chapter 9 lec 1

Gases: Properties and Pressure

Characteristics and Composition of Gases

Gases are one of the fundamental states of matter, notable for their ability to fill any container and mix uniformly. Only a few elements exist naturally as gases at standard conditions, but many molecular compounds are also gaseous.

• Common elemental gases: Hydrogen (H2), Nitrogen (N2), Oxygen (O2), Fluorine (F2), Chlorine (Cl2), and noble gases (He, Ne, Ar, Kr, Xe, Rn).

• Common molecular gases: HCl, HCN, NH3, CO2, CO, N2O, CH4.

• Atmospheric composition: ~78% N2, ~21% O2, ~1% other gases (CO2, Ar, etc.).

• Toxic gases: HCN, CO, H2S, NO2, O3, SO2, SO3 (dangerous at high concentrations).

Physical Properties of Gases

• Shape and Volume: Gases assume both the shape and volume of their container.

• Compressibility: Gases are highly compressible due to large intermolecular distances.

• Density: Gases have much lower and more variable densities than liquids or solids.

• Mixing: Gases mix evenly and completely in any proportion, forming homogeneous mixtures.

Gas Pressure

Gas pressure results from the constant motion of gas molecules colliding with surfaces. Atmospheric pressure is the force exerted by the weight of air above a surface.

• Definition: Pressure is force per unit area.

$\text{Pressure} = \frac{\text{Force}}{\text{Area}}$

• SI unit: Pascal (Pa), where $1\ \text{Pa} = 1\ \text{N}/\text{m}^2$ and $1\ \text{N} = 1\ \text{kg}\cdot\text{m}/\text{s}^2$.

• Other units: mm Hg (torr), atm, bar, psi.

Unit	Equivalent
1 atm	760 mm Hg = 760 torr = 101,325 Pa = 1.01325 × 105 Pa = 1.01325 × 102 kPa = 29.92 in Hg = 14.7 psi
1 torr	133.322 Pa
1 bar	1 × 105 Pa

Measuring Pressure: Barometers and Manometers

• Barometer: Measures atmospheric pressure using a column of mercury (Hg). Standard atmospheric pressure supports a 760 mm column of Hg at 0°C at sea level.

• Manometer: Measures gas pressure in a closed or open system. The height of the liquid column is inversely proportional to the liquid's density.

Example: Calculating pressure from a column of Hg 130.0 cm high (density = 13.595 g/cm3):

$P = \text{height} \times \text{density} \times g$

$P = 1.30\ \text{m} \times 13.595\ \text{g/cm}^3 \times (100\ \text{cm}/1\ \text{m})^3 \times (1\ \text{kg}/1000\ \text{g}) \times 9.80665\ \text{m/s}^2$

$P = 1.7332 \times 10^5\ \text{Pa} = 1.711\ \text{atm}$

Atmospheric Pressure and Altitude

• Atmospheric pressure decreases with altitude due to the thinning of the air.

• 50% of the atmosphere is within 6.4 km (4.0 mi) of Earth's surface; 90% within 16 km (10 mi); 99% within 32 km (20 mi).

Pressure Conversions

Example: Convert 23.8 mm Hg to kPa:

$23.8\ \text{mm Hg} \times \frac{1\ \text{atm}}{760\ \text{mm Hg}} \times \frac{1.01325 \times 10^2\ \text{kPa}}{1\ \text{atm}} = 3.17\ \text{kPa}$

Gas Laws: Relationships Between Pressure, Volume, Temperature, and Amount

Overview of Gas Variables

Four variables describe the state of a gas: pressure (P), volume (V), temperature (T), and amount in moles (n). Knowing any three allows calculation of the fourth.

Boyle’s Law: Pressure-Volume Relationship

At constant temperature, the volume of a fixed amount of gas is inversely proportional to its pressure.

• Mathematical form: $P \propto \frac{1}{V}$ or $PV = k$ (k = constant at constant T and n)

• Two-state form: $P_1V_1 = P_2V_2$

Example: A gas at 4.52 atm and 7.25 L is reduced to 1.21 atm. What is the new volume?

$4.52\ \text{atm} \times 7.25\ \text{L} = 1.21\ \text{atm} \times V_2$

$V_2 = \frac{4.52 \times 7.25}{1.21} = 27.1\ \text{L}$

Charles’s Law: Volume-Temperature Relationship

At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature (in Kelvin).

• Mathematical form: $V \propto T$ or $\frac{V}{T} = k$

• Two-state form: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$ (T in K)

Example: A gas has a volume of 2.80 L at unknown T. At 0°C (273.15 K), its volume is 2.57 L. What was the initial T?

$\frac{2.80}{T_1} = \frac{2.57}{273.15}$ $T_1 = \frac{2.80}{2.57} \times 273.15 = 297.6\ \text{K} = 24.4^\circ\text{C}$

Gay-Lussac’s Law: Pressure-Temperature Relationship

At constant volume, the pressure of a fixed amount of gas is directly proportional to its absolute temperature.

• Mathematical form: $P \propto T$ or $\frac{P}{T} = k$

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

Avogadro’s Law: Volume-Amount Relationship

At constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of gas present.

• Mathematical form: $V \propto n$ or $\frac{V}{n} = k$

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

Example: 4.65 L of He contains 0.225 mol. To reach 6.48 L, how many more moles are needed?

$n_2 = \frac{6.48}{4.65} \times 0.225 = 0.314\ \text{mol}$ $0.314 - 0.225 = 0.089\ \text{mol}$ to be added

Combining Gas Laws: The Ideal Gas Law

By combining Boyle’s, Charles’s, and Avogadro’s laws, we obtain the ideal gas law, which describes the behavior of an ideal gas:

$PV = nRT$

• P: Pressure (atm)

• V: Volume (L)

• n: Moles of gas

• R: Gas constant ($0.082057\ \text{L-atm}/\text{mol-K}$)

• T: Temperature (K)

Standard Temperature and Pressure (STP): 0°C (273.15 K) and 1 atm. At STP, 1 mol of any ideal gas occupies 22.414 L.

$R = \frac{(1\ \text{atm})(22.414\ \text{L})}{(1\ \text{mol})(273.15\ \text{K})} = 0.082057\ \text{L-atm}/\text{mol-K}$

Applications of the Ideal Gas Law

• Finding Volume: $V = \frac{nRT}{P}$

• Finding Pressure: $P = \frac{nRT}{V}$

• Finding Moles: $n = \frac{PV}{RT}$

Example: What volume does 0.845 mol N2 occupy at 1.37 atm and 315 K?

$V = \frac{0.845 \times 0.0821 \times 315}{1.37} = 15.9\ \text{L}$

Example: An 8.50 L tire contains 0.552 mol gas at 32°C (305 K). What is the pressure?

$P = \frac{0.552 \times 0.0821 \times 305}{8.50} = 1.63\ \text{atm}$ $1.63\ \text{atm} \times 14.7\ \text{psi}/\text{atm} = 24.0\ \text{psi}$

Example: A 3.24 L basketball at 24.3 psi and 25°C. How many moles of gas?

$n = \frac{(24.3\ \text{psi})(1\ \text{atm}/14.7\ \text{psi})(3.24\ \text{L})}{0.0821 \times 298} = 0.219\ \text{mol}$

Combined Gas Law

When a gas sample undergoes changes in P, V, and T, but n is constant:

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

Example: A gas bubble at depth (P = 7.6 atm, T = 1°C, V = 3.45 mL) rises to the surface (P = 1.0 atm, T = 25°C). What is the new volume?

$V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{7.6 \times 3.45 \times 298}{274 \times 1.0} = 28.5\ \text{mL}$

Density and Molar Mass of Gases

Density from the Ideal Gas Law

The ideal gas law can be rearranged to solve for the density (d) or molar mass (MW) of a gas:

$n = \frac{m}{MW}$ $PV = nRT = \frac{mRT}{MW}$ $\Rightarrow d = \frac{m}{V} = \frac{P \times MW}{RT}$ $\Rightarrow MW = \frac{dRT}{P}$

Example: Find the density of N2 at 125°C and 755 mm Hg.

$d = \frac{(755/760) \times 28.01}{0.0821 \times 398} = 0.852\ \text{g/L}$

Example: A gas has a density of 1.43 g/L at 23°C and 0.789 atm. Find its molar mass.

$MW = \frac{1.43 \times 0.0821 \times 296}{0.789} = 44.0\ \text{g/mol}$

Example: A gas sample: mass = 0.311 g, V = 0.225 L, T = 55°C (328 K), P = 886 mm Hg. Find MW.

$n = \frac{(886/760) \times 0.225}{0.0821 \times 328} = 0.00974\ \text{mol}$ $MW = \frac{0.311}{0.00974} = 31.9\ \text{g/mol}$

Stoichiometry Involving Gases

Volume of Gaseous Reactants and Products

Gas laws can be used to relate the volumes of gases involved in chemical reactions, especially at constant T and P.

Example: Synthesis of methanol:

$\text{CO}(g) + 2\text{H}_2(g) \rightarrow \text{CH}_3\text{OH}(g)$

• To synthesize 35.7 g CH3OH at 355 K and 738 mm Hg, how much H2 (L) is needed?

$35.7\ \text{g} \times \frac{1\ \text{mol}}{32.04\ \text{g}} = 1.114\ \text{mol CH}_3\text{OH}$ $1.114\ \text{mol} \times \frac{2\ \text{mol H}_2}{1\ \text{mol CH}_3\text{OH}} = 2.228\ \text{mol H}_2$ $V = \frac{nRT}{P} = \frac{2.228 \times 0.0821 \times 355}{738/760} = 66.9\ \text{L}$

At STP: $2.228\ \text{mol} \times 22.4\ \text{L/mol} = 49.9\ \text{L}$

Example: Combustion of glucose:

$\text{C}_6\text{H}_{12}\text{O}_6(aq) + 6\text{O}_2(g) \rightarrow 6\text{CO}_2(g) + \text{H}_2\text{O}(l)$

• 10.0 g glucose consumed at 37°C and 1.00 atm. Find volume of CO2 produced.

$10.0\ \text{g} \times \frac{1\ \text{mol}}{180.162\ \text{g}} = 0.0555\ \text{mol}$ $0.0555\ \text{mol} \times 6 = 0.333\ \text{mol CO}_2$ $V = \frac{0.333 \times 0.0821 \times 310.15}{1.00} = 8.48\ \text{L}$

Summary Table: Gas Laws and Their Equations

Law	Relationship	Equation	Variables Constant
Boyle's Law	P ∝ 1/V	$P_1V_1 = P_2V_2$	n, T
Charles's Law	V ∝ T	$\frac{V_1}{T_1} = \frac{V_2}{T_2}$	n, P
Gay-Lussac's Law	P ∝ T	$\frac{P_1}{T_1} = \frac{P_2}{T_2}$	n, V
Avogadro's Law	V ∝ n	$\frac{V_1}{n_1} = \frac{V_2}{n_2}$	P, T
Combined Gas Law	---	$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$	n
Ideal Gas Law	---	$PV = nRT$	---

Key Takeaways

• Gases are compressible, low-density, and fill their containers.

• Gas pressure is measured in several units; conversions are essential.

• Gas laws relate P, V, T, and n; the ideal gas law unifies these relationships.

• Density and molar mass of gases can be calculated using the ideal gas law.

• Stoichiometry with gases often uses molar volume at STP and the ideal gas law for non-standard conditions.