The most important equation in gas chemistry:

$PV = nRT$

Where:

• $P$ = pressure (usually atm, but can be kPa, mmHg, torr)
• $V$ = volume (usually L)
• $n$ = number of moles
• $R$ = ideal gas constant
• $T$ = temperature (must be in Kelvin!)

Gas Constant (R) Values

Most common:

$R = 0.0821 \frac{\text{L·atm}}{\text{mol·K}}$

Other values (less common):

$R = 8.314 \frac{\text{J}}{\text{mol·K}} = 8.314 \frac{\text{kPa·L}}{\text{mol·K}}$

$R = 62.4 \frac{\text{L·mmHg}}{\text{mol·K}}$

Choose R based on units of pressure and volume!

Standard Temperature and Pressure (STP)

Definition:

• Temperature: 0°C = 273.15 K
• Pressure: 1 atm = 101.3 kPa = 760 mmHg = 760 torr

At STP, one mole of any ideal gas occupies:

$V = 22.4 \text{ L}$

This is a useful conversion factor!

Using PV = nRT

The ideal gas law connects all four variables

If you know any 3, you can find the 4th:

Example problems:

1. Know P, V, T → find n (moles)
2. Know n, T, P → find V (volume)
3. Know n, V, P → find T (temperature)
4. Know n, V, T → find P (pressure)

Key conversions:

• Temperature: °C + 273.15 = K
• Pressure: 1 atm = 760 mmHg = 101.3 kPa
• Volume: Usually given in L (if not, convert!)

Gas Density and Molar Mass

Density of a gas can be related to molar mass

Starting with PV = nRT:

$n = \frac{m}{M}$

Where:

• $m$ = mass (g)
• $M$ = molar mass (g/mol)

Substitute into ideal gas law:

$PV = \frac{m}{M}RT$

Rearrange for density:

$\frac{m}{V} = \frac{PM}{RT}$

$d = \frac{PM}{RT}$

Key relationships:

• Higher molar mass → Higher density (at same T and P)
• Higher temperature → Lower density (at same P and M)
• Higher pressure → Higher density (at same T and M)

Why this is useful:

• Calculate molar mass from density measurements
• Predict relative densities of different gases
• Explain why hot air rises (lower density)

Combined Gas Law

When amount of gas (n) is constant:

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

This combines:

• Boyle's Law: $P_1V_1 = P_2V_2$ (constant T and n)
• Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$ (constant P and n)
• Gay-Lussac's Law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ (constant V and n)

Use when:

• Gas undergoes change in conditions
• Amount of gas stays the same
• Need to find final state from initial state

Special cases:

Constant temperature (isothermal):

$P_1V_1 = P_2V_2$ (Boyle's Law)

Constant pressure (isobaric):

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$ (Charles's Law)

Constant volume (isochoric):

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$ (Gay-Lussac's Law)

Dalton's Law of Partial Pressures

In a mixture of gases, each gas exerts pressure independently

Total pressure = Sum of partial pressures:

$P_{total} = P_1 + P_2 + P_3 + ...$

For each gas in the mixture:

$P_i = \chi_i \cdot P_{total}$

Where:

• $P_i$ = partial pressure of gas i
• $\chi_i$ = mole fraction of gas i
• $P_{total}$ = total pressure

Mole fraction:

$\chi_i = \frac{n_i}{n_{total}}$

Key concept: Each gas behaves as if it alone occupies the entire volume

Application: Collecting Gas Over Water

Common lab technique:

• Gas produced in reaction bubbles through water
• Collected gas is mixture of desired gas + water vapor

Pressure relationship:

$P_{total} = P_{gas} + P_{H_2O}$

Where:

• $P_{total}$ = atmospheric pressure
• $P_{gas}$ = partial pressure of collected gas
• $P_{H_2O}$ = vapor pressure of water (depends on temperature)

To find amount of dry gas:

$P_{gas} = P_{total} - P_{H_2O}$

Then use ideal gas law with $P_{gas}$:

$n_{gas} = \frac{P_{gas}V}{RT}$

Water vapor pressure at common temperatures:

Kinetic Molecular Theory (KMT)

Model explaining ideal gas behavior

Five Postulates

1. Gas particles are in constant, random motion

• Particles move in straight lines until collision
• Collisions are perfectly elastic (no energy lost)

2. Gas particles have negligible volume

• Volume of particles << volume of container
• Most of gas volume is empty space

3. No attractive or repulsive forces between particles

• Particles don't interact except during collisions
• Move independently

4. Average kinetic energy proportional to absolute temperature

$KE_{avg} = \frac{3}{2}RT$

Or for a single molecule:

$KE = \frac{1}{2}mv^2 = \frac{3}{2}k_BT$

Where:

• $m$ = mass of particle
• $v$ = velocity
• $k_B$ = Boltzmann constant = $R/N_A$

Higher temperature → Higher average KE → Faster particles

5. Collisions are perfectly elastic

• Total kinetic energy conserved
• Energy transferred between particles
• No energy lost to heat, sound, etc.

Implications of KMT

Temperature and KE:

• At same T, all gases have same average KE
• Lighter gases move faster (same KE, less mass)

Root-mean-square speed:

$v_{rms} = \sqrt{\frac{3RT}{M}}$

Where:

• $R$ = 8.314 J/(mol·K)
• $T$ = temperature (K)
• $M$ = molar mass (kg/mol)

Relationships:

• $v_{rms} \propto \sqrt{T}$ (higher T → faster)
• $v_{rms} \propto \frac{1}{\sqrt{M}}$ (lighter → faster)

At same temperature:

• He atoms move faster than O₂ molecules
• All have same average KE

Maxwell-Boltzmann Distribution

Not all particles have same speed!

Distribution of molecular speeds:

• Some particles very slow
• Most particles near average
• Some particles very fast
• Shape depends on temperature and molar mass

Effects of temperature:

• Higher T → distribution shifts right (faster average)
• Higher T → distribution broadens (wider range)
• Peak becomes lower and flatter

Effects of molar mass:

• Lower M → distribution shifts right (faster)
• Lower M → distribution broadens

Effusion and Diffusion

Graham's Law of Effusion

Effusion: Gas escapes through tiny hole into vacuum

Rate of effusion inversely proportional to square root of molar mass:

$\frac{\text{rate}_1}{\text{rate}_2} = \sqrt{\frac{M_2}{M_1}}$

Lighter gases effuse faster

Example: He effuses faster than N₂

$\frac{\text{rate}_{He}}{\text{rate}_{N_2}} = \sqrt{\frac{28}{4}} = \sqrt{7} \approx 2.65$

Helium effuses 2.65 times faster than nitrogen

Diffusion

Diffusion: Gas spreads through space or through another gas

Also follows Graham's Law (approximately):

• Lighter gases diffuse faster
• Same mathematical relationship

Example: Smell of perfume spreading across room

Why faster with lighter gases?

• Lighter → higher average velocity (from KMT)
• Higher velocity → spreads faster

Real Gases vs. Ideal Gases

When Do Real Gases Deviate from Ideal Behavior?

Ideal gas law assumes:

1. No volume (particles are points)
2. No intermolecular forces

Real gases deviate when:

1. High pressure

• Gas particles forced close together
• Volume of particles becomes significant
• $V_{real} > V_{ideal}$

2. Low temperature

• Particles move slowly
• IMFs have more effect
• Particles attracted to each other
• $P_{real} < P_{ideal}$

General rule:

Real gases behave most ideally at:

• Low pressure (particles far apart)
• High temperature (high KE overcomes IMFs)

Real gases deviate most at:

• High pressure (particles close)
• Low temperature (IMFs significant)

Van der Waals Equation

Correction to ideal gas law for real gases:

$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$

Pressure correction: $\frac{an^2}{V^2}$

• Accounts for attractive forces
• Reduces effective pressure
• $a$ = strength of IMFs (larger for polar molecules)

Volume correction: $nb$

• Accounts for volume of particles
• Reduces available volume
• $b$ = size of particles (larger for bigger molecules)

Constants a and b:

• Specific to each gas
• Larger a → stronger IMFs
• Larger b → larger molecules

When van der Waals ≈ Ideal:

• Low pressure (correction terms become small)
• High temperature (KE >> IMFs)

Gas Stoichiometry

Using ideal gas law in chemical reactions

Mole Relationships

For reactions involving gases:

$aA\text{(g)} + bB\text{(g)} \rightarrow cC\text{(g)} + dD\text{(g)}$

Mole ratios from balanced equation:

• At same T and P, volume ratios = mole ratios
• This is Avogadro's Law

Example:

$2H2\text{(g)} + O2\text{(g)} \rightarrow 2H2O\text{(g)}$

At same T and P:

• 2 volumes H₂ + 1 volume O₂ → 2 volumes H₂O
• 2 moles H₂ + 1 mole O₂ → 2 moles H₂O

Problem-Solving Strategy

Given: Conditions of gas (P, V, T) and reaction

Steps:

1. Use PV = nRT to find moles of given gas
2. Use stoichiometry (mole ratios) to find moles of desired gas
3. Use PV = nRT to find conditions of desired gas

Or use directly:

$\frac{P_1V_1}{n_1T_1} = \frac{P_2V_2}{n_2T_2}$

With $\frac{n_1}{n_2}$ from stoichiometry

Summary of Gas Laws

Key Problem-Solving Tips

Always convert:

• Temperature to Kelvin (K = °C + 273.15)
• Pressure to match R (usually atm)
• Volume to match R (usually L)

Choose correct R:

• Match units to pressure and volume
• Most common: 0.0821 L·atm/(mol·K)

Partial pressures:

• Use mole fractions
• Account for water vapor when collecting gas

Real vs ideal:

• Real gases → use at low P, high T
• Deviations at high P, low T

Gas stoiometry:

• Convert to moles first
• Use mole ratios
• Convert back to gas conditions

Find: (a) Mole fractions, (b) Partial pressures

Part (a): Calculate mole fractions

Step 1: Find total moles

$n_{total} = n_{He} + n_{Ne} + n_{Ar}$

$n_{total} = 2.00 + 3.00 + 5.00 = 10.00 \text{ mol}$

Step 2: Calculate mole fraction of each gas

Mole fraction formula:

$\chi_i = \frac{n_i}{n_{total}}$

For helium:

$\chi_{He} = \frac{2.00 \text{ mol}}{10.00 \text{ mol}} = 0.200$

For neon:

$\chi_{Ne} = \frac{3.00 \text{ mol}}{10.00 \text{ mol}} = 0.300$

For argon:

$\chi_{Ar} = \frac{5.00 \text{ mol}}{10.00 \text{ mol}} = 0.500$

Check: Sum of mole fractions should equal 1

$\chi_{He} + \chi_{Ne} + \chi_{Ar} = 0.200 + 0.300 + 0.500 = 1.000$ ✓

Answer (a):

$\boxed{\chi_{He} = 0.200, \quad \chi_{Ne} = 0.300, \quad \chi_{Ar} = 0.500}$

Part (b): Calculate partial pressures

Step 3: Use Dalton's Law of Partial Pressures

Relationship between mole fraction and partial pressure:

$P_i = \chi_i \cdot P_{total}$

For helium:

$P_{He} = \chi_{He} \cdot P_{total}$

$P_{He} = (0.200)(800 \text{ mmHg})$

$P_{He} = 160 \text{ mmHg}$

For neon:

$P_{Ne} = \chi_{Ne} \cdot P_{total}$

$P_{Ne} = (0.300)(800 \text{ mmHg})$

$P_{Ne} = 240 \text{ mmHg}$

For argon:

$P_{Ar} = \chi_{Ar} \cdot P_{total}$

$P_{Ar} = (0.500)(800 \text{ mmHg})$

$P_{Ar} = 400 \text{ mmHg}$

Check: Sum of partial pressures should equal total pressure

$P_{total} = P_{He} + P_{Ne} + P_{Ar}$

$P_{total} = 160 + 240 + 400 = 800 \text{ mmHg}$ ✓

Answer (b):

$\boxed{P_{He} = 160 \text{ mmHg}, \quad P_{Ne} = 240 \text{ mmHg}, \quad P_{Ar} = 400 \text{ mmHg}}$

Summary Table:

Key insights:

1. Mole fraction = fraction of total moles

   • Dimensionless (no units)
   • Always between 0 and 1
   • Sum = 1.000

2. Partial pressure proportional to mole fraction

   • Gas with most moles → highest partial pressure
   • Ar has 50% of moles → 50% of pressure

3. Each gas behaves independently (Dalton's Law)

   • He exerts 160 mmHg as if alone in container
   • Ne exerts 240 mmHg as if alone in container
   • Ar exerts 400 mmHg as if alone in container
   • Total = sum of all partial pressures

Conceptual understanding:

Why does this work?

From ideal gas law for each gas:

$P_i = \frac{n_iRT}{V}$

Total pressure:

$P_{total} = \frac{n_{total}RT}{V}$

Ratio:

$\frac{P_i}{P_{total}} = \frac{n_i}{n_{total}} = \chi_i$

Therefore: $P_i = \chi_i \cdot P_{total}$

This is Dalton's Law!

(c) Mole fraction of oxygen: χ_O₂ = n_O₂ / n_total = 0.300 / 1.00 = 0.300

Note: χ_O₂ = P_O₂ / P_total = 0.738 / 2.46 = 0.300 ✓

• H₂ gas (what we want)
• H₂O vapor (from evaporation)

Total pressure = sum of partial pressures (Dalton's Law):

$P_{total} = P_{H_2} + P_{H_2O}$

Step 2: Solve for partial pressure of H₂

$P_{H_2} = P_{total} - P_{H_2O}$

$P_{H_2} = 745 \text{ mmHg} - 23.8 \text{ mmHg}$

$P_{H_2} = 721.2 \text{ mmHg}$

Answer (a):

$\boxed{P_{H_2} = 721.2 \text{ mmHg} = 721 \text{ mmHg}}$

This is the pressure of the DRY hydrogen gas

Part (b): Moles of H₂

Step 3: Convert units for ideal gas law

Temperature:

$T = 25°C + 273.15 = 298.15 \text{ K} \approx 298 \text{ K}$

Pressure: Convert mmHg to atm

$P_{H_2} = 721.2 \text{ mmHg} \times \frac{1 \text{ atm}}{760 \text{ mmHg}}$

$P_{H_2} = 0.9489 \text{ atm}$

Volume: Convert mL to L

$V = 250.0 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.2500 \text{ L}$

Step 4: Apply ideal gas law

$PV = nRT$

$n = \frac{PV}{RT}$

Use: $R = 0.0821 \frac{\text{L·atm}}{\text{mol·K}}$

Step 5: Calculate moles

$n_{H_2} = \frac{(0.9489 \text{ atm})(0.2500 \text{ L})}{(0.0821 \frac{\text{L·atm}}{\text{mol·K}})(298 \text{ K})}$

$n_{H_2} = \frac{0.2372}{24.46}$

$n_{H_2} = 0.00970 \text{ mol}$

Answer (b):

$\boxed{n_{H_2} = 0.00970 \text{ mol} = 9.70 \times 10^{-3} \text{ mol}}$

Or: $\boxed{0.0097 \text{ mol}}$ (2 sig figs)

Part (c): Mass of H₂

Step 6: Convert moles to mass

Molar mass of H₂:

$M_{H_2} = 2 \times 1.008 = 2.016 \text{ g/mol}$

Mass formula:

$m = n \times M$

$m_{H_2} = (0.00970 \text{ mol})(2.016 \text{ g/mol})$

$m_{H_2} = 0.0196 \text{ g}$

$m_{H_2} = 19.6 \text{ mg}$

Answer (c):

$\boxed{m_{H_2} = 0.0196 \text{ g} = 19.6 \text{ mg}}$

Or: $\boxed{0.020 \text{ g}}$ (2 sig figs)

Summary of Results:

Key Concepts and Explanations:

1. Why subtract water vapor pressure?

When collecting gas over water:

• Water evaporates into collection container
• Total pressure includes both H₂ and H₂O vapor
• Must subtract P(H₂O) to get P(dry gas)

Diagram of setup:

• Gas bubbles through water
• Collected in inverted tube over water
• Total P = P(H₂) + P(H₂O vapor)

2. Water vapor pressure depends on temperature

Higher temperature → more evaporation → higher vapor pressure

3. Why use P(H₂) not P(total) in ideal gas law?

Ideal gas law for H₂ only:

$P_{H_2}V = n_{H_2}RT$

If we used P(total):

• Would calculate moles of (H₂ + H₂O)
• Wrong answer!

Must use partial pressure of H₂ alone

4. Common mistakes to avoid:

❌ Using P(total) instead of P(H₂) ❌ Forgetting to convert °C to K ❌ Forgetting to convert mL to L ❌ Using wrong vapor pressure for temperature

5. Check reasonableness:

At STP: 1 mol H₂ = 22.4 L

Our conditions: ~0.01 mol in 0.25 L

Expected: $\frac{0.01 \text{ mol}}{0.25 \text{ L}} = 0.04$ mol/L

At STP: $\frac{1 \text{ mol}}{22.4 \text{ L}} = 0.045$ mol/L

Close! ✓ (Our P slightly lower, T slightly higher than STP)

Mass check:

• 0.01 mol × 2 g/mol = 0.02 g
• Our answer: 0.0196 g ✓

Everything checks out!

Application: This is how chemists measure amount of gas produced in reactions (like H₂ from metal + acid).