Ideal Gas Law and Gas Properties

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

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Ideal Gas Law and Gas Properties

The Ideal Gas Law

The most important equation in gas chemistry:

PV=nRTPV = nRT

Where:

  • PP = pressure (usually atm, but can be kPa, mmHg, torr)
  • VV = volume (usually L)
  • nn = number of moles
  • RR = ideal gas constant
  • TT = temperature (must be in Kelvin!)

Gas Constant (R) Values

Most common:

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

Other values (less common):

R=8.314Jmol\cdotpK=8.314kPa\cdotpLmol\cdotpKR = 8.314 \frac{\text{J}}{\text{mol·K}} = 8.314 \frac{\text{kPa·L}}{\text{mol·K}}

R=62.4L\cdotpmmHgmol\cdotpKR = 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 LV = 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=mMn = \frac{m}{M}

Where:

  • mm = mass (g)
  • MM = molar mass (g/mol)

Substitute into ideal gas law:

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

Rearrange for density:

mV=PMRT\frac{m}{V} = \frac{PM}{RT}

d=PMRTd = \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:

P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

This combines:

  • Boyle's Law: P1V1=P2V2P_1V_1 = P_2V_2 (constant T and n)
  • Charles's Law: V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2} (constant P and n)
  • Gay-Lussac's Law: P1T1=P2T2\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):

P1V1=P2V2P_1V_1 = P_2V_2 (Boyle's Law)

Constant pressure (isobaric):

V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2} (Charles's Law)

Constant volume (isochoric):

P1T1=P2T2\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:

Ptotal=P1+P2+P3+...P_{total} = P_1 + P_2 + P_3 + ...

For each gas in the mixture:

Pi=χiPtotalP_i = \chi_i \cdot P_{total}

Where:

  • PiP_i = partial pressure of gas i
  • χi\chi_i = mole fraction of gas i
  • PtotalP_{total} = total pressure

Mole fraction:

χi=nintotal\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:

Ptotal=Pgas+PH2OP_{total} = P_{gas} + P_{H_2O}

Where:

  • PtotalP_{total} = atmospheric pressure
  • PgasP_{gas} = partial pressure of collected gas
  • PH2OP_{H_2O} = vapor pressure of water (depends on temperature)

To find amount of dry gas:

Pgas=PtotalPH2OP_{gas} = P_{total} - P_{H_2O}

Then use ideal gas law with PgasP_{gas}:

ngas=PgasVRTn_{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

KEavg=32RTKE_{avg} = \frac{3}{2}RT

Or for a single molecule:

KE=12mv2=32kBTKE = \frac{1}{2}mv^2 = \frac{3}{2}k_BT

Where:

  • mm = mass of particle
  • vv = velocity
  • kBk_B = Boltzmann constant = R/NAR/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:

vrms=3RTMv_{rms} = \sqrt{\frac{3RT}{M}}

Where:

  • RR = 8.314 J/(mol·K)
  • TT = temperature (K)
  • MM = molar mass (kg/mol)

Relationships:

  • vrmsTv_{rms} \propto \sqrt{T} (higher T → faster)
  • vrms1Mv_{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:

rate1rate2=M2M1\frac{\text{rate}_1}{\text{rate}_2} = \sqrt{\frac{M_2}{M_1}}

Lighter gases effuse faster

Example: He effuses faster than N₂

rateHerateN2=284=72.65\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
  • Vreal>VidealV_{real} > V_{ideal}

2. Low temperature

  • Particles move slowly
  • IMFs have more effect
  • Particles attracted to each other
  • Preal<PidealP_{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:

(P+an2V2)(Vnb)=nRT\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT

Pressure correction: an2V2\frac{an^2}{V^2}

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

Volume correction: nbnb

  • Accounts for volume of particles
  • Reduces available volume
  • bb = 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:

\ceaA(g)+bB(g)>cC(g)+dD(g)\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:

\ce2H2(g)+O2(g)>2H2O(g)\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:

  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:

P1V1n1T1=P2V2n2T2\frac{P_1V_1}{n_1T_1} = \frac{P_2V_2}{n_2T_2}

With n1n2\frac{n_1}{n_2} from stoichiometry

Summary of Gas Laws

| Law | Equation | Constant | Relationship | |-----|----------|----------|--------------| | Boyle's | P1V1=P2V2P_1V_1 = P_2V_2 | T, n | Inverse (P↑V↓) | | Charles's | V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2} | P, n | Direct (T↑V↑) | | Gay-Lussac's | P1T1=P2T2\frac{P_1}{T_1} = \frac{P_2}{T_2} | V, n | Direct (T↑P↑) | | Avogadro's | V1n1=V2n2\frac{V_1}{n_1} = \frac{V_2}{n_2} | P, T | Direct (n↑V↑) | | Combined | P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} | n | - | | Ideal | PV=nRTPV = nRT | - | All variables | | Dalton's | PT=P1+P2+...P_T = P_1 + P_2 + ... | - | Partial pressures | | Graham's | r1r2=M2M1\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 1easy

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)+273.15T(K) = T(°C) + 273.15

T=25+273.15=298.15 KT = 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=0.0821L\cdotpatmmol\cdotpKR = 0.0821 \frac{\text{L·atm}}{\text{mol·K}}

Step 3: Apply ideal gas law

PV=nRTPV = nRT

Solve for n:

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

Step 4: Substitute values

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

n=7.5024.5n = \frac{7.50}{24.5}

n=0.306 moln = 0.306 \text{ mol}

Answer:

n=0.306 mol N2\boxed{n = 0.306 \text{ mol N}_2}

Or: 0.31 mol\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: 5.00 L22.4 L/mol0.22\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: 0.22×1.5/1.090.300.22 \times 1.5 / 1.09 \approx 0.30 mol ✓

Units check:

atmLL\cdotpatmmol\cdotpKK=atmLmol\cdotpKL\cdotpatmK=mol\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:

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

2Problem 2medium

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:

  • nHen_{He} = 2.00 mol
  • nNen_{Ne} = 3.00 mol
  • nArn_{Ar} = 5.00 mol
  • PtotalP_{total} = 800 mmHg

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


Part (a): Calculate mole fractions

Step 1: Find total moles

ntotal=nHe+nNe+nArn_{total} = n_{He} + n_{Ne} + n_{Ar}

ntotal=2.00+3.00+5.00=10.00 moln_{total} = 2.00 + 3.00 + 5.00 = 10.00 \text{ mol}

Step 2: Calculate mole fraction of each gas

Mole fraction formula:

χi=nintotal\chi_i = \frac{n_i}{n_{total}}

For helium:

χHe=2.00 mol10.00 mol=0.200\chi_{He} = \frac{2.00 \text{ mol}}{10.00 \text{ mol}} = 0.200

For neon:

χNe=3.00 mol10.00 mol=0.300\chi_{Ne} = \frac{3.00 \text{ mol}}{10.00 \text{ mol}} = 0.300

For argon:

χAr=5.00 mol10.00 mol=0.500\chi_{Ar} = \frac{5.00 \text{ mol}}{10.00 \text{ mol}} = 0.500

Check: Sum of mole fractions should equal 1

χHe+χNe+χAr=0.200+0.300+0.500=1.000\chi_{He} + \chi_{Ne} + \chi_{Ar} = 0.200 + 0.300 + 0.500 = 1.000

Answer (a):

χHe=0.200,χNe=0.300,χAr=0.500\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:

Pi=χiPtotalP_i = \chi_i \cdot P_{total}

For helium:

PHe=χHePtotalP_{He} = \chi_{He} \cdot P_{total}

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

PHe=160 mmHgP_{He} = 160 \text{ mmHg}

For neon:

PNe=χNePtotalP_{Ne} = \chi_{Ne} \cdot P_{total}

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

PNe=240 mmHgP_{Ne} = 240 \text{ mmHg}

For argon:

PAr=χArPtotalP_{Ar} = \chi_{Ar} \cdot P_{total}

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

PAr=400 mmHgP_{Ar} = 400 \text{ mmHg}

Check: Sum of partial pressures should equal total pressure

Ptotal=PHe+PNe+PArP_{total} = P_{He} + P_{Ne} + P_{Ar}

Ptotal=160+240+400=800 mmHgP_{total} = 160 + 240 + 400 = 800 \text{ mmHg}

Answer (b):

PHe=160 mmHg,PNe=240 mmHg,PAr=400 mmHg\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:

  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:

Pi=niRTVP_i = \frac{n_iRT}{V}

Total pressure:

Ptotal=ntotalRTVP_{total} = \frac{n_{total}RT}{V}

Ratio:

PiPtotal=nintotal=χi\frac{P_i}{P_{total}} = \frac{n_i}{n_{total}} = \chi_i

Therefore: Pi=χiPtotalP_i = \chi_i \cdot P_{total}

This is Dalton's Law!

3Problem 3hard

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):

Ptotal=PH2+PH2OP_{total} = P_{H_2} + P_{H_2O}

Step 2: Solve for partial pressure of H₂

PH2=PtotalPH2OP_{H_2} = P_{total} - P_{H_2O}

PH2=745 mmHg23.8 mmHgP_{H_2} = 745 \text{ mmHg} - 23.8 \text{ mmHg}

PH2=721.2 mmHgP_{H_2} = 721.2 \text{ mmHg}

Answer (a):

PH2=721.2 mmHg=721 mmHg\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 K298 KT = 25°C + 273.15 = 298.15 \text{ K} \approx 298 \text{ K}

Pressure: Convert mmHg to atm

PH2=721.2 mmHg×1 atm760 mmHgP_{H_2} = 721.2 \text{ mmHg} \times \frac{1 \text{ atm}}{760 \text{ mmHg}}

PH2=0.9489 atmP_{H_2} = 0.9489 \text{ atm}

Volume: Convert mL to L

V=250.0 mL×1 L1000 mL=0.2500 LV = 250.0 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.2500 \text{ L}

Step 4: Apply ideal gas law

PV=nRTPV = nRT

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

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

Step 5: Calculate moles

nH2=(0.9489 atm)(0.2500 L)(0.0821L\cdotpatmmol\cdotpK)(298 K)n_{H_2} = \frac{(0.9489 \text{ atm})(0.2500 \text{ L})}{(0.0821 \frac{\text{L·atm}}{\text{mol·K}})(298 \text{ K})}

nH2=0.237224.46n_{H_2} = \frac{0.2372}{24.46}

nH2=0.00970 moln_{H_2} = 0.00970 \text{ mol}

Answer (b):

nH2=0.00970 mol=9.70×103 mol\boxed{n_{H_2} = 0.00970 \text{ mol} = 9.70 \times 10^{-3} \text{ mol}}

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


Part (c): Mass of H₂

Step 6: Convert moles to mass

Molar mass of H₂:

MH2=2×1.008=2.016 g/molM_{H_2} = 2 \times 1.008 = 2.016 \text{ g/mol}

Mass formula:

m=n×Mm = n \times M

mH2=(0.00970 mol)(2.016 g/mol)m_{H_2} = (0.00970 \text{ mol})(2.016 \text{ g/mol})

mH2=0.0196 gm_{H_2} = 0.0196 \text{ g}

mH2=19.6 mgm_{H_2} = 19.6 \text{ mg}

Answer (c):

mH2=0.0196 g=19.6 mg\boxed{m_{H_2} = 0.0196 \text{ g} = 19.6 \text{ mg}}

Or: 0.020 g\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:

PH2V=nH2RTP_{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: 0.01 mol0.25 L=0.04\frac{0.01 \text{ mol}}{0.25 \text{ L}} = 0.04 mol/L

At STP: 1 mol22.4 L=0.045\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).