One equation, many uses
Chapter 2.8 covered the gas laws and the ideal gas law. AP topic 3.4 asks you to use PV = nRT quickly and accurately: convert among units, find molar mass from density, handle partial pressures in gas mixtures, and read graphs and particle diagrams of gas behavior.
PV = nRT
The AP equation sheet gives R = 8.314 J/(mol·K) = 0.08206 L·atm/(mol·K) = 62.36 L·torr/(mol·K). The pressure unit decides which R to use; volume goes in liters, and temperature must be in kelvins (T = t + 273.15). At STP (0 °C and 1 atm), 1 mol of an ideal gas occupies 22.4 L.
How the pressure responds when one condition changes and the others stay fixed can be read straight from the equation: P is proportional to n and T and inversely proportional to V.
Molar mass from density
Substitute n = m/M into PV = nRT and use density d = m/V:
At the same temperature and pressure, the density of a gas is proportional to its molar mass. Measure the density of an unknown gas and you can calculate its molar mass, and from that, suggest what the gas is.
Gas mixtures: partial pressures and mole fractions
In a mixture, each gas fills the whole container as if the others weren’t there, and the pressure it exerts on its own is its partial pressure分压. The total pressure is the sum of the partial pressures (Dalton’s law):
X is the mole fraction摩尔分数. In one container every gas has the same T and V, so the ratio of partial pressures equals the ratio of amounts in moles, which is also the ratio of particle counts in a particle diagram.
If the total pressure in this container is 1.50 atm, the mole fraction of the red gas is 4/12 = 1/3, so its partial pressure is 1/3 × 1.50 = 0.500 atm; the blue gas exerts 1.00 atm.
When two containers holding different gases are joined by a valve and the valve is opened, each gas spreads through the combined volume. At constant temperature, P₁V₁ = P₂V₂ for each gas gives its partial pressure:
Pressure changes from reactions
In a rigid container at constant temperature, P ∝ n (the total amount of gas). A reaction that reduces the number of gas molecules lowers the pressure; one that increases it raises the pressure:
When 2.0 mol of CO and 1.0 mol of react completely, the gas goes from 3.0 mol to 2.0 mol, so once it returns to the original temperature the pressure is 2/3 of what it was. If the container has a freely moving piston instead (constant pressure and temperature), the volume changes: V ∝ n, and the piston moves in. Look only at the amount of gas: mass is conserved in the reaction and doesn’t affect the pressure, and the size of the molecules themselves doesn’t matter for an ideal gas.
Collecting a gas over water
A gas collected over water is mixed with water vapor (2.8.5): P(gas) = P(atm) − P(water vapor). To find the amount of gas collected, you need the volume of gas, the water temperature (to look up the vapor pressure of water and to convert to K), and the atmospheric pressure. The mass of the water in the collection vessel isn’t needed.
Graphs of gas relationships
| Held constant | Relationship | Straight line through the origin |
|---|---|---|
| n, T | P ∝ 1/V (Boyle’s law) | P against 1/V |
| n, P | V ∝ T (T in K) | V against T |
| n, V | P ∝ T (T in K) | P against T |
| T, V | P ∝ n | P against n |
With temperature in degrees Celsius on the horizontal axis, V against t is still a straight line, but it doesn’t pass through the origin; extrapolated, it reaches zero volume at −273.15 °C.
Example 1: Identifying a gas from its density
An unknown gas has a density of 1.25 g/L at 0 °C and 1.00 atm. Find its molar mass and suggest what the gas might be.
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Derive a formula with density from PV = nRT. Substitute n = m/M, then use d = m/V:
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Substitute. T = 273.15 K, pressure in atm, and R = 0.08206 L·atm/(mol·K):
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Suggest an identity. Common gases with molar masses near 28 g/mol include (28.02), CO (28.01), and (28.05).
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Note. Molar mass alone can’t decide among them; other properties are needed, such as whether the gas burns.
Example 2: Opening a valve
A 2.00 L container holds He at 1.50 atm, and a 3.00 L container holds at 0.500 atm. They are joined by a valve, at the same constant temperature. After the valve is opened, find each partial pressure, the total pressure, and the mole fraction of He.
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Each gas fills the total volume. The total volume is 2.00 + 3.00 = 5.00 L. For each gas, n and T are constant, so P₁V₁ = P₂V₂:
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Total pressure. By Dalton’s law:
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Mole fraction of He. At the same temperature n ∝ PV, so use the data from before the valve was opened:
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Check. Working back with P_A = X_A·P_total: 0.667 × 0.900 = 0.600 atm, the same as in step 1.
Example 3: Molar mass by the Dumas method
The molar mass of a colorless, volatile liquid is found by the Dumas method. A little liquid is put in a weighed flask covered with foil that has a pinhole, and the flask is placed in boiling water; once the liquid has vaporized and the excess vapor has escaped through the hole, the flask is filled with vapor. The flask is removed and cooled, the vapor condenses, and the flask is weighed again. Data: flask volume 251.0 mL, bath temperature 99.8 °C, atmospheric pressure 748 mmHg, condensed liquid 0.696 g. Elemental analysis gives the empirical formula . Find the molar mass and the molecular formula.
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Be clear about what is measured. The vapor in the flask is at the bath temperature and atmospheric pressure, and its volume is the flask volume; the mass of liquid that condenses on cooling is the mass of that vapor.
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Put the data in consistent units.
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Find the moles of vapor.
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Find the molar mass.
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From empirical to molecular formula. The formula mass of is 3 × 12.01 + 7 × 1.008 = 43.09:
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Think about error. If some liquid had not yet vaporized when the flask was removed, the measured mass would be too high, while n, calculated from PV/RT, would be unaffected, so M would come out too high. For error analysis, decide which measured quantity is off and in which direction, then see where it appears in the formula.
- Constant volume, temperature raised from 300 K to 450 K
P ∝ T: 1.5 times as large.
- Constant temperature, volume changed from 2 L to 4 L
P ∝ 1/V: halved.
- Constant volume and temperature, more of the same gas added
P ∝ n.
- Constant volume and temperature, He added: the partial pressure of the original gas
That gas's n, V, and T are unchanged, so its partial pressure is unchanged; the total pressure rises.
- Constant volume and temperature, He added: the total pressure
The total amount of gas increases.
- Volume halved while the absolute temperature doubles
P ∝ T/V: 4 times as large.
- Volume doubled while the absolute temperature doubles
T/V is unchanged.
- Half the gas let out while the volume is halved
n/V is unchanged.
- Constant volume, temperature lowered from 27 °C to −73 °C
From 300 K to 200 K, so P becomes 2/3 as large.
- Constant volume, temperature raised from 10 °C to 20 °C
From 283 K to 293 K: only about 3.5% higher, not double.
Common mistakes
Mistake 1: “Put the Celsius temperature into PV = nRT.” Use absolute temperature. Going from 10 °C to 20 °C raises the pressure by only about 3.5%, not by a factor of 2.
Mistake 2: “Adding another gas lowers the partial pressure of the gas already there.” At constant volume and temperature, the amount of the original gas hasn’t changed, so neither has its partial pressure. Its mole fraction falls, and the total pressure rises.
Mistake 3: “At the same temperature and pressure, equal masses of gases have equal volumes.” At the same temperature and pressure, equal amounts in moles have equal volumes (Avogadro’s law). 1 g of occupies 16 times the volume of 1 g of .
Quick check
0.500 mol of a gas is in a 4.00 L container at 27 °C. Which expression gives its pressure in atm?
P = nRT/V with T in K; with pressure in atm and volume in L, R = 0.08206 L·atm/(mol·K). A uses the Celsius temperature; B swaps n and V; C uses R in J/(mol·K), which doesn't give atm.
A particle diagram of a gas mixture shows 3 red particles and 9 blue particles, and the total pressure is 2.40 atm. Find the partial pressure of the red gas.
X(red) = 3/12 = 0.25, so P(red) = 0.25 × 2.40 = 0.600 atm. The blue gas exerts 1.80 atm (B).
A container holds equal masses of He and Ne at a total pressure of 1.20 atm. The partial pressure of He is closest to
For equal masses, moles are inversely proportional to molar mass: n(He) : n(Ne) = 1/4.00 : 1/20.18 ≈ 5.0 : 1, so X(He) ≈ 5.0/6.0 = 0.83 and P(He) ≈ 0.83 × 1.20 ≈ 1.00 atm. B treats the mass fraction as if it were the mole fraction.
A gas has a density of 1.96 g/L at STP (0 °C and 1 atm). It is most likely
M = dRT/P = 1.96 × 0.08206 × 273.15 ÷ 1.00 ≈ 43.9 g/mol; or, since 1 mol of gas occupies 22.4 L at STP, 1.96 g/L × 22.4 L/mol ≈ 43.9 g/mol, closest to (44.01 g/mol).
A rigid container holds . At constant temperature, some He is added. How does the partial pressure of change, and why?
A partial pressure depends only on that gas's own n, V, and T. The fact in A is true (the mole fraction does fall), but the total pressure rises at the same time, so their product is unchanged. D: the volume of the molecules themselves is negligible for an ideal gas.
In a rigid container, 2.0 mol of CO and 1.0 mol of react completely to form , and the container returns to its original temperature. How does the pressure change?
At constant temperature and volume, P ∝ n(gas). A: mass is conserved, but pressure depends on the number of gas molecules. C: the size of the molecules themselves doesn't affect the pressure of an ideal gas. D miscounts the moles.
At ground level (1.00 atm, 300 K) a balloon has a volume of 2.0 L. It rises to where the pressure is 0.50 atm and the temperature is 250 K (the pressure inside equals the pressure outside). Its volume is closest to
V₂ = V₁ × (P₁/P₂) × (T₂/T₁) = 2.0 × (1.00/0.50) × (250/300) ≈ 3.3 L. Halving the pressure doubles the volume, while the cooling only multiplies it by 0.83.
A sample of is collected over water, and you want to find the amount of in moles. Which measurement is NOT needed?
n = PV/RT with P(H₂) = P(atm) − P(water vapor). The volume, the temperature (also used to look up the vapor pressure of water), and the atmospheric pressure are all needed; the mass of the water isn't.
For a fixed amount of gas at constant temperature, which graph is a straight line through the origin?
P = nRT × (1/V), so P is proportional to 1/V. P against V is a hyperbola; V against t is straight but doesn't pass through the origin (and here the temperature is constant anyway); PV is constant, so PV against P is a horizontal line.
At constant pressure, a sample of gas is heated from 100 °C to 200 °C. By what factor does its volume change?
V ∝ T in kelvins: 473.15 ÷ 373.15 ≈ 1.27. A comes from using Celsius temperatures.
AP-style free response
A 5.00 L rigid container holds 0.200 mol of and 0.100 mol of at 27 °C.
(a) Calculate the total pressure in the container.
(b) Calculate the partial pressure of .
(c) At constant temperature, 0.100 mol of He is added to the container. Calculate the new total pressure, and describe how the partial pressure and mole fraction of change.
(d) Explain at the particle level why the partial pressure of doesn’t change when He is added.
Model answer and scoring
(a) n(total) = 0.300 mol and T = 300.15 K. P = nRT/V = 0.300 × 0.08206 × 300.15 ÷ 5.00 ≈ 1.48 atm.
(b) X() = 0.100 ÷ 0.300 = 1/3, so P() = 1/3 × 1.48 ≈ 0.493 atm (or directly from P = nRT/V).
(c) n(total) = 0.400 mol, so P(total) = 0.400 × 0.08206 × 300.15 ÷ 5.00 ≈ 1.97 atm. The partial pressure of is unchanged, still about 0.493 atm; its mole fraction falls from 1/3 to 0.100 ÷ 0.400 = 0.250.
(d) The number of molecules, the volume, and the temperature are all unchanged, so the molecules hit the walls just as often and as hard as before. In an ideal gas, He and molecules don’t interact, so the He only adds its own share of the pressure.
Scoring: (a) temperature converted to K, about 1.48 atm; (b) about 0.493 atm; (c) about 1.97 atm, with the partial pressure unchanged and the mole fraction falling to 0.250; (d) states that n, V, and T of are unchanged, so its collisions with the walls are unchanged.
Summary
- PV = nRT: temperature in K, and R chosen to match the pressure unit; P is proportional to n and T and inversely proportional to V.
- Molar mass from density: M = dRT/P.
- Gas mixtures: P(total) = sum of the partial pressures; P_A = X_A·P(total); in one container, the ratio of partial pressures = the ratio of amounts = the ratio of particle counts.
- In a rigid container at constant temperature, P ∝ n: a reaction that reduces the number of gas molecules lowers the pressure; with a free piston, the volume changes instead.
- P against 1/V, V against T (in K), P against T (in K), and P against n are all straight lines through the origin.
AP tip: In calculations, watch three things: convert temperature to K, match R to the pressure unit, and keep significant figures consistent with the data. Conceptual questions often ask how the pressure (or a partial pressure) changes when one condition changes: write P = nRT/V first, then see which quantities changed.