Ideal Gas Law and Gas Properties

Master the ideal gas law, gas law calculations, partial pressures, kinetic molecular theory, and real gas behavior.

Ideal Gas Law and Gas Properties

The Ideal Gas Law

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:

Know P, V, T → find n (moles)
Know n, T, P → find V (volume)
Know n, V, P → find T (temperature)
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:

| Temp (°C) | P(H₂O) mmHg | |-----------|-------------| | 20 | 17.5 | | 25 | 23.8 | | 30 | 31.8 | | 35 | 42.2 |

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:

No volume (particles are points)
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:

$\ce{aA(g) + bB(g) -> cC(g) + dD(g)}$

Mole ratios from balanced equation:

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

Example:

$\ce{2H2(g) + O2(g) -> 2H2O(g)}$

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:

Use PV = nRT to find moles of given gas
Use stoichiometry (mole ratios) to find moles of desired gas
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

| Law | Equation | Constant | Relationship | |-----|----------|----------|--------------| | Boyle's | $P_1V_1 = P_2V_2$ | T, n | Inverse (P↑V↓) | | Charles's | $\frac{V_1}{T_1} = \frac{V_2}{T_2}$ | P, n | Direct (T↑V↑) | | Gay-Lussac's | $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ | V, n | Direct (T↑P↑) | | Avogadro's | $\frac{V_1}{n_1} = \frac{V_2}{n_2}$ | P, T | Direct (n↑V↑) | | Combined | $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ | n | - | | Ideal | $PV = nRT$ | - | All variables | | Dalton's | $P_T = P_1 + P_2 + ...$ | - | Partial pressures | | Graham's | $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$ | - | Effusion rate |

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

📚 Practice Problems

1Problem 1medium

❓ Question:

A sample of nitrogen gas occupies 5.00 L at 25°C and 1.50 atm. (a) How many moles of N₂ are present? (b) What is the mass of the nitrogen gas? (c) If the temperature is increased to 100°C at constant pressure, what will be the new volume?

💡 Show Solution

Solution:

Given: V = 5.00 L, T = 25°C = 298 K, P = 1.50 atm, R = 0.08206 L·atm/(mol·K)

(a) Moles using PV = nRT: n = PV/(RT) = (1.50 atm)(5.00 L) / [(0.08206)(298 K)] n = 7.50 / 24.45 = 0.307 mol N₂

(b) Mass: Molar mass of N₂ = 2(14.01) = 28.02 g/mol Mass = 0.307 mol × 28.02 g/mol = 8.60 g

(c) New volume (Charles's Law at constant P): V₁/T₁ = V₂/T₂ T₂ = 100°C = 373 K V₂ = V₁(T₂/T₁) = 5.00 L × (373 K / 298 K) V₂ = 6.25 L

2Problem 2easy

❓ Question:

A sample of nitrogen gas occupies 5.00 L at 25°C and 1.50 atm. How many moles of nitrogen are present?

💡 Show Solution

Solution:

Given:

Volume (V) = 5.00 L
Temperature (T) = 25°C
Pressure (P) = 1.50 atm

Find: Number of moles (n)

Step 1: Convert temperature to Kelvin

$T(K) = T(°C) + 273.15$

$T = 25 + 273.15 = 298.15 \text{ K}$

Can round to 298 K for calculations

Step 2: Identify the appropriate gas constant

Units we have:

P in atm ✓
V in L ✓
T in K ✓

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

Step 3: Apply ideal gas law

$PV = nRT$

Solve for n:

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

Step 4: Substitute values

$n = \frac{(1.50 \text{ atm})(5.00 \text{ L})}{(0.0821 \frac{\text{L·atm}}{\text{mol·K}})(298 \text{ K})}$

$n = \frac{7.50}{24.5}$

$n = 0.306 \text{ mol}$

Answer:

$\boxed{n = 0.306 \text{ mol N}_2}$

Or: $\boxed{0.31 \text{ mol}}$ (2 sig figs)

Check reasonableness:

At STP (1 atm, 273 K):

1 mole occupies 22.4 L
Our conditions: Higher P (1.50 atm) and higher T (298 K)

Higher P → compressed → less volume per mole Higher T → expanded → more volume per mole

Net effect: Roughly 0.3 mol in 5 L seems reasonable

Alternative check using proportions:

At STP: 1 mol in 22.4 L

Our gas: $\frac{5.00 \text{ L}}{22.4 \text{ L/mol}} \approx 0.22$ mol at STP

But P is 1.5× higher → compress → more moles fit And T is ~1.09× higher → expand → fewer moles fit

Net: $0.22 \times 1.5 / 1.09 \approx 0.30$ mol ✓

Units check:

$\frac{\text{atm} \cdot \text{L}}{\frac{\text{L·atm}}{\text{mol·K}} \cdot \text{K}} = \frac{\text{atm} \cdot \text{L} \cdot \text{mol·K}}{\text{L·atm} \cdot \text{K}} = \text{mol}$ ✓

Key concepts:

Always convert temperature to Kelvin
Match R units to given units
Use PV = nRT when relating P, V, T, and n
Check reasonableness against STP values

3Problem 3medium

❓ Question:

💡 Show Solution

Solution:

Given: V = 5.00 L, T = 25°C = 298 K, P = 1.50 atm, R = 0.08206 L·atm/(mol·K)

(a) Moles using PV = nRT: n = PV/(RT) = (1.50 atm)(5.00 L) / [(0.08206)(298 K)] n = 7.50 / 24.45 = 0.307 mol N₂

(b) Mass: Molar mass of N₂ = 2(14.01) = 28.02 g/mol Mass = 0.307 mol × 28.02 g/mol = 8.60 g

(c) New volume (Charles's Law at constant P): V₁/T₁ = V₂/T₂ T₂ = 100°C = 373 K V₂ = V₁(T₂/T₁) = 5.00 L × (373 K / 298 K) V₂ = 6.25 L

4Problem 4hard

❓ Question:

A mixture of gases contains 0.500 mol N₂, 0.300 mol O₂, and 0.200 mol CO₂ in a 10.0 L container at 27°C. (a) Calculate the total pressure. (b) Calculate the partial pressure of each gas. (c) What is the mole fraction of oxygen?

💡 Show Solution

Solution:

Given: n_N₂ = 0.500 mol, n_O₂ = 0.300 mol, n_CO₂ = 0.200 mol V = 10.0 L, T = 27°C = 300 K, R = 0.08206 L·atm/(mol·K)

(a) Total pressure (Dalton's Law): n_total = 0.500 + 0.300 + 0.200 = 1.00 mol

P_total = n_total RT/V = (1.00)(0.08206)(300) / 10.0 P_total = 2.46 atm

(b) Partial pressures: Using P_i = (n_i/n_total) × P_total or P_i = n_i RT/V

P_N₂ = (0.500/1.00) × 2.46 = 1.23 atm P_O₂ = (0.300/1.00) × 2.46 = 0.738 atm P_CO₂ = (0.200/1.00) × 2.46 = 0.492 atm

Check: 1.23 + 0.738 + 0.492 = 2.46 atm ✓

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

5Problem 5hard

❓ Question:

💡 Show Solution

Solution:

Given: n_N₂ = 0.500 mol, n_O₂ = 0.300 mol, n_CO₂ = 0.200 mol V = 10.0 L, T = 27°C = 300 K, R = 0.08206 L·atm/(mol·K)

(a) Total pressure (Dalton's Law): n_total = 0.500 + 0.300 + 0.200 = 1.00 mol

P_total = n_total RT/V = (1.00)(0.08206)(300) / 10.0 P_total = 2.46 atm

(b) Partial pressures: Using P_i = (n_i/n_total) × P_total or P_i = n_i RT/V

P_N₂ = (0.500/1.00) × 2.46 = 1.23 atm P_O₂ = (0.300/1.00) × 2.46 = 0.738 atm P_CO₂ = (0.200/1.00) × 2.46 = 0.492 atm

Check: 1.23 + 0.738 + 0.492 = 2.46 atm ✓

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

6Problem 6medium

❓ Question:

A mixture of gases contains 2.00 mol He, 3.00 mol Ne, and 5.00 mol Ar at a total pressure of 800 mmHg. Calculate: (a) the mole fraction of each gas, (b) the partial pressure of each gas.

💡 Show Solution

Solution:

Given:

$n_{He}$ = 2.00 mol
$n_{Ne}$ = 3.00 mol
$n_{Ar}$ = 5.00 mol
$P_{total}$ = 800 mmHg

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:

| Gas | Moles | Mole Fraction | Partial Pressure | |-----|-------|---------------|------------------| | He | 2.00 mol | 0.200 (20%) | 160 mmHg | | Ne | 3.00 mol | 0.300 (30%) | 240 mmHg | | Ar | 5.00 mol | 0.500 (50%) | 400 mmHg | | Total | 10.00 mol | 1.000 | 800 mmHg |

Key insights:

Mole fraction = fraction of total moles
- Dimensionless (no units)
- Always between 0 and 1
- Sum = 1.000
Partial pressure proportional to mole fraction
- Gas with most moles → highest partial pressure
- Ar has 50% of moles → 50% of pressure
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!

7Problem 7hard

❓ Question:

Hydrogen gas is collected over water at 25°C and 745 mmHg atmospheric pressure. The volume of gas collected is 250.0 mL. The vapor pressure of water at 25°C is 23.8 mmHg. (a) What is the partial pressure of the dry hydrogen gas? (b) How many moles of H₂ were collected? (c) What mass of H₂ was collected?

💡 Show Solution

Solution:

Given:

Temperature = 25°C
Total pressure (atmospheric) = 745 mmHg
Volume collected = 250.0 mL
Vapor pressure of H₂O at 25°C = 23.8 mmHg

Find: (a) Partial pressure of H₂, (b) Moles of H₂, (c) Mass of H₂

Part (a): Partial pressure of dry H₂

Step 1: Understand the situation

Gas collected over water contains:

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:

| Quantity | Value | |----------|-------| | (a) Partial pressure H₂ | 721 mmHg (0.949 atm) | | (b) Moles H₂ | 0.00970 mol | | (c) Mass H₂ | 0.0196 g (19.6 mg) |

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

| Temperature | P(H₂O) | |-------------|--------| | 20°C | 17.5 mmHg | | 25°C | 23.8 mmHg | | 30°C | 31.8 mmHg |

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).

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