6.5-6.7 : Gases – Ideal Gas Law, Molar Volume, Density, and Partial Pressure

Chapter 6. Gases

Introduction

This chapter covers the fundamental properties and laws governing gases, including the Ideal Gas Law, molar volume, density calculations, and partial pressures in mixtures. These concepts are essential for understanding the behavior of gases in chemical reactions and laboratory settings.

Ideal Gas Law

Definition and Variables

The Ideal Gas Law is a fundamental equation that relates the pressure, volume, temperature, and amount of a gas. It combines several simple gas laws into one comprehensive relationship.

• P: Pressure (in atm)

• V: Volume (in L)

• n: Number of moles (in mol)

• T: Temperature (in K)

• R: Universal gas constant ($R = 0.08206\ \frac{L\cdot atm}{mol\cdot K}$)

The equation is:

$PV = nRT$

Standard Temperature and Pressure (STP)

STP is a reference point used in gas calculations:

• Standard Pressure: 1 atm

• Standard Temperature: 273.15 K

At STP, the Ideal Gas Law can be rearranged to solve for volume:

$V = \frac{nRT}{P}$

Molar Volume of a Gas

Definition and Calculation

The molar volume is the volume occupied by one mole of a gas at STP. This value is useful for stoichiometric calculations involving gases.

• At STP, 1 mole of any ideal gas occupies 22.4 L.

Example calculation:

$V = \frac{1.00\ mol \times 0.08206\ \frac{L\cdot atm}{mol\cdot K} \times 273.15\ K}{1.00\ atm} = 22.4\ L$

Density of a Gas

Relationship to Molar Mass and Volume

The density of a gas can be deduced from its molar mass and molar volume:

$\text{density} = \frac{\text{molar mass}\ (g/mol)}{\text{molar volume}\ (L/mol)}$

Example for helium:

$d_{He} = \frac{4.00\ g/mol}{22.4\ L/mol} = 0.179\ g/L$

• The density of a gas is directly proportional to its molar mass.

• The greater the molar mass, the more dense the gas.

Molar Mass Determination Using the Ideal Gas Law

Method and Example

The molar mass of a gas can be determined experimentally by measuring its mass, volume, pressure, and temperature, then applying the Ideal Gas Law.

1. Convert temperature to Kelvin: $T(K) = T(°C) + 273.15$

2. Calculate moles using the Ideal Gas Law: $n = \frac{PV}{RT}$

3. Calculate molar mass: $\text{Molar Mass} = \frac{\text{mass (g)}}{n (mol)}$

Example: A sample of unknown gas has a mass of 3.45 g, volume of 0.597 L, temperature of 75.00 °C, and pressure of 1.5 atm.

• Convert temperature: $75.00\ ^\circ C + 273.15 = 348.15\ K$

• Calculate moles: $n = \frac{(1.5\ atm)(0.597\ L)}{(0.08206\ \frac{L\cdot atm}{mol\cdot K})(348.15\ K)} = 0.03134\ mol$

• Calculate molar mass: $\text{Molar Mass} = \frac{3.45\ g}{0.03134\ mol} = 110\ g/mol$

Partial Pressure and Gas Mixtures

Dalton's Law of Partial Pressures

In a mixture of gases, each gas exerts a pressure independently of the others. The pressure due to any individual gas is called its partial pressure.

• Dalton's Law: The total pressure of a mixture is the sum of the partial pressures of each component.

$P_{total} = P_A + P_B + P_C + \ldots$

Calculating Partial Pressure

The partial pressure of a gas can be calculated using its mole fraction:

• Mole fraction ($x_A$): $x_A = \frac{n_A}{n_{total}}$

• Partial pressure: $P_A = x_A P_{total}$

Example: If the mole fraction of $N_2$ in air is 0.78 and the total pressure is 1 atm:

$P_{N_2} = 0.78 \times 1\ atm = 0.78\ atm$

Stoichiometry of Gaseous Reactions

Volume Relationships at STP

In chemical reactions involving gases, volumes can be related to moles using the molar volume at STP.

• At STP, $1\ mol = 22.4\ L$

• Stoichiometric calculations can use volume ratios directly for gases at the same temperature and pressure.

Example: Synthesis of Methanol

Given the reaction: $CO(g) + 2\ H_2(g) \rightarrow CH_3OH(g)$

To find the volume of $H_2$ needed to synthesize 35.7 g of methanol at 355 K and 738 mmHg:

1. Convert mass of methanol to moles: $35.7\ g \rightarrow 1.1142\ mol$

2. Use stoichiometry: $1.1142\ mol\ CH_3OH \times \frac{2\ mol\ H_2}{1\ mol\ CH_3OH} = 2.2284\ mol\ H_2$

3. Convert pressure to atm: $738\ mmHg \times \frac{1\ atm}{760\ mmHg} = 0.97105\ atm$

4. Use Ideal Gas Law to solve for volume: $V = \frac{nRT}{P}$ $V = \frac{2.2284\ mol \times 0.08206\ \frac{L\cdot atm}{mol\cdot K} \times 355\ K}{0.97105\ atm} = 66.9\ L$

Summary Table: Key Gas Law Relationships

Law/Concept	Equation	Key Variables	Application
Ideal Gas Law	$PV = nRT$	P, V, n, T, R	Relates all major gas properties
Molar Volume at STP	$V = 22.4\ L$ (for 1 mol at STP)	n, STP	Volume-mole conversions
Density of a Gas	$\text{density} = \frac{\text{molar mass}}{\text{molar volume}}$	Molar mass, molar volume	Finding gas density
Dalton's Law	$P_{total} = \sum P_i$	Partial pressures	Mixtures of gases
Mole Fraction	$x_A = \frac{n_A}{n_{total}}$	n (moles)	Calculating partial pressure

Additional info: These notes expand on the original slides and text to provide full academic context, definitions, and worked examples for each major concept in the chapter.