How do I know whether to use the combined gas law or PV = nRT?

Count the sets of conditions. If the problem describes a gas changing — before and after, initial and final, at the surface and at altitude — you have two states and the combined gas law applies. If it describes a single situation and mentions moles, grams, or molar mass, use PV = nRT. A tell-tale sign for the combined gas law is that the amount of gas is never given and never changes, so it cancels out of the comparison.

Which value of R should I use?

Match R to the pressure unit in your data. Use 0.0821 with atmospheres, 8.314 with kilopascals, and 62.4 with mmHg or torr. All three require liters, kelvin, and moles no matter which one you pick. If your data mix pressure units, convert everything to one unit first, then choose the matching R.

Why can't I just use Celsius if I use it consistently on both sides?

Because the gas laws are proportionalities that require a scale starting at true zero. Celsius zero is the freezing point of water, not the absence of molecular motion, so ratios of Celsius temperatures are meaningless. Worse, T sits in a denominator in the combined gas law, so 0\ ^ would mean dividing by zero and any temperature below freezing would predict a negative volume.

Where does the 22.4 L per mole at STP number come from?

It is just PV = nRT evaluated at standard conditions. Substituting n = 1\ mol, P = 1\ atm, T = 273\ K, and R = 0.0821 gives V = 22.4\ L. Because it is a consequence of the ideal gas law rather than a separate rule, it applies only at STP — at any other temperature or pressure you must recalculate. It is still a handy benchmark for checking whether an answer is physically reasonable.

CHEM-7.4

The Combined & Ideal Gas Laws

Master the combined gas law and PV = nRT: pick the right R, convert to kelvin, match your units, and solve for pressure, volume, temperature, or moles.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on The Combined & Ideal Gas Laws, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start The Combined & Ideal Gas Laws free

What this lesson covers

You already know Boyle's, Charles's, and Gay-Lussac's laws, each holding two variables fixed while two others trade off. Real gas samples rarely cooperate that neatly — a weather balloon rising through the atmosphere changes pressure, volume, and temperature all at once. The combined gas law bundles those three relationships into one equation, and the ideal gas law goes one step further by bringing in the amount of gas, in moles.

This lesson shows you how to decide which equation a problem calls for, how to choose the right value of the gas constant

R

, and why every temperature must be in kelvin before it touches an equation. The math is short; the care is in the setup. Most mistakes in this topic are not algebra errors at all — they are unit errors, and they are completely preventable once you know where to look.

The Combined Gas Law: One Equation for Changing Conditions

The combined gas law merges Boyle's, Charles's, and Gay-Lussac's laws into a single relationship for a fixed amount of gas:

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

Read it as a statement that the quantity

\frac{PV}{T}

stays constant for a sealed sample. If the gas is compressed, its pressure must rise or its temperature must fall to keep the ratio balanced.

Each of the earlier laws is hiding inside this one. If temperature is held constant,

T_1 = T_2

cancels and you are left with

P_1V_1 = P_2V_2

, which is Boyle's law. If pressure is constant, you get

\frac{V_1}{T_1} = \frac{V_2}{T_2}

, Charles's law. If volume is constant — a rigid steel tank, for instance — you get

\frac{P_1}{T_1} = \frac{P_2}{T_2}

, Gay-Lussac's law. That means you never have to memorize four equations. Memorize one and cancel whatever does not change.

Because the combined gas law compares two states of the same sample, units only have to be consistent between the two sides, not any particular set. If

V_1

is in milliliters,

V_2

comes out in milliliters. If

P_1

is in kPa, use kPa for

P_2

. The one exception is temperature:

T

appears in a denominator, so a Celsius value of zero or a negative Celsius value would break the math entirely. Temperature must always be absolute.

A useful habit is to solve for the unknown symbolically first. For

V_2

V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1}

Then check the two ratios against physical sense. If pressure dropped,

\frac{P_1}{P_2}

should be greater than 1 and the volume should grow.

Why Temperature Must Be in Kelvin

Gas laws describe proportionality, and proportionality only works from a true zero. The Celsius scale places its zero at the freezing point of water, an arbitrary reference. Absolute zero,

-273.15\ ^\circ\text{C}

, is where the kinetic-molecular model says particle motion stops, so it is the only zero that makes "twice the temperature means twice the average kinetic energy" a true statement.

Convert with

T(\text{K}) = T(^\circ\text{C}) + 273.15

, or

+273

when the data have three significant figures.

Here is the concrete damage Celsius does. Take a gas at

10.0\ ^\circ\text{C}

that is warmed to

20.0\ ^\circ\text{C}

at constant pressure. Using Celsius numbers you would predict the volume doubles. Using kelvin,

283.15\ \text{K} \to 293.15\ \text{K}

, the volume grows by about 3.5 percent. The Celsius answer is not slightly off; it is wrong by a factor of nearly 30.

Situation	Celsius value	Kelvin value	What Celsius does to the math
Ice water	$0\ ^\circ\text{C}$	$273\ \text{K}$	Division by zero
Dry ice	$-78\ ^\circ\text{C}$	$195\ \text{K}$	Negative volume predicted
Room temp	$25\ ^\circ\text{C}$	$298\ \text{K}$	Ratios badly distorted

The reverse error also shows up: students convert to kelvin, solve correctly, and then report a final temperature of

350

without noting the unit, or subtract

273

when the question wanted kelvin. Write the unit on every line, and if a problem gives Celsius data it usually expects a Celsius answer — convert back at the end, after the algebra is finished.

The Ideal Gas Law and Choosing R

The combined gas law compares two states, but it cannot tell you how much gas you have. The ideal gas law can:

PV = nRT

Here

n

is moles and

R

is the universal gas constant. Use

PV = nRT

whenever a problem involves one set of conditions and mentions moles, grams, or molar mass. Use the combined gas law when a sample changes from one set of conditions to another.

R

is one physical constant written in different units, and the units you pick must match the units in your data.

Value of $R$	Units	Use when pressure is in
$0.0821$	$\frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}$	atmospheres
$8.314$	$\frac{\text{L}\cdot\text{kPa}}{\text{mol}\cdot\text{K}}$	kilopascals
$62.4$	$\frac{\text{L}\cdot\text{mmHg}}{\text{mol}\cdot\text{K}}$	mmHg or torr

Every version demands volume in liters, temperature in kelvin, and amount in moles. Milliliters must become liters (

\div 1000

); grams must become moles (divide by molar mass).

A quick sanity check: at standard temperature and pressure,

273\ \text{K}

and

1\ \text{atm}

, one mole of an ideal gas occupies

V = \frac{nRT}{P} = \frac{(1)(0.0821)(273)}{1} \approx 22.4\ \text{L}

If your answer for roughly one mole near room conditions comes out near

0.0224

22{,}400

, you dropped or added a factor of 1000 somewhere in a volume conversion.

The law is called ideal for a reason: it assumes particles have no volume and no attractions. Real gases follow it closely at ordinary temperatures and moderate pressures, and deviate at very high pressure or very low temperature, where particles are crowded and attractions matter.

Setting Up Problems Without Getting Lost

A reliable routine handles almost every problem in this topic.

First, list what you are given with units, and mark the unknown. Second, decide which equation fits: two states means the combined gas law, one state with an amount of gas means

PV = nRT

. Third, convert — kelvin always, liters for

PV = nRT

, and pressures matched to your chosen

R

. Fourth, rearrange algebraically before substituting numbers. Fifth, check whether the answer moved in the direction physics says it should.

The most common places students go wrong:

Forgetting that a rigid or sealed container fixes volume. Words like "steel cylinder," "rigid tank," or "bulb" mean

V_1 = V_2

, so those terms cancel.

Mixing pressure units within one problem — using

760\ \text{mmHg}

for the initial pressure and

0.85\ \text{atm}

for the final. Convert one to the other first. Remember

1\ \text{atm} = 760\ \text{mmHg} = 760\ \text{torr} = 101.3\ \text{kPa}

.

Using

PV = nRT

for a problem where moles never appear. If the amount of gas is unchanged and unmentioned, the combined gas law is faster and needs no constant.

Inverting a ratio. If a gas is cooled at constant pressure, the volume must shrink; if your answer grew, you multiplied by

\frac{T_1}{T_2}

instead of

\frac{T_2}{T_1}

.

Leaving mass as grams.

PV = nRT

has no slot for grams. Convert with

n = \frac{m}{M}

, where

M

is molar mass in grams per mole. Substituting that in gives

PV = \frac{m}{M}RT

, which is how these problems connect to finding an unknown gas's molar mass.

Key terms

Combined gas law.: The relationship $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ , valid for a fixed amount of gas moving between two sets of conditions.
Ideal gas law.: $PV = nRT$ , relating pressure, volume, moles, and absolute temperature of a gas at a single set of conditions.
Universal gas constant (R).: The proportionality constant in $PV = nRT$ ; equal to $0.0821\ \frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}$ , $8.314\ \frac{\text{L}\cdot\text{kPa}}{\text{mol}\cdot\text{K}}$ , or $62.4\ \frac{\text{L}\cdot\text{mmHg}}{\text{mol}\cdot\text{K}}$ .
Absolute temperature.: Temperature measured on the Kelvin scale, whose zero is absolute zero; found from $T(\text{K}) = T(^\circ\text{C}) + 273.15$ .
Ideal gas.: A model gas whose particles occupy no volume and exert no attractive forces on one another; real gases approach this behavior at low pressure and high temperature.
STP.: Standard temperature and pressure, $273\ \text{K}$ and $1\ \text{atm}$ , at which one mole of an ideal gas occupies $22.4\ \text{L}$ .
Molar volume.: The volume occupied by one mole of a gas at stated conditions; $22.4\ \text{L/mol}$ at STP.
Rigid container.: A vessel whose volume cannot change, making $V_1 = V_2$ so the volume terms cancel from the combined gas law.

Worked example

A weather balloon holds 4.50 L of helium at 1.05 atm and 22 °C. It rises until the surrounding pressure is 0.720 atm and the temperature is −8 °C. (a) What is the new volume of the balloon? (b) How many moles of helium are in the balloon?

Part (a) — two states, so use the combined gas law.

List and convert.

P_1 = 1.05\ \text{atm}

V_1 = 4.50\ \text{L}

T_1 = 22 + 273 = 295\ \text{K}

P_2 = 0.720\ \text{atm}

V_2 = ?

T_2 = -8 + 273 = 265\ \text{K}

. Both pressures are already in atm, so no pressure conversion is needed.

Rearrange before substituting:

V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1}

Substitute:

V_2 = 4.50\ \text{L} \times \frac{1.05}{0.720} \times \frac{265}{295}

V_2 = 4.50 \times 1.458 \times 0.8983 = 5.89\ \text{L}

Sense check: the pressure dropped a lot (expansion) while the temperature dropped only a little (slight contraction). Expansion should win, and it did — 5.89 L is larger than 4.50 L.

Part (b) — one state with an amount of gas, so use $PV = nRT$ .

Use the ground-level conditions and

R = 0.0821\ \frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}

because pressure is in atm.

n = \frac{PV}{RT} = \frac{(1.05\ \text{atm})(4.50\ \text{L})}{(0.0821)(295\ \text{K})} = \frac{4.725}{24.22} = 0.195\ \text{mol}

Verify with the high-altitude conditions, since the balloon is sealed and

n

cannot change:

n = \frac{(0.720)(5.89)}{(0.0821)(265)} = \frac{4.24}{21.76} = 0.195\ \text{mol}

The two agree, which confirms part (a). That cross-check is worth doing whenever a problem gives you both states.

Practice questions

A sealed 10.0 L flask contains 2.00 mol of nitrogen gas at 27 °C. What is the pressure inside the flask?

0.164 atm
0.493 atm
4.93 atm
44.3 atm

Answer: 4.93 atm

Only one set of conditions is given and moles are named, so use

PV = nRT

. Convert the temperature first:

27 + 273 = 300\ \text{K}

. Using

R = 0.0821\ \frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}

P = \frac{nRT}{V} = \frac{(2.00)(0.0821)(300)}{10.0} = \frac{49.26}{10.0} = 4.93\ \text{atm}

. The value 44.3 atm comes from leaving the temperature in Celsius and multiplying incorrectly, and 0.493 atm comes from a misplaced decimal in the division. A quick reality check helps: 2 moles squeezed into 10 L is denser than the 22.4 L per mole at STP, so the pressure should be several atmospheres, not a fraction of one.

A rigid steel cylinder of oxygen reads 15.0 atm at 20.0 °C. It is left in a hot truck and warms to 65.0 °C. Calculate the new pressure, and explain why you do not need to know the cylinder's volume.

Answer: About 17.3 atm.

Start from the combined gas law,

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

. The word rigid means the cylinder cannot expand, so

V_1 = V_2

and both volume terms cancel, leaving

\frac{P_1}{T_1} = \frac{P_2}{T_2}

— Gay-Lussac's law emerging from the general equation. Because volume cancels algebraically, its actual value never matters. Convert temperatures:

T_1 = 293\ \text{K}

and

T_2 = 338\ \text{K}

. Then

P_2 = P_1 \times \frac{T_2}{T_1} = 15.0 \times \frac{338}{293} = 17.3\ \text{atm}

. Heating raised the pressure, which matches the kinetic-molecular picture: faster particles strike the walls harder and more often in the same space.

A student solving a gas problem writes

PV = nRT

with

P = 745\ \text{mmHg}

V = 250.\ \text{mL}

T = 25\ ^\circ\text{C}

, and

R = 0.0821\ \frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}

. Identify every unit problem and give the corrected values.

Answer: Three problems: pressure must be converted to atm (0.980 atm), volume to liters (0.250 L), and temperature to kelvin (298 K).

The chosen

R

carries units of

\frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}

, so every quantity substituted into the equation must use those same units. Pressure:

745\ \text{mmHg} \times \frac{1\ \text{atm}}{760\ \text{mmHg}} = 0.980\ \text{atm}

. Volume:

250.\ \text{mL} \times \frac{1\ \text{L}}{1000\ \text{mL}} = 0.250\ \text{L}

. Temperature:

25 + 273 = 298\ \text{K}

. An equally valid alternative is to keep the pressure in mmHg and switch to

R = 62.4\ \frac{\text{L}\cdot\text{mmHg}}{\text{mol}\cdot\text{K}}

, but volume and temperature still need converting. With the corrections,

n = \frac{(0.980)(0.250)}{(0.0821)(298)} = 0.0100\ \text{mol}

FAQ

How do I know whether to use the combined gas law or PV = nRT?: Count the sets of conditions. If the problem describes a gas changing — before and after, initial and final, at the surface and at altitude — you have two states and the combined gas law applies. If it describes a single situation and mentions moles, grams, or molar mass, use $PV = nRT$ . A tell-tale sign for the combined gas law is that the amount of gas is never given and never changes, so it cancels out of the comparison.
Which value of R should I use?: Match $R$ to the pressure unit in your data. Use $0.0821$ with atmospheres, $8.314$ with kilopascals, and $62.4$ with mmHg or torr. All three require liters, kelvin, and moles no matter which one you pick. If your data mix pressure units, convert everything to one unit first, then choose the matching $R$ .
Why can't I just use Celsius if I use it consistently on both sides?: Because the gas laws are proportionalities that require a scale starting at true zero. Celsius zero is the freezing point of water, not the absence of molecular motion, so ratios of Celsius temperatures are meaningless. Worse, $T$ sits in a denominator in the combined gas law, so $0\ ^\circ\text{C}$ would mean dividing by zero and any temperature below freezing would predict a negative volume.
Where does the 22.4 L per mole at STP number come from?: It is just $PV = nRT$ evaluated at standard conditions. Substituting $n = 1\ \text{mol}$ , $P = 1\ \text{atm}$ , $T = 273\ \text{K}$ , and $R = 0.0821$ gives $V = 22.4\ \text{L}$ . Because it is a consequence of the ideal gas law rather than a separate rule, it applies only at STP — at any other temperature or pressure you must recalculate. It is still a handy benchmark for checking whether an answer is physically reasonable.

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