F. Gases and Atmospheric Chemistry · Gases and Atmospheric ChemistryThe Ideal Gas Law: PV = nRT
F3.5 F3.4 F2.3About 70 minutes
Learning goals- I can explain how the ideal gas law combines the simple gas laws.
- I can use PV = nRT to find pressure, volume, amount or temperature.
- I can choose between the combined gas law and the ideal gas law.
- I can calculate the molar mass and density of a gas from experimental data.
Words to know
ideal gas law
PV = nRT: the relationship among pressure, volume, amount and temperature for an ideal gas.
universal gas constant (R)
8.314 kPa·L/(mol·K), the same for all gases.
Dumas method
A way to find the molar mass of a volatile liquid by measuring the mass of its vapour that fills a flask of known volume at known T and P.
gas density
Mass per unit volume of a gas, d = PM/RT, usually in g/L.
One law to rule them all
You now know four relationships: V ∝ 1/P (Boyle), V ∝ T (Charles), P ∝ T (Gay-Lussac) and V ∝ n (Avogadro). Each one needs two variables held constant. But what if you want to know the pressure inside a scuba tank that holds 85 mol of air at 20 °C? Or the molar mass of a mystery gas? For problems about one gas in one set of conditions, chemists use a single equation that links all four variables at once: the ideal gas law.
PV = nRT
Combine the proportionalities: V ∝ (1/P), V ∝ T and V ∝ n, so V ∝ nT/P. Turn the proportion into an equation by putting in a constant, R:
PV = nRT
- P = pressure in kPa
- V = volume in L
- n = amount in mol
- T = temperature in K
- R = the universal gas constant = 8.314 kPa·L/(mol·K)
With these units, R always has this value, for every gas. (You may see R = 8.314 J/(mol·K) in physics or in SCH4U; it's the same number, because 1 kPa·L = 1 J.)
Quick check: for 1.000 mol at STP, V = nRT/P = (1.000)(8.314)(273.15) ÷ 101.325 = 22.4 L. The molar volume from the last lesson is just a special case of the ideal gas law.
Combined gas law or ideal gas law?
- If the gas changes from one set of conditions to another (two states, "before and after") and n is constant → combined gas law.
- If you have one set of conditions and need P, V, n or T, or anything involving moles, mass or molar mass → ideal gas law.
Always convert to kPa, L and K before substituting into PV = nRT.
Worked exampleExample 1: Finding volume
What volume do 2.00 mol of oxygen occupy at 25 °C and 150 kPa?
- n = 2.00 mol, T = 25 + 273 = 298 K, P = 150 kPa
- V = nRT ÷ P
- V = (2.00 mol)(8.314 kPa·L/(mol·K))(298 K) ÷ 150 kPa
- Units: mol × kPa·L/(mol·K) × K ÷ kPa = L ✓
- V = 33.0 L
Worked exampleExample 2: Pressure in a scuba tank
A 12.0 L scuba tank holds 85.0 mol of air at 20 °C. What is the pressure in the tank, in kPa and in atm?
- T = 293 K
- P = nRT ÷ V = (85.0)(8.314)(293) ÷ 12.0 = 1.73 × 10⁴ kPa
- In atm: 17 255 ÷ 101.325 = 170 atm (keep the unrounded pressure)
About 1.73 × 10⁴ kPa (170 atm). That's why scuba tanks are thick steel or aluminium.
Finding molar mass and density
Since n = m ÷ M, you can substitute into PV = nRT:
PV = (m ÷ M)RT, so M = mRT ÷ PV
This lets you find the molar mass of an unknown gas or volatile liquid from measurements of mass, pressure, volume and temperature. In the classic Dumas method, a small amount of volatile liquid is put in a flask with a tiny hole and heated in boiling water. The liquid vaporizes and pushes the air out; the extra vapour escapes through the hole until the flask is full of vapour at the bath temperature and room pressure. The flask is cooled, the vapour condenses, and the mass of the condensed liquid is measured.
Rearranging another way gives gas density: d = m ÷ V = PM ÷ RT. A gas is denser when it is cold, at high pressure, or made of heavier molecules.
Worked exampleExample 3: Molar mass by the Dumas method
A 0.500 L flask is filled with the vapour of an unknown liquid at 100 °C and 101.3 kPa. After cooling, the condensed vapour has a mass of 0.949 g. Find the molar mass. Could the liquid be acetone, C₃H₆O?
- T = 373 K, P = 101.3 kPa, V = 0.500 L, m = 0.949 g
- M = mRT ÷ PV = (0.949)(8.314)(373) ÷ [(101.3)(0.500)]
- M = 2943 ÷ 50.65 = 58.1 g/mol
- M(C₃H₆O) = 3(12.01) + 6(1.01) + 16.00 = 58.09 g/mol ✓
58.1 g/mol, which matches acetone.
Worked exampleExample 4: Cold air is dense air
Compare the density of air (M = 28.97 g/mol) at 101.3 kPa on a −30 °C winter day and a 30 °C summer day.
- d = PM ÷ RT
- Winter: d = (101.3)(28.97) ÷ [(8.314)(243)] = 1.45 g/L
- Summer: d = (101.3)(28.97) ÷ [(8.314)(303)] = 1.16 g/L
Winter air is about 25 % denser (1.45 g/L versus 1.16 g/L). That's one reason aircraft need shorter take-off runs on cold days, and why hot-air balloons fly more easily in cold weather.
Hands-on activityHow much gas is in a bottle of pop?
You need: an unopened 2 L bottle of carbonated water or pop, a kitchen scale (reading to 1 g), a thermometer or weather app, a large bowl.
Safety: open the bottle slowly over the bowl or sink; shaken pop can spray. Do not point the cap at anyone.
- Weigh the unopened bottle with its cap. Record the mass and the room temperature.
- Open it slowly. Pour out about half into a jug (you can drink it later if it's yours!). Weigh the jug and the bottle; together they still contain all the liquid. Leave both uncovered for a day, stirring now and then until they're completely flat.
- Weigh the bottle (with its cap) and the jug again. The difference in total mass (minus the mass of the jug, which you weigh empty) is roughly the mass of CO₂ that escaped.
- Calculate n = m ÷ 44.01 g/mol, then V = nRT ÷ P with P ≈ 101 kPa.
What to look for: a loss of several grams, which becomes several litres of CO₂ gas, more than the volume of the bottle itself. Sources of error: some water also evaporates overnight (making the CO₂ estimate too high), and a kitchen scale reads only to the nearest gram. You could reduce the water error by covering the containers loosely with a paper towel.
Just for funWhy should you never trust an ideal gas?
Because it doesn't take up any space, it has no attractions, and it claims every collision was perfectly elastic. Sounds like an excuse!
Your turn
1Write the ideal gas law and give R with its units.
PV = nRT; R = 8.314 kPa·L/(mol·K).
2Find the pressure of 0.250 mol of gas in a 2.00 L container at 27 °C.
P = (0.250)(8.314)(300) ÷ 2.00 = 312 kPa
3How many moles of gas are in 5.0 L at 120 kPa and 22 °C?
n = PV ÷ RT = (120)(5.0) ÷ [(8.314)(295)] = 0.24 mol
4What volume do 2.50 mol of gas occupy at 35 °C and 101.3 kPa?
V = (2.50)(8.314)(308) ÷ 101.3 = 63.2 L
5At what temperature (in °C) would 0.500 mol of gas occupy 10.0 L at 100 kPa?
T = PV ÷ nR = (100)(10.0) ÷ [(0.500)(8.314)] = 241 K = −32 °C
6Show that 1.00 mol of an ideal gas at STP occupies 22.4 L.
V = nRT ÷ P = (1.00)(8.314)(273.15) ÷ 101.325 = 22.4 L
More practice: warm-up
1What does R stand for in PV = nRT?
The universal gas constant.
2In PV = nRT with R = 8.314, what units must P, V and T be in?
3Rearrange PV = nRT to solve for n.
4Rearrange PV = nRT to solve for T.
5Which law should you use for a gas that changes from one set of conditions to another with no gas added or removed?
More practice: on your own
1A 1.20 g sample of a gas occupies 0.750 L at 98.0 kPa and 22 °C. Find its molar mass and identify it from He, Ne, Ar, Kr.
M = mRT ÷ PV = (1.20)(8.314)(295) ÷ [(98.0)(0.750)] = 40.0 g/mol. It is argon (39.95 g/mol).
2Calculate the density of carbon dioxide at 25 °C and 100 kPa.
d = PM ÷ RT = (100)(44.01) ÷ [(8.314)(298)] = 1.78 g/L
3What volume is needed to hold 3.0 mol of gas at 27 °C and 250 kPa?
V = (3.0)(8.314)(300) ÷ 250 = 30 L (29.9 L)
4What volume does 5.0 g of methane (CH₄) occupy at −20 °C and 101.3 kPa?
n = 5.0 ÷ 16.05 = 0.312 mol. V = (0.312)(8.314)(253) ÷ 101.3 = 6.5 L
5Explain using d = PM/RT why a helium balloon rises in air.
At the same P and T, density is proportional to molar mass. Helium (4.00 g/mol) is much less dense than air (about 29 g/mol), so the buoyant force from the air is greater than the weight of the balloon.
Challenge
1In a Dumas experiment, a 0.500 L flask at 100 °C and 101.3 kPa holds vapour that condenses to 1.95 g of liquid. Find the molar mass. Is the liquid more likely to be chloroform (CHCl₃) or ethanol (C₂H₅OH)?
M = (1.95)(8.314)(373) ÷ [(101.3)(0.500)] = 119 g/mol. Chloroform: 12.01 + 1.01 + 3(35.45) = 119.4 g/mol. Ethanol is 46.08 g/mol. It's chloroform.
2When a bottle of pop goes flat, 7.0 g of CO₂ escapes at 22 °C and 101.3 kPa. What volume of gas is this?
n = 7.0 ÷ 44.01 = 0.159 mol. V = (0.159)(8.314)(295) ÷ 101.3 = 3.9 L, about twice the volume of a 2 L bottle.
3Starting from PV = nRT and n = m/M, derive the formula d = PM/RT, then use it to find the molar mass of a gas whose density is 2.86 g/L at STP.
PV = (m/M)RT → PM = (m/V)RT → d = m/V = PM/RT. Rearranged: M = dRT/P = (2.86)(8.314)(273) ÷ 101.3 = 64.1 g/mol (it could be SO₂, 64.07 g/mol).
Quiz
1. Which units must be used with R = 8.314 kPa·L/(mol·K)?
2. 2.00 mol of gas at 300 K occupy 10.0 L. The pressure is about:
3. Which expression gives the molar mass of a gas?
4. At the same temperature and pressure, which gas has the greatest density?
5. The ideal gas law is the best choice when you need to:
6. Air is denser on a cold winter day than on a hot summer day at the same pressure because:
Big idea (F3.5, F3.4, F2.3): the ideal gas law, PV = nRT, relates all four gas variables in one state. It also gives molar mass (M = mRT/PV) and density (d = PM/RT).
How to help: ask your teen to write the units beside every number before substituting, and to check that the units cancel to the unit of the answer. This catches mL-versus-L and °C-versus-K errors.
Common difficulties:
- Units: mL instead of L, °C instead of K, atm instead of kPa.
- Choosing the law: one state → ideal gas law; two states → combined gas law.
- Algebra: rearranging PV = nRT. Practise solving for each variable.
Looking ahead: in Grade 12 Chemistry (SCH4U) and physics, R is often written 8.314 J/(mol·K) and used in thermodynamics. It is the same constant.
Safety: open carbonated drinks slowly over a sink.