$

• To atm: Use the conversion $1\ \text{bar} \approx 0.9869\ \text{atm}$.
  $13.8\ \text{bar} \times \frac{0.9869\ \text{atm}}{1\ \text{bar}} \approx 13.61\ \text{atm}$

2. Basic Gas Laws

Description: These laws describe the relationships between pressure ($P$), volume ($V$), temperature ($T$), and the amount of gas ($n$, in moles) when one or more variables are kept constant.

• Boyle's Law ($P<em>1V</em>1 = P<em>2V</em>2$): At constant moles and temperature, pressure and volume are inversely related. If one goes up, the other goes down.

• Charles's Law ($\frac{V<em>1}{T</em>1} = \frac{V<em>2}{T</em>2}$): At constant moles and pressure, volume and temperature (in Kelvin) are directly related. If one goes up, the other goes up. Remember to always use Kelvin for temperature! To convert $^{\circ}\text{C}$ to K, add 273.15.

• Avogadro's Law ($V \propto n$): At constant pressure and temperature, volume and moles are directly related.

Practice Question 3: Charles's Law Application
A balloon contains $2.0\ \text{L}$ of air at $27^{\circ}\text{C}$. If the temperature is increased to $227^{\circ}\text{C}$ at constant pressure, what is the new volume of the balloon?

Solution:

• Convert temperatures to Kelvin:
  $T<em>1 = 27^{\circ}\text{C} + 273.15 = 300.15\ \text{K}$ $T</em>2 = 227^{\circ}\text{C} + 273.15 = 500.15\ \text{K}$

• Apply Charles's Law:
  $\frac{V<em>1}{T</em>1} = \frac{V<em>2}{T</em>2} \implies V<em>2 = V</em>1 \times \frac{T<em>2}{T</em>1}$
  $V_2 = 2.0\ \text{L} \times \frac{500.15\ \text{K}}{300.15\ \text{K}} \approx 3.33\ \text{L}$

3. Ideal Gas Law

Description: This is the most comprehensive gas law, combining Boyle's, Charles's, and Avogadro's laws into one equation: $PV = nRT$. It's used to relate pressure, volume, moles, and temperature for an ideal gas. The gas constant $R$ changes depending on the units used for $P$ and $V$. Common values: $R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$, or $R = 0.0821\ \mathrm{L\,atm\,mol^{-1}\,K^{-1}}$, or $R = 0.0831\ \mathrm{L\,bar\,mol^{-1}\,K^{-1}}$.

Practice Question 4: Ideal Gas Law Calculation
What volume does $0.50\ \text{mol}$ of oxygen gas ($\text{O}_2$) occupy at $25^{\circ}\text{C}$ and $1.0\ \text{atm}$ pressure?

Solution:

• Identify knowns and unknowns:

  • $n = 0.50\ \text{mol}$

  • $T = 25^{\circ}\text{C} = 25 + 273.15 = 298.15\ \text{K}$

  • $P = 1.0\ \text{atm}$

  • $R = 0.0821\ \mathrm{L\,atm\,mol^{-1}\,K^{-1}}$ (chosen because P is in atm and we want V in L)

  • $V = ?$

• Rearrange Ideal Gas Law to solve for V:
  $V = \frac{nRT}{P}$

• Substitute values and calculate:
  V = \frac{(0.50\ \text{mol})(0.0821\ \mathrm{L\,atm\,mol^{-1}\,K^{-1}})(298.15\ \text{K})}{1.0\ \text{atm}} = 12.23\ \text{L}}

4. Gas Density and Molar Mass

Description: The Ideal Gas Law can be rearranged to find the density ($\rho = \frac{\text{mass}}{\text{volume}}$) or molar mass ($M$) of a gas, since $n = \frac{m}{M}$. The key formula for density is $\rho = \frac{PM}{RT}$. This means at constant temperature and pressure, a gas with a larger molar mass will be denser.

Practice Question 5: Gas Density Comparison
At the same temperature and pressure, which gas has the greatest density among $\text{SO}<em>2$ ($M = 64.13\ \text{g/mol}$), $\text{N}</em>2$ ($M = 28.01\ \text{g/mol}$), and $\text{HCl}$ ($M = 36.46\ \text{g/mol}$)?

Solution:

• From the formula $\rho = \frac{PM}{RT}$, at constant $P$ and $T$, density ($\rho$) is directly proportional to molar mass ($M$). Therefore, the gas with the highest molar mass will have the greatest density.

• Comparing molar masses: $\text{SO}<em>2$ ($64.13\ \text{g/mol}$), $\text{N}</em>2$ ($28.01\ \text{g/mol}$), $\text{HCl}$ ($36.46\ \text{g/mol}$).

• Answer: $\text{SO}_2$ has the greatest density because it has the largest molar mass.

5. Mixtures of Gases and Partial Pressures

Description: In a mixture of non-reacting gases, each gas exerts its own pressure independently. The total pressure of the mixture is the sum of the partial pressures of all individual gases.

• Dalton's Law of Partial Pressures: $P<em>{\text{total}} = P</em>a + P<em>b + P</em>c + \dots$ where $P<em>a, P</em>b, P_c$ are the partial pressures of each gas.

• Mole Fraction ($\chi<em>i$): The mole fraction of a gas ($\chi</em>i$) is its moles ($n<em>i$) divided by the total moles ($n</em>{\text{total}}$) in the mixture: $\chi<em>i = \frac{n</em>i}{n_{\text{total}}}$.

• Partial Pressure from Mole Fraction: The partial pressure of a gas can also be found by multiplying its mole fraction by the total pressure: $P<em>i = \chi</em>i P_{\text{total}}$.

• Collecting Gases over Water: When a gas is collected over water, the total pressure includes the partial pressure of the collected gas plus the partial pressure of water vapor ($P<em>{\text{H}</em>2\text{O}}$), which depends on temperature: $P<em>{\text{total}} = P</em>{\text{gas}} + P<em>{\text{H}</em>2\text{O}}$.

Practice Question 6: Dalton's Law Application
A container holds a mixture of nitrogen ($\text{N}<em>2$) and oxygen ($\text{O}</em>2$). The total pressure is $1.5\ \text{atm}$. If the partial pressure of nitrogen ($P<em>{\text{N}</em>2}$) is $0.9\ \text{atm}$, what is the partial pressure of oxygen ($P<em>{\text{O}</em>2}$)?

Solution:

• According to Dalton's Law: $P<em>{\text{total}} = P</em>{\text{N}<em>2} + P</em>{\text{O}_2}$

• $1.5\ \text{atm} = 0.9\ \text{atm} + P<em>{\text{O}</em>2}$

• $P<em>{\text{O}</em>2} = 1.5\ \text{atm} - 0.9\ \text{atm} = 0.6\ \text{atm}$

6. Gases in Chemical Reactions (Stoichiometry)

Description: The Ideal Gas Law ($PV = nRT$) is a powerful tool to relate the volume of a gas to the number of moles. This connection allows us to perform stoichiometry problems involving gaseous reactants or products, just like we would with masses and molar masses.

Practice Question 7: Stoichiometry with Gases
Methanol ($\text{CH}<em>3\text{OH}$) can be synthesized from carbon monoxide and hydrogen gas by the reaction: $\text{CO(g)} + 2\ \text{H}</em>2\text{(g)} \to \text{CH}<em>3\text{OH(g)}$. If $35.7\ \text{g}$ of methanol are produced at $P = 984\ \text{mbar}$ and $T = 355\ \text{K}$, what volume of hydrogen gas was consumed? (Molar mass of $\text{CH}</em>3\text{OH} = 32.04\ \text{g/mol}$)

Solution:

1. Convert methanol mass to moles:
   $n<em>{\text{CH}</em>3\text{OH}} = \frac{35.7\ \text{g}}{32.04\ \text{g/mol}} = 1.114\ \text{mol}$

2. Use stoichiometry to find moles of hydrogen:
   From the balanced equation, $2\ \text{mol}\ \text{H}<em>2$ are needed for $1\ \text{mol}\ \text{CH}</em>3\text{OH}$.
   $n<em>{\text{H}</em>2} = 1.114\ \text{mol}\ \text{CH}<em>3\text{OH} \times \frac{2\ \text{mol}\ \text{H}</em>2}{1\ \text{mol}\ \text{CH}<em>3\text{OH}} = 2.228\ \text{mol}\ \text{H}</em>2$

3. Convert pressure to bar (for R constant compatibility):
   $P = 984\ \text{mbar} = 0.984\ \text{bar}$

4. Use Ideal Gas Law ($PV = nRT$) to find volume of hydrogen:
   Select $R = 0.0831\ \mathrm{L\,bar\,mol^{-1}\,K^{-1}}$
   $V = \frac{n<em>{\text{H}</em>2}RT}{P} = \frac{(2.228\ \text{mol})(0.0831\ \mathrm{L\,bar\,mol^{-1}\,K^{-1}})(355\ \text{K})}{0.984\ \text{bar}}$
   $V \approx 66.8\ \text{L}$

7. Kinetic Molecular Theory (KMT)

Description: KMT explains why gases behave the way they do based on the motion and properties of individual gas particles. It provides a theoretical foundation for the ideal gas law.

• Postulates:

  1. Gas particles are extremely small and far apart (mostly empty space).

  2. The average kinetic energy ($\langle KE \rangle$) of particles is directly proportional to the absolute temperature ($T$) in Kelvin ($\langle KE \rangle = \frac{3}{2} RT$ per mole).

  3. Collisions between particles and with container walls are perfectly elastic (no energy loss).

• Molecular Velocities:

  • Particles move at various speeds. The root mean square speed ($u<em>{\mathrm{rms}}$) is a way to describe the average speed and is given by $u</em>{\mathrm{rms}} = \sqrt{\frac{3RT}{M}}$, where $M$ is the molar mass in $\text{kg/mol}$.

  • Lighter gases move faster at the same temperature.

  • Higher temperatures lead to faster average speeds and a broader distribution of speeds.

• Mean Free Path, Diffusion, Effusion:

  • Mean free path: The average distance a particle travels between collisions.

  • Diffusion: The process where gas spreads out to fill a volume, moving from high concentration to low concentration.

  • Effusion: The process where gas particles escape through a tiny hole into a vacuum.

• Graham's Law of Effusion: Lighter gases effuse faster. This law quantifies the ratio of effusion rates for two gases ($A$ and $B$):
  $\frac{\text{rate}<em>A}{\text{rate}</em>B} = \sqrt{\frac{M<em>B}{M</em>A}}$ (where $M$ is molar mass).

Practice Question 8: Root Mean Square Speed
Calculate the root mean square speed ($u<em>{\mathrm{rms}}$) for a hydrogen molecule ($\text{H}</em>2$) at $25^{\circ}\text{C}$. (Molar mass of $\text{H}_2 \approx 2.016\ \text{g/mol}$)

Solution:

1. Convert temperature to Kelvin:
   $T = 25^{\circ}\text{C} + 273.15 = 298.15\ \text{K}$

2. Convert molar mass to $\text{kg/mol}$:
   $M = 2.016\ \text{g/mol} = 0.002016\ \text{kg/mol}$

3. Use the gas constant $R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$ (because J = $\text{kg}\cdot\text{m}^2\text{s}^{-2}$, which works with kg and m/s).

4. Apply the $u<em>{\mathrm{rms}}$ formula:
   $u</em>{\mathrm{rms}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3(8.314\ \mathrm{J\,mol^{-1}\,K^{-1}})(298.15\ \text{K})}{0.002016\ \text{kg/mol}}}$
   $u_{\mathrm{rms}} \approx \sqrt{3.69 \times 10^6\ \text{m}^2/\text{s}^2} \approx 1921\ \text{m/s}$

Practice Question 9: Graham's Law of Effusion
If neon gas ($\text{Ne}$, $M = 20.18\ \text{g/mol}$) effuses at a rate of $1.5\ \text{mol/min}$, what is the effusion rate of argon gas ($\text{Ar}$, $M = 39.95\ \text{g/mol}$) under the same conditions?

Solution:

1. Apply Graham's Law:
   $\frac{\text{rate}<em>{\text{Ne}}}{\text{rate}</em>{\text{Ar}}} = \sqrt{\frac{M<em>{\text{Ar}}}{M</em>{\text{Ne}}}}$

2. Substitute known values:
   $\frac{1.5\ \text{mol/min}}{\text{rate}_{\text{Ar}}} = \sqrt{\frac{39.95\ \text{g/mol}}{20.18\ \text{g/mol}}} = \sqrt{1.9796} \approx 1.407$

3. Solve for $\text{rate}<em>{\text{Ar}}$:
   $\text{rate}</em>{\text{Ar}} = \frac{1.5\ \text{mol/min}}{1.407} \approx 1.07\ \text{mol/min}$

8. Real Gases

Description: The Ideal Gas Law works well under "normal" conditions (moderate temperatures and pressures). However, real gases deviate from ideal behavior, especially at high pressures or low temperatures. This is because two assumptions of KMT break down:

1. Finite volume of gas particles: Ideal gases assume particles have no volume. In reality, they do, and at high pressures, this occupied volume becomes significant, reducing the "empty" space available.

2. Intermolecular forces: Ideal gases assume no attractive or repulsive forces between particles. At low temperatures or high densities, these forces become significant, affecting how particles collide and exert pressure.

• van der Waals Equation: This equation empirically corrects for both effects: $\left(P + \frac{a n^2}{V^2}\right)(V - nb) = nRT$

  • The term $+\frac{a n^2}{V^2}$ corrects for intermolecular attractions (they reduce observed pressure).

  • The term $(V - nb)$ corrects for the finite volume of particles (they reduce the available volume).

  • $a$ and $b$ are constants specific to each gas.

• Deviations ($PV/RT$):

  • We use the ratio $\frac{PV}{RT}$ (also called the compressibility factor) to diagnose real gas behavior.

  • If $\frac{PV}{RT} = 1$, the gas behaves ideally.

  • If \frac{PV}{RT} > 1 (positive deviation), the finite particle volume effect is dominant (gas takes up more space than ideal).

  • If \frac{PV}{RT} < 1 (negative deviation), intermolecular attractions are dominant (gas takes up less space than ideal due to particles pulling each other closer).

Practice Question 10: Real Gas Behavior
Which of the following conditions would cause a real gas to deviate most negatively from ideal behavior (i.e., \frac{PV}{RT} < 1)?
(a) High temperature and high pressure
(b) Low temperature and low pressure
(c) High temperature and low pressure
(d) Low temperature and high pressure

Solution:

• Negative deviation (\frac{PV}{RT} < 1) occurs when intermolecular attractive forces are dominant.

• Intermolecular forces become most significant when particles are close together and moving slowly.

• Low temperature means particles move slowly, allowing attractive forces to have more time to act.

• High pressure means particles are forced closer