The Ideal Gas Law — Print · Grade 11 Science

Chapter: States of Matter & Gases

The Ideal Gas Law

Learning goals

Readings for this unit

Before this unit

Moles, kelvin, pressure units, and rearranging equations.

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Start with something to explain

Does the same gas volume always contain the same number of particles?

Illustrative evidence case. Two equal-size vessels have different absolute pressures or temperatures.

A gas constant is only usable with its matching unit systemA gas constant is only usable with its matching unit systemPV = nRT variableMeaningCompatible exampleunitPAbsolute pressureatmVGas volumeLnAmount of gasmolTAbsolute temperatureKFor these units: R ≈ 0.0821 L·atm/(mol·K).
Original unit organizer with rounded school constants; the four-variable law assumes ideal-gas behavior.

Notice and wonder

Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.

Does the same gas volume always contain the same number of particles?

Your first model

Write the P, V, T and n relationship before equating their entity counts.

Build the language from the idea

ideal gas law
The equation PV = nRT, which relates the pressure, volume, moles, and Kelvin temperature of a gas in a single snapshot.
ideal gas constant (R)
The proportionality constant in PV = nRT; equal to 0.0821 L·atm/(mol·K), or 8.314 J/(mol·K) in SI units.
mole (n)
A count of 6.02 × 10²³ particles; the amount of gas, symbolized n, in the ideal gas law.
STP
Standard Temperature and Pressure: 0°C (273 K) and 1 atm, the reference conditions for comparing gases.
molar volume
The volume occupied by exactly one mole of a gas; it is 22.4 L at STP for any ideal gas.
Avogadro's Law
Equal volumes of gases at the same temperature and pressure contain equal numbers of particles; volume is directly proportional to moles (V ∝ n).
ideal gas
A hypothetical gas that perfectly obeys the KMT assumptions; real gases behave most ideally at low pressure and high temperature.

The combined gas law is perfect for a gas that *changes* from one condition to another, but it has a blind spot: it never tells you how *much* gas you actually have. The ideal gas law closes that gap by adding the amount of gas, in moles, right into the equation. In one compact statement — PV = nRT — it ties together all four of a gas's defining quantities: pressure, volume, the number of moles, and the Kelvin temperature. Instead of comparing a 'before' and 'after,' it describes a single gas sample at a single moment.

Each letter earns its place. P is pressure, V is volume, n is the number of moles, T is the Kelvin temperature, and R is the ideal gas constant — the fixed number that makes the units balance. R comes in two common flavors, and picking the right one is half the battle: use R = 0.0821 L·atm/(mol·K) when pressure is in atmospheres and volume is in liters, or R = 8.314 J/(mol·K) when you are working in SI energy units. Match your units to your R, and the arithmetic takes care of itself.

A gas constant is only usable with its matching unit systemA gas constant is only usable with its matching unit systemPV = nRT variableMeaningCompatible exampleunitPAbsolute pressureatmVGas volumeLnAmount of gasmolTAbsolute temperatureKFor these units: R ≈ 0.0821 L·atm/(mol·K).
Original unit organizer with rounded school constants; the four-variable law assumes ideal-gas behavior.

Solving with PV = nRT follows a dependable rhythm. Decide which of the four variables you are missing, rearrange the equation to isolate it, and substitute — always with temperature in kelvin and units matched to your R. Want pressure? P = nRT/V. Want the number of moles hiding in a tank? n = PV/RT. Because the law describes one snapshot, you do not need a 'before' and 'after' at all; you just need three of the four quantities to find the fourth.

The ideal-gas equation gives amount before chemical identity gives massThe ideal-gas equation gives amount before chemical identity gives massP = 1.00 atm; V = 2.46 L; T = 300 Kn = PV/RT = (1.00)(2.46)/(0.0821×300)n ≈ 0.100 molTo find mass, multiply by the specified gas’s molar mass.
Original selected near-0.100-mol calculation; constants and input values are rounded for school use.

An ideal gas at 273.15 K and 1 atm has molar volume approximately 22.4 L/mol using the stated constants. Some standard-condition conventions instead use 100 kPa, giving a different value, so always state pressure and temperature rather than rely on the initials STP alone. Molar volume is a condition-dependent bridge between amount and gas volume, not the definition of a mole or a fixed value at room temperature.

Why should oxygen, helium, and carbon dioxide — molecules of wildly different mass — all crowd into the same 22.4 liters? The answer is Avogadro's Law: equal volumes of gases at the same temperature and pressure contain equal *numbers* of particles. What fills space is not the mass of the molecules but how many of them there are and how fast they move, and PV = nRT depends only on the count n, never on the identity of the gas. That is why the 'n' in the equation, not the chemical formula, is what matters.

One honest caveat rounds out the picture: no gas is truly ideal. Real molecules do take up a little space, and they do feel faint attractions for one another — the very assumptions KMT ignores. Those effects are negligible under everyday conditions, so the ideal gas law is an excellent approximation for the gases you will meet. But squeeze a gas to very high pressure or chill it toward its boiling point at very low temperature, and the molecules get close enough that their volume and attractions start to matter, and real gases begin to drift away from the law's tidy predictions. That drift is exactly where the next topic — the forces between molecules — takes over.

An ideal model has a measurable domain of approximationAn ideal model has a measurable domain of approximationCompressibility Z = PV/(nRT)0.80.911.1123456Selected increasing density indexZ = 1 is ideal; deviations require a more detailed account.
Original invented nonideal trend for reasoning only; the horizontal density index is not a real gas’s calibrated pressure or density scale.

Model check: 22.4L/mol assumes about 273 K and 1atm; it is not universal at room temperature. V=nRT/P=24.63L in the ideal model. Match R to atmosphere and liter units.

Build your understanding

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Part 1 · Connect all four gas variables

The ideal gas law PV=nRT relates pressure, volume, amount and absolute temperature. The value of R depends on the unit representation: 0.0821 L·atm/(mol·K) is compatible with liters and atmospheres, while 8.314 J/(mol·K) matches pascals and cubic meters through Pa·m³=J. Using kPa and liters requires the corresponding consistent form. A correct equation with mismatched input units gives a wrong result, so cancel units before relying on the calculator display.

A gas constant is only usable with its matching unit systemA gas constant is only usable with its matching unit systemPV = nRT variableMeaningCompatible exampleunitPAbsolute pressureatmVGas volumeLnAmount of gasmolTAbsolute temperatureKFor these units: R ≈ 0.0821 L·atm/(mol·K).
Original unit organizer with rounded school constants; the four-variable law assumes ideal-gas behavior.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Why can inserting liters into an SI Pa–m³ calculation create a factor-of-1000 error?

Visual reasoning — after your attempt

One liter is 0.001 m³. The volume must match the R and pressure unit system before substitution.

PV=nRT links pressure, volume, amount, and absolute temperature for the ideal model. Choose R with units matching P and V. The equation is a relation among state variables, not a statement that every gas always behaves ideally. List each supplied quantity before rearranging.

For R=0.08206 L·atm·mol⁻¹·K⁻¹, use liters, atmospheres, moles, and Kelvin. Mixing kPa with that R without conversion produces a unit mismatch.

Pause and explain

How do you choose the value and units of R?

Check your reasoning after answering

Choose a consistent unit set for pressure and volume and use the matching gas constant. Verify units cancel to the requested quantity.

Part 2 · Solve for amount and use molar mass separately

Solving PV=nRT for n provides mole amount, not mass. To obtain mass, use the named gas’s molar mass afterward. Samples of different ideal gases with the same P, V and T have equal mole amounts in this model but can weigh differently because their molecules have different masses. A molar-volume shortcut is valid only at its stated pressure and temperature; “one mole is 22.4 L” is not a definition of the mole and is not universal.

The ideal-gas equation gives amount before chemical identity gives massThe ideal-gas equation gives amount before chemical identity gives massP = 1.00 atm; V = 2.46 L; T = 300 Kn = PV/RT = (1.00)(2.46)/(0.0821×300)n ≈ 0.100 molTo find mass, multiply by the specified gas’s molar mass.
Original selected near-0.100-mol calculation; constants and input values are rounded for school use.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Would replacing the gas by a heavier species at the same ideal P, V and T change n or its mass?

Visual reasoning — after your attempt

The calculated n stays the same, while mass changes with molar mass.

Amount n=PV/RT can be converted to mass using m=nM. Gas identity enters through molar mass for this conversion; ideal-gas volume at fixed P,T,n does not directly depend on molecular mass. Specify purity before using one molar mass for a sample.

Two gases with one mole each at the same ideal P and T occupy equal volumes despite different masses.

Pause and explain

Does a heavier gas always occupy greater volume at the same ideal amount, pressure, and temperature?

Check your reasoning after answering

No. Ideal volume depends on n,T,P, not molar mass. The corresponding mass differs between species.

Part 3 · Derive density rather than memorize a disconnected rule

Substituting n=m/Mₘ into the ideal gas law gives ρ=PMₘ/(RT). Gas density therefore depends on pressure and temperature as well as chemical identity. Rearranging to Mₘ=ρRT/P can estimate a molar mass when the sample is sufficiently ideal and its composition is appropriate. A mixture’s result can represent an average rather than one pure species. State the gas conditions and uncertainty before interpreting a density as unique substance identification.

Density connects amount and mass without replacing either quantityDensity connects amount and mass without replacing either quantityn = m/MₘPV = (m/Mₘ)RTρ = m/V = PMₘ/(RT)Mₘ = ρRT/PA density comparison must keep P, T and units compatible.
Original algebraic derivation; it does not assume liquid density follows the same temperature law.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

If temperature increases at fixed pressure for the same ideal gas, what happens to its density?

Visual reasoning — after your attempt

It decreases inversely with absolute temperature because the amount per volume decreases under the model.

Substitute n=m/M into PV=nRT to obtain density m/V=PM/RT. This shows how pressure, temperature, and identity affect ideal gas density. The expression needs consistent units and conditions; a denser sample does not necessarily imply more moles when volume differs.

At fixed P and T, a larger molar mass gives a larger ideal density. Heating at fixed pressure lowers density in the model.

Pause and explain

Why should density comparisons state temperature and pressure?

Check your reasoning after answering

Both influence m/V in the ideal relation. Unmatched conditions can hide or reverse an identity-based comparison.

Part 4 · Identify nonideal and measurement limits

The compressibility factor Z=PV/(nRT) equals one for an ideal gas. Real intermolecular attractions and finite molecular volume can produce departures, with the direction depending on conditions. A universal claim that real gases always have pressure lower than ideal ignores competing effects. Low density and temperatures sufficiently far from condensation often improve the approximation, but a numerical correction needs relevant data or a more detailed equation of state. Use observed departures to refine the model rather than erase inconvenient points.

An ideal model has a measurable domain of approximationAn ideal model has a measurable domain of approximationCompressibility Z = PV/(nRT)0.80.911.1123456Selected increasing density indexZ = 1 is ideal; deviations require a more detailed account.
Original invented nonideal trend for reasoning only; the horizontal density index is not a real gas’s calibrated pressure or density scale.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Why does a Z value different from one not by itself identify a unique chemical species?

Visual reasoning — after your attempt

It records deviation under particular conditions; composition, temperature, density and interactions can all affect the result.

High density and attractions near condensation can cause nonideal behavior. A collection over water includes water-vapor pressure; the target dry-gas pressure is not automatically the total pressure. Leaks and temperature differences also affect inferred amount.

If total pressure is 100 kPa with supplied water-vapor pressure 3 kPa, the dry gas contributes 97 kPa in the partial-pressure model.

Pause and explain

Why can using total wet-gas pressure overestimate dry-gas amount?

Check your reasoning after answering

The total includes water vapor. Assigning all pressure to the target gas incorrectly attributes the vapor’s contribution to that gas.

Apply the whole unit · Calculate gas amount and mass

Illustrative dry ideal CO₂ sample: P=1.00 atm,V=2.46 L,T=300 K. Use R=0.08206 L·atm·mol⁻¹·K⁻¹ and M(CO₂)=44.0 g/mol.

Your task

Calculate moles, mass, and density, show unit cancellation, and explain what would need adjustment if the same record described gas collected over water.

What your response should show

  • Choose the appropriate chemical model or relationship and state its conditions.
  • Show substitutions, units, particle or charge counts, and intermediate reasoning as relevant.
  • Use supplied evidence to justify the conclusion; distinguish observation from inference.
  • Identify a meaningful limitation, uncertainty, or next measurement.
Response guidance — after your attempt

  1. n=PV/RT=2.46/(0.08206×300)≈0.0999 mol.

  2. m=nM≈4.40 g.

  3. Density=m/V≈1.79 g/L.

The amount is about 0.100 mol and mass about 4.40 g under the dry ideal-gas assumption. Use unrounded n for subsequent arithmetic. For wet collection, subtract the supplied water-vapor partial pressure before assigning P to CO₂; no correction can be calculated without that additional value. Purity and temperature equilibrium also matter.

Scientific references and source-model foundations

Parent / teacher teaching notes

Require units on R and a dry/wet gas distinction. Use supplied records rather than improvised gas-generation or pressure equipment. Ask the learner to explain each particle symbol, axis, unit and arrow before calculating. Distinguish hypothetical models and calculated curves from measured records. Public OpenSciEd excerpts here are selected foundation prompts, not complete high-school teacher editions or an endorsement. All required advanced explanations and practice are provided locally.

The key idea

PV=nRT connects an ideal gas’s absolute pressure, volume, amount and temperature using compatible units. Molar volume depends on the stated pressure and temperature; 22.4 L/mol is an approximate 273.15 K, 1 atm value.

Ideal Gas Law

PV = nRT

P = pressure, V = volume, n = moles, R = ideal gas constant, T = kelvin temperature.

Ideal Gas Constant (R)

R = 0.0821 L·atm/(mol·K)

use with P in atm, V in liters, T in kelvin; R = 8.314 J/(mol·K) when working in SI units (Pa, m³).

Molar Volume at STP

1 mol of gas = 22.4 L at STP

STP = 0°C (273 K) and 1 atm; volume (L) = moles × 22.4, and moles = volume ÷ 22.4.

Avogadro's Law

V₁/n₁ = V₂/n₂

at constant temperature and pressure, volume is directly proportional to the number of moles of gas.

Gas Density & Molar Mass

M = (m·R·T)/(P·V) or density = M ÷ 22.4 (at STP)

M = molar mass (g/mol), m = sample mass (g); lets you find a gas's molar mass from a measured sample or its STP density.

Worked examples

Example 1

How many liters does 2.0 mol of an ideal gas occupy at 1.0 atm and 300 K?
  1. Solve PV = nRT for volume: V = nRT/P.
  2. Use R = 0.0821 L·atm/(mol·K) since pressure is in atm: V = (2.0 × 0.0821 × 300) / 1.0.
  3. V = 49.26 / 1.0 ≈ 49.3 L.
Answer: ≈ 49.3 L

Example 2 · Worked application

Find volume of 0.50 mol ideal gas at 300 K and 100 kPa using R=8.31 L·kPa/(mol·K).
  1. Rearrange V=nRT/P.
  2. V=0.50×8.31×300/100=12.465 L.
  3. Round to about two significant figures.
Answer: About 12 L.
Check the reasoning

Match the units of pressure and R.

Example 3

How many liters does 0.50 mol of gas occupy at STP?
  1. At STP, one mole of any ideal gas occupies 22.4 L (the molar volume).
  2. Multiply moles by the molar volume: 0.50 mol × 22.4 L/mol.
  3. = 11.2 L.
Answer: 11.2 L

Example 4 · Find and fix the mistake

Context: Find volume of 0.50 mol ideal gas at 300 K and 100 kPa using R=8.31 L·kPa/(mol·K). A student claims: “Any numerical R works regardless of the pressure and volume units.” Evaluate the claim and write a corrected explanation.
  1. Test the claim against the stated conditions: Match the units of pressure and R.
  2. R has unit-dependent numerical forms; choose one compatible with all quantities.
Answer: R has unit-dependent numerical forms; choose one compatible with all quantities.
Check the reasoning

Match the units of pressure and R.

Example 5 · Reason from the visual model

Density connects amount and mass without replacing either quantityDensity connects amount and mass without replacing either quantityn = m/MₘPV = (m/Mₘ)RTρ = m/V = PMₘ/(RT)Mₘ = ρRT/PA density comparison must keep P, T and units compatible.
Original algebraic derivation; it does not assume liquid density follows the same temperature law.
If temperature increases at fixed pressure for the same ideal gas, what happens to its density?
  1. Substituting n=m/Mₘ into the ideal gas law gives ρ=PMₘ/(RT). Gas density therefore depends on pressure and temperature as well as chemical identity. Rearranging to Mₘ=ρRT/P can estimate a molar mass when the sample is sufficiently ideal and its composition is appropriate. A mixture’s result can represent an average rather than one pure species. State the gas conditions and uncertainty before interpreting a density as unique substance identification.
  2. It decreases inversely with absolute temperature because the amount per volume decreases under the model.
Answer: It decreases inversely with absolute temperature because the amount per volume decreases under the model.
Check the reasoning

Original algebraic derivation; it does not assume liquid density follows the same temperature law.

Avoid these mistakes

Check your understanding

Find V for 1 mol at 300 K and 1atm using R=.0821.

Discuss the answer

V=nRT/P=24.63L in the ideal model. Match R to atmosphere and liter units.

Return to the phenomenon

Return to the initial case. Does the same gas volume always contain the same number of particles? Revise your drawing or explanation using two named ideas and one stated limitation.

Compare explanations after your attempt

Equal volume alone is insufficient. Equal ideal-gas P, V and T gives equal mole amount; changing pressure or temperature changes the amount compatible with that volume.

What should we investigate next?

Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.

Investigate, calculate and revise

Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.

Activity 1 · Return to the opening case

Assigned reading and preparation

Use the opening phenomenon, unit definitions and explained Parts 1–4.

Your task

Return to the initial case. Does the same gas volume always contain the same number of particles? Revise your drawing or explanation using two named ideas and one stated limitation.

Success criteria

  • Use the stated observations and appropriate units.
  • Connect a labeled model with the chemical mechanism.
  • State assumptions and an evidence limitation.
  • Explain a substantive revision or next test.
Hint

Keep species identity, material transfer and energy account distinct.

Response guidance — discuss after your attempt

Equal volume alone is insufficient. Equal ideal-gas P, V and T gives equal mole amount; changing pressure or temperature changes the amount compatible with that volume.

Activity 2 · Transfer to a changed case

Assigned reading and preparation

Cover the worked answer and use the explained models, relationships and notebook.

Your task

A gas has density 1.20 g/L at 300 K and 1.00 atm. Using R=0.0821 L·atm/(mol·K), estimate molar mass and explain why identity remains uncertain.

Success criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.
Hint

Name known quantities and conditions, show intermediate steps and check conservation.

Response guidance — discuss after your attempt

Mₘ=ρRT/P=1.20×0.0821×300/1.00≈29.6 g/mol. The estimate assumes ideal behavior and suitable composition; several substances or mixtures can have a similar average molar mass.

Words to know

ideal gas law
The equation PV = nRT, which relates the pressure, volume, moles, and Kelvin temperature of a gas in a single snapshot.
ideal gas constant (R)
The proportionality constant in PV = nRT; equal to 0.0821 L·atm/(mol·K), or 8.314 J/(mol·K) in SI units.
mole (n)
A count of 6.02 × 10²³ particles; the amount of gas, symbolized n, in the ideal gas law.
STP
Standard Temperature and Pressure: 0°C (273 K) and 1 atm, the reference conditions for comparing gases.
molar volume
The volume occupied by exactly one mole of a gas; it is 22.4 L at STP for any ideal gas.
Avogadro's Law
Equal volumes of gases at the same temperature and pressure contain equal numbers of particles; volume is directly proportional to moles (V ∝ n).
ideal gas
A hypothetical gas that perfectly obeys the KMT assumptions; real gases behave most ideally at low pressure and high temperature.

The idea nobody believed for fifty years

In 1811, the Italian scientist Amedeo Avogadro proposed that equal volumes of any gases, at the same temperature and pressure, hold equal numbers of particles. It was exactly right — and almost entirely ignored for half a century, partly because chemists couldn't accept that elements like oxygen traveled as two-atom molecules. Not until Stanislao Cannizzaro championed the idea at the 1860 Karlsruhe Congress did it finally take hold, untangling decades of confusion about atomic weights. The 'n' in PV = nRT rests on an insight that was correct long before the world was ready for it.

What 22.4 liters actually looks like — and the fine print

One mole of gas at STP fills 22.4 liters, which is bigger than it sounds: picture a cube about 28 cm on each side, roughly the volume of three basketballs. Inside that space swarm 6.02 × 10²³ molecules, whatever the gas. But read the fine print — that magic number only holds at STP (0°C and 1 atm). Warm the same mole to room temperature and it swells to about 24.5 liters, so 22.4 L is a reference point, not a universal constant to plug in everywhere.

Where the constant R actually comes from

Why is R the same number for helium, oxygen, and carbon dioxide alike? Because it is built from two of nature's most fundamental constants: R = N_A × k_B, Avogadro's number multiplied by the Boltzmann constant. Boltzmann's constant links the energy of a single particle to temperature; multiply by how many particles are in a mole and you get the energy behavior of a mole. That is why R is called the universal gas constant — it doesn't care what the gas is made of, only how many particles are moving and how warm they are.

Collecting a gas over water in the lab

When a reaction produces a gas, chemists often catch it by bubbling it up into an inverted jar of water — but the gas you collect is never pure. It shares the space with water vapor, and by Dalton's Law of partial pressures, the total pressure is the gas plus the water vapor added together. To find the true amount of your gas with PV = nRT, you have to look up the water's vapor pressure at that temperature and subtract it out first. It is a neat reminder that in the real world, gases mix and share the pressure.

When the ideal gas law starts to lie

PV = nRT rests on two convenient fictions: that molecules take up no space and never attract one another. Under everyday conditions those lies are harmless, but crank the pressure very high or the temperature very low and the molecules crowd close enough that their real volume and their mutual attractions suddenly matter. Johannes van der Waals fixed this with a corrected equation that adds a term for the space molecules occupy and a term for the pull between them. That pull is precisely the intermolecular forces of the next topic — the reason every real gas eventually condenses into a liquid.

In the real world

Automobile airbags use the ideal gas law by design: engineers calculate exactly how many moles of nitrogen gas a chemical reaction must release to inflate the bag to the right volume and pressure in a few thousandths of a second.

Exit check

Find V for 1 mol at 300 K and 1atm using R=.0821.

Check your explanation after answering

V=nRT/P=24.63L in the ideal model. Match R to atmosphere and liter units.

Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.

Problem 1

In the ideal gas law PV = nRT, what does n represent?
  1. the mass in grams
  2. the number of collisions
  3. the number of moles of gas
  4. the number of neutrons
Answer and explanation

the number of moles of gas

n is the amount of gas measured in moles.

Problem 2

Which value of R should you use when pressure is in atm, volume in liters, and temperature in kelvin?
  1. 8.314 J/(mol·K)
  2. 22.4 L/mol
  3. 6.02 × 10²³
  4. 0.0821 L·atm/(mol·K)
Answer and explanation

0.0821 L·atm/(mol·K)

R = 0.0821 L·atm/(mol·K) is the value that matches atm, liters, and kelvin.

Problem 3

What volume does 3.0 mol of an ideal gas occupy at STP?
Answer and explanation

67.2 L

Volume = moles × 22.4 L/mol = 3.0 × 22.4 = 67.2 L.

Problem 4

How many moles of gas are present in a 11.2 L sample at STP?
Answer and explanation

0.5 mol

Moles = volume ÷ 22.4 L/mol = 11.2 ÷ 22.4 = 0.50 mol.

Problem 5

A 10.0 L container holds 2.0 mol of gas at 300 K. What is the pressure, in atm?
Answer and explanation

4.93 atm

P = nRT/V = (2.0 × 0.0821 × 300) / 10.0 = 49.26 / 10.0 ≈ 4.93 atm.

Problem 6

Explain why 1 mole of oxygen gas and 1 mole of helium gas occupy the same volume at STP, even though oxygen molecules are much heavier.
Answer and explanation

By Avogadro's Law, equal volumes at the same temperature and pressure hold equal numbers of particles, regardless of their mass. The ideal gas law depends only on the number of moles (n), not the type or weight of the gas, so one mole of any ideal gas fills 22.4 L at STP.

Problem 7

Find pressure for 1.0 mol in 10 L at 300 K using the same R. Show the reasoning, not just the conclusion.
Hints

Match the units of pressure and R.

Answer and explanation

P=nRT/V=1.0×8.31×300/10=249.3 kPa, about 250 kPa.

Problem 8

Why do real gases depart from the ideal model especially at high pressure and low temperature?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Finite particle volume and intermolecular attractions become more important under those conditions.

Problem 9

Density connects amount and mass without replacing either quantityDensity connects amount and mass without replacing either quantityn = m/MₘPV = (m/Mₘ)RTρ = m/V = PMₘ/(RT)Mₘ = ρRT/PA density comparison must keep P, T and units compatible.
Original algebraic derivation; it does not assume liquid density follows the same temperature law.
Interpret the supplied visual. If temperature increases at fixed pressure for the same ideal gas, what happens to its density?
Answer and explanation

It decreases inversely with absolute temperature because the amount per volume decreases under the model.

Substituting n=m/Mₘ into the ideal gas law gives ρ=PMₘ/(RT). Gas density therefore depends on pressure and temperature as well as chemical identity. Rearranging to Mₘ=ρRT/P can estimate a molar mass when the sample is sufficiently ideal and its composition is appropriate. A mixture’s result can represent an average rather than one pure species. State the gas conditions and uncertainty before interpreting a density as unique substance identification.

It decreases inversely with absolute temperature because the amount per volume decreases under the model.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the chemistry relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 10

A gas has density 1.20 g/L at 300 K and 1.00 atm. Using R=0.0821 L·atm/(mol·K), estimate molar mass and explain why identity remains uncertain.
Answer and explanation

Mₘ=ρRT/P=1.20×0.0821×300/1.00≈29.6 g/mol. The estimate assumes ideal behavior and suitable composition; several substances or mixtures can have a similar average molar mass.

Mₘ=ρRT/P=1.20×0.0821×300/1.00≈29.6 g/mol. The estimate assumes ideal behavior and suitable composition; several substances or mixtures can have a similar average molar mass.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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