10.3 Additional Gas Laws
PRE-HEALTH PREP
Dalton's Law of Partial Pressure, Gas Density & Graham's Law of Effusion
General Chemistry 10.3
Dalton's Law of Partial Pressures describes how the total pressure of a gas mixture equals the sum of each component's partial pressure, determined by its mole fraction. Gas density under ideal conditions is calculated from pressure, molar mass, and temperature using d = PM/RT, while the molar volume at standard temperature and pressure (STP) provides the useful conversion factor of 22.4 L/mol for stoichiometric calculations. Graham's Law of Effusion establishes that lighter gas molecules escape through small openings faster than heavier ones, with the effusion rate ratio equal to the inverse square root of the molar masses.
| ADDITIONAL GAS LAWS | |
|---|---|
| Dalton's Law of Partial Pressures | Ptotal = PA + PB + PC … PA = XAPTotal XA = mole fraction A |
| Molar Volume at STP (STP: T = 273 K, P = 1 atm) | At STP, 1 mole of gas has a volume of 22.4 L. |
| Gas Density | d = PM / RT (M = molar mass) |
| Graham's Law of Effusion | Rate1 / Rate2 = √(M2 / M1) (r = rate of effusion, M = molar mass) |
Dalton's Law of Partial Pressures
What Is Dalton's Law?
Dalton's Law of Partial Pressures states that the total pressure exerted by a mixture of gases equals the sum of the individual partial pressures of each gas. Each gas contributes a portion of the total pressure, and that portion depends entirely on how much of that gas is present relative to the total amount in the mixture.
The key quantity here is the mole fraction — the fraction of the total moles contributed by a particular gas. If a gas accounts for half the moles in the mixture, it accounts for half the total pressure. The relationship is direct and proportional.
Why molecular size doesn't affect partial pressure: Under most conditions, a gas is composed of mostly empty space. Whether you're comparing N₂, O₂, or CO₂, the size of the molecules is largely irrelevant. Every gas has an equal opportunity to contribute to the pressure — the only factor that matters is the number of moles present.
How to Calculate Partial Pressure from Mole Fraction
Consider a container holding 10 moles of N₂, 8 moles of O₂, and 2 moles of CO₂, with a total pressure of 50 atm. That gives 20 total moles of gas.
Example: Calculating Partial Pressures from Mole Fractions
Nitrogen (N₂): 10 out of 20 total moles → mole fraction = 1/2 → partial pressure = 0.5 × 50 atm = 25 atm
Oxygen (O₂): 8 out of 20 total moles → mole fraction = 8/20 = 0.4 → partial pressure = 0.4 × 50 atm = 20 atm
Carbon dioxide (CO₂): 2 out of 20 total moles → mole fraction = 1/10 = 0.1 → partial pressure = 0.1 × 50 atm = 5 atm
Check: 25 + 20 + 5 = 50 atm ✓
Dalton's Law Formulas
Dalton's Law of Partial Pressures is expressed in two equivalent forms:
Total pressure as a sum of partial pressures:
Ptotal = P1 + P2 + P3 + …
Individual partial pressure from mole fraction:
Pi = χi × Ptotal
where χi (chi) is the mole fraction of gas i, defined as the moles of that gas divided by the total moles in the mixture.
Tip: If the logic makes sense to you, there's no need to memorize the equation. Half the moles means half the pressure — it follows directly from the concept. For the last gas in a mixture, you can always find its partial pressure by subtracting the other partial pressures from the total.
STP and the Molar Volume of a Gas
What Is STP in Chemistry?
STP stands for standard temperature and pressure, corresponding to the following specific conditions:
| Condition | Value |
|---|---|
| Temperature | 273 K (0°C) |
| Pressure | 1 atm |
Why Molar Volume at STP Is 22.4 L/mol
One mole of any ideal gas at STP occupies a volume of 22.4 liters. This value comes directly from the ideal gas law:
V = nRT / P = (1 mol)(0.08206 L·atm/mol·K)(273 K) / (1 atm) = 22.4 L
This molar volume is independent of the identity of the gas. Half a mole at STP occupies 11.2 L. Ten moles at STP occupy 224 L. The proportionality is direct and linear for all ideal gases.
Note: The 22.4 L/mol shortcut applies only at STP. At any other temperature or pressure, you must use the full ideal gas law to determine volume.
Gas Stoichiometry at STP: Volume of Gas Produced
Example: Magnesium Reacting with Hydrochloric Acid
Mg + 2 HCl → H₂ + MgCl₂
If 48.6 g of magnesium reacts with excess HCl at STP, what volume of H₂ gas is produced?
Step 1: Convert grams of Mg to moles.
48.6 g ÷ 24.3 g/mol = 2.00 mol Mg
Step 2: Apply the mole-to-mole ratio.
The balanced equation shows a 1:1 ratio of Mg to H₂, so 2.00 mol Mg produces 2.00 mol H₂.
Step 3: Convert moles of gas to volume at STP.
2.00 mol × 22.4 L/mol = 44.8 L of H₂
Tip: Because STP appears so frequently in gas-phase stoichiometry, treating 22.4 L/mol as a conversion factor in dimensional analysis saves significant time. If conditions are not STP, calculate moles first and then use PV = nRT to find volume.
Gas Density and Molar Mass
The Gas Density Formula: d = PM/RT
Density is defined as mass divided by volume. For an ideal gas, density can be expressed directly in terms of pressure, molar mass, and temperature using the gas density formula:
d = PM / RT
This formula is derived from the ideal gas law. Start with PV = nRT and rearrange to isolate n/V:
PV = nRT → n/V = P/RT
This gives moles per volume, not mass per volume. Multiplying both sides by the molar mass M converts moles to mass:
nM/V = PM/RT
Since moles times molar mass equals mass, the left side becomes m/V, which is density. The result is a formula relating gas density to 3 variables: pressure, molar mass, and temperature.
Worked Example: Calculating Gas Density with d = PM/RT
Example: What is the density of helium gas at 2.0 atm and 273 K?
d = PM / RT
d = (2.0 atm)(4.0 g/mol) / (0.08206 L·atm/mol·K)(273 K)
d = 8.0 / 22.4 = 0.357 g/L
Unit check: Atmospheres cancel, moles cancel, kelvins cancel → grams per liter ✓
Perspective: A gas density of 0.357 g/L is remarkably low compared to liquid water at 1000 g/L. Gas molecules are spread far apart with mostly empty space between them, which is why gas densities are orders of magnitude smaller than liquid or solid densities.
Alternative Method: Gas Density from the Ideal Gas Law Directly
If the density formula is unavailable, you can calculate gas density from first principles using the ideal gas law. The key insight is that density is an intensive property — it doesn't depend on sample size. You can therefore choose any convenient amount of gas for your calculation.
Example: Helium Density Using the Ideal Gas Law (Without the Formula)
Step 1: Choose a convenient sample size — 1 mole of helium.
Step 2: Determine the mass.
1 mol He × 4.0 g/mol = 4.0 g
Step 3: Calculate the volume using the ideal gas law.
V = nRT/P = (1 mol)(0.08206)(273 K) / (2.0 atm) = 11.2 L
Step 4: Divide mass by volume.
4.0 g / 11.2 L = 0.357 g/L ✓
Tip: Both methods yield identical results. The direct formula (d = PM/RT) is faster, but the ideal gas law approach works even when the formula isn't provided on your equation sheet. Just remember to choose a sample size — one mole is almost always the most convenient choice.
Graham's Law of Effusion
What Is Effusion? (Effusion vs. Diffusion)
Effusion is the process by which gas molecules escape through a tiny opening (a narrow slit or pinhole) into a vacuum or region of lower pressure. It is often confused with diffusion, but the two are distinct: diffusion involves gases spreading through one another, while effusion involves escape through a physical opening. Graham's Law applies specifically to effusion.
Real-world example: An easy-to-visualize case of effusion occurs when you get a flat tire. If you run over a nail and it punctures your tire, when you remove the nail the tire begins to deflate as the gases inside exit through the puncture hole — that's effusion.
Why Do Lighter Gases Effuse Faster?
The reasoning begins with kinetic energy. At a given temperature, all gas molecules have the same average kinetic energy, regardless of their identity. Temperature alone determines average kinetic energy.
However, equal kinetic energy does not mean equal velocity. Recall that kinetic energy is defined as:
KE = ½mv²
If two gases share the same kinetic energy but one has a much larger molar mass, the heavier gas must move slower to compensate. The lighter gas, with less mass, moves faster on average. Because lighter molecules have higher average velocities, they encounter the opening more frequently and escape faster. That is the physical basis of Graham's Law of Effusion.
Analogy: The Volkswagen and the Semi-Truck
Imagine placing the same engine in both a Volkswagen and a fully loaded semi-truck. Both vehicles receive the same energy output, but they won't travel at the same speed. The lighter Volkswagen accelerates to a much higher velocity, while the heavier semi-truck moves slowly. Gas molecules behave the same way — lighter molecules move faster at any given temperature.
Graham's Law Formula
Graham's Law of Effusion relates the effusion rates of 2 gases to their molar masses:
Rate1 / Rate2 = √(M2 / M1)
Notice the inverse relationship: the lighter gas (smaller M) appears in the denominator under the square root, producing a ratio greater than 1 — confirming that the lighter gas effuses faster.
Common Mistake: Students often place the molar masses in the wrong order. Remember that the molar masses are in the opposite ratio to the effusion rates. If Gas 1 is the lighter gas and you want to know how many times faster it effuses, put the heavier gas's molar mass on top.
Worked Example: H₂ vs. O₂ Effusion Rates
Example: How many times faster does H₂ effuse compared to O₂?
RateH₂ / RateO₂ = √(MO₂ / MH₂)
= √(32 g/mol ÷ 2 g/mol)
= √16 = 4
Hydrogen effuses 4 times faster than oxygen. Despite being 16 times lighter in molar mass, H₂ escapes only 4 times faster due to the square root relationship in Graham's Law.
Important distinction: Graham's Law gives you a ratio of effusion rates, not the absolute rate of either gas. If asked "What is the rate of effusion of hydrogen?" the answer is "It cannot be determined from Graham's Law alone." But if asked "How does the effusion rate of hydrogen compare to oxygen?" you can confidently say it is 4 times faster. Don't forget that both H₂ and O₂ are diatomic when looking up their molar masses.
van der Waals Equation for Real Gases
Two of the major assumptions made for ideal gases are that there are no attractive forces between molecules and that the molecules themselves have negligible volume. van der Waals attempted to correct for the inaccuracies in both of these assumptions in what is now referred to as the van der Waals equation of state.
You'll note that this has similarities with the Ideal Gas Law but with the addition of two terms into the equation. These two terms contain the van der Waals constants, 'a' and 'b,' which are determined empirically for a gas and are published in tables for a variety of gases.
The first additional term to appear in the van der Waals equation accounts for the attractive intermolecular forces that all gases have, which are more significant at lower temperatures and for gases with larger attractive forces.
The constant 'a': The greater the attractive forces between atoms/molecules, the greater this correction will need to be, leading to a greater magnitude of this van der Waals constant. A gas that has lower attractive forces is said to behave more ideally and would be expected to have a smaller value for this constant.
The second additional term to appear in the van der Waals equation accounts for the volume occupied by the atoms/molecules themselves. While the ideal gas law assumes that this volume is negligible, this is less and less true as pressure increases and the molecules are pressed closer and closer together.
The constant 'b': The greater the volume of the atoms/molecules, the greater this correction will need to be, leading to a greater magnitude of this van der Waals constant. A gas with a smaller volume is said to behave more ideally and would be expected to have a smaller value for this constant.
More accurate equations of state such as the virial equation of state are more commonly used today than the van der Waals equation. But the van der Waals equation was one of the earliest attempts to model real gases and generally gives more accurate calculations of pressure, volume, or temperature than the ideal gas law.
| Constant | Corrects For | Larger Value Means… |
|---|---|---|
| a | Intermolecular attractive forces | Stronger attractive forces; less ideal behavior |
| b | Molecular volume | Larger molecules; less ideal behavior |
9 Key Takeaways
- Dalton's Law of Partial Pressures states that the total pressure of a gas mixture equals the sum of each gas's partial pressure: Ptotal = P1 + P2 + P3 + …
- The partial pressure of an individual gas equals its mole fraction multiplied by the total pressure: Pi = χi × Ptotal.
- STP (standard temperature and pressure) is defined as 273 K and 1 atm. One mole of any ideal gas at STP occupies 22.4 L — the molar volume at STP.
- The molar volume at STP (22.4 L/mol) serves as a convenient conversion factor in gas-phase stoichiometry problems, but only at STP conditions.
- The gas density formula d = PM/RT calculates ideal gas density from pressure, molar mass, and temperature. Alternatively, pick a 1-mole sample and divide its mass by its volume from the ideal gas law.
- Graham's Law of Effusion states that the effusion rate is inversely proportional to the square root of the molar mass: lighter gases effuse faster than heavier gases.
- Graham's Law yields a ratio of effusion rates, not absolute rates. For H₂ vs. O₂, hydrogen effuses 4 times faster despite being 16 times lighter in molar mass, because the relationship involves a square root.
- The van der Waals equation corrects ideal gas behavior for intermolecular attractive forces (constant a) and finite molecular volume (constant b): (P + an²/V²)(V − nb) = nRT.
- Real gases deviate most from ideal behavior at high pressures and low temperatures, where intermolecular forces and molecular volume become significant.