AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/3/3-4)
Unit 3 · Topic 3.4
3.4 Ideal Gas Law
The ideal gas law, PV = nRT, ties together a gas's pressure, volume, amount in moles and Kelvin temperature. In a mixture, each gas exerts its own partial pressure, and those add to the total pressure (Dalton's law).
Key terms
- ideal gas law
- gas constant (R)
- partial pressure
- Dalton's law
- mole fraction
- Kelvin temperature
PV = nRT
P is pressure, V is volume, n is moles of gas, T is temperature in kelvins and R is the gas constant. The units of R decide the units of everything else.
| Value of R | Use with |
|---|---|
| 0.08206 L·atm/(mol·K) | P in atm, V in L |
| 8.314 J/(mol·K) | energy calculations |
Units to check every time
- Temperature must be in kelvins: K = °C + 273.15. Using °C can give zero or negative values.
- Pressure conversions: 1 atm = 760 mm Hg = 760 torr.
- Volume in liters: 1 L = 1000 mL.
- At STP (273.15 K and 1.0 atm), one mole of an ideal gas takes up 22.4 L. Use this only when the conditions really are STP.
How the variables relate
Hold some variables constant and you get simple relationships. At constant n and T, P and V are inversely proportional: halve the volume and the pressure doubles. A graph of P against V is a curve that drops steeply then levels off. At constant n and P, V is directly proportional to T in kelvins: a graph of V against T is a straight line that would pass through 0 K. At constant n and V, P is directly proportional to T. At constant P and T, V is directly proportional to n.
For one sample changing conditions, the equation sheet gives P₁V₁/T₁ = P₂V₂/T₂. Cancel anything that doesn't change.
You can also rearrange PV = nRT to find molar mass. Since n = m/M and density D = m/V, M = DRT/P.
Mixtures: Dalton's law of partial pressures
In a mixture of ideal gases, each gas acts as if it were alone in the container. The pressure it would exert alone is its partial pressure. The total pressure is the sum: P(total) = P(A) + P(B) + P(C) + …
Each gas's share of the total pressure equals its share of the moles. The mole fraction of gas A is X(A) = moles of A ÷ total moles, and P(A) = X(A) × P(total). Partial pressure depends on the number of particles, not on what the gas is: 1 mole of helium and 1 mole of xenon exert the same partial pressure in the same container.
A common lab use: gas collected by bubbling it into an upside-down bottle of water is mixed with water vapor. The pressure inside equals the gas's partial pressure plus the vapor pressure of water at that temperature, so subtract water's vapor pressure to get the pressure of the gas you collected.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Straight PV = nRT
What volume does 0.500 mol of N₂ occupy at 25 °C and 1.20 atm?
Show the solutionHide the solution
- Step 1: Convert temperature: 25 + 273.15 = 298.15 K.
- Step 2: Rearrange: V = nRT/P.
- Step 3: V = (0.500 mol)(0.08206 L·atm/(mol·K))(298.15 K) ÷ 1.20 atm = 10.2 L.
Answer: 10.2 L
- Example 2Calculator allowed
Partial pressures in a mixture
A 10.0 L flask at 27 °C contains 4.00 g of He and 16.0 g of O₂. Find the partial pressure of each gas, the total pressure and the mole fraction of He.
Show the solutionHide the solution
- Step 1: Moles: He = 4.00 ÷ 4.003 = 0.999 mol; O₂ = 16.0 ÷ 32.00 = 0.500 mol.
- Step 2: T = 27 + 273 = 300 K.
- Step 3: P(He) = nRT/V = (0.999)(0.08206)(300) ÷ 10.0 = 2.46 atm.
- Step 4: P(O₂) = (0.500)(0.08206)(300) ÷ 10.0 = 1.23 atm.
- Step 5: P(total) = 2.46 + 1.23 = 3.69 atm. X(He) = 0.999 ÷ 1.499 = 0.666. Check: 0.666 × 3.69 = 2.46 atm.
Answer: P(He) = 2.46 atm, P(O₂) = 1.23 atm, P(total) = 3.69 atm, X(He) = 0.666
- Example 3Calculator allowed
Using Celsius by mistake (classic trap)
A 2.00 L balloon at 25 °C is warmed to 50 °C at constant pressure. What is its new volume?
Show the solutionHide the solution
- Step 1: At constant P and n, V₁/T₁ = V₂/T₂, with T in kelvins.
- Step 2: T₁ = 298.15 K and T₂ = 323.15 K.
- Step 3: V₂ = 2.00 L × (323.15 ÷ 298.15) = 2.17 L.
- Step 4: Using Celsius would say the temperature doubled and give 4.00 L, which is far too large. Going from 25 °C to 50 °C raises the kelvin temperature by only about 8%.
Answer: 2.17 L
Common mistakes
- Using °C instead of K.
- Mixing units with R: using 0.08206 with pressure in torr or volume in mL.
- Using 22.4 L/mol when the gas isn't at STP.
- Thinking a heavier gas exerts a larger partial pressure. Partial pressure depends on moles, not molar mass.
On the exam
- Gas law calculations often appear inside stoichiometry problems, such as finding the volume of gas a reaction produces. Show the equation, your values with units and the answer with units.
- Expect qualitative questions about what happens to one variable when another changes, or which graph shows a given relationship.
Connected topics
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Check yourself
4 questions on 3.4 Ideal Gas Law. Pick an answer to see if you got it, and why.
A 4.00 L container holds a gas at 1.50 atm and 27 °C. Approximately how many moles of gas are in the container? (R = 0.08206 L·atm/(mol·K))
A rigid container holds 2.0 mol of N₂, 1.0 mol of O₂ and 1.0 mol of Ar. The total pressure is 2.00 atm. What is the partial pressure of O₂?
A sample of gas in a cylinder with a movable piston is heated from 27 °C to 327 °C while the pressure and the amount of gas stay constant. By what factor does the volume of the gas change?
A gas has a density of 2.86 g/L at 0 °C and 1.00 atm. Which of the following could the gas be? (R = 0.08206 L·atm/(mol·K))
0 of 4 answered