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How do pressure, volume and temperature of a gas relate, and how do you use the combined gas law?

The gas laws: use the combined gas law to relate the pressure, volume and Kelvin temperature of a fixed mass of gas, with STP from Table A.

A focused Regents Chemistry answer on the gas laws: the qualitative pressure-volume and volume-temperature relationships, the combined gas law from Table T, the use of Kelvin temperature, and STP values from Table A, with a worked calculation.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-13

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What this topic is asking
The qualitative relationships
The combined gas law
STP and units
Equal volumes, equal numbers
Try this

What this topic is asking

The Core Curriculum asks you to relate the pressure, volume and temperature of a fixed mass of gas, qualitatively and through the combined gas law. The Regents provides the combined gas law on Table T and the STP values on Table A, so this topic is about setting up and solving the relationship with Kelvin temperatures.

The qualitative relationships

Both follow from the kinetic molecular theory. Squeezing a gas into a smaller volume makes the particles strike the walls more often, raising the pressure. Heating a gas makes the particles move faster, so they push the walls out (at constant pressure) and the volume grows.

The combined gas law

This single equation contains Boyle's law (hold $T$ constant) and Charles's law (hold $P$ constant) as special cases. To use it, identify the initial conditions ( $P_1, V_1, T_1$ ) and the final conditions, convert any Celsius temperatures to Kelvin, then solve for the unknown by rearranging.

STP and units

Because pressure appears on both sides of the equation, its units cancel as long as you use the same unit throughout, so you do not have to convert kPa to atm (or vice versa) provided both pressures match.

Equal volumes, equal numbers

A useful idea behind the gas laws is Avogadro's hypothesis: equal volumes of gases at the same temperature and pressure contain equal numbers of particles. This is why gas behavior does not depend on the identity of the gas in the ideal model, and why the combined gas law works the same way for any gas. It also explains why one mole of any gas occupies the same volume at STP, a result the Regents may use when relating moles of a gas to its volume. The practical consequence is that you can compare the amounts of two gases just by comparing their volumes, as long as the temperature and pressure are the same.

Using the combined gas law

A gas has a volume of $2.0$ L at $300$ K and $101.3$ kPa. What is its volume at $150$ K and $101.3$ kPa?

step 1 Write the combined gas law

$\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$ .

step 2 Note the constant pressure

Pressure is unchanged ( $101.3$ kPa both times), so it cancels, leaving $\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$ (Charles's law).

step 3 Solve for the new volume

$V_2 = V_1 \times \dfrac{T_2}{T_1} = 2.0\ \text{L} \times \dfrac{150}{300} = 1.0$ L.

step 4 Sense-check

Halving the Kelvin temperature at constant pressure should halve the volume, and $1.0$ L is half of $2.0$ L, so the result is consistent.

Try this

Q1. A gas at $200$ kPa and $4.0$ L is compressed to $2.0$ L at constant temperature. Find the new pressure. [2 points]

Cue. $P_1 V_1 = P_2 V_2$ , so $P_2 = \dfrac{200 \times 4.0}{2.0} = 400$ kPa.

Q2. State the standard temperature and standard pressure values from Table A. [1 point]

Cue. $273$ K and $101.3$ kPa.

Exam-style practice questions

Practice questions written in the style of NYSED exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Regents (Part C style)3 marksA gas occupies a volume of

4.0

L at a pressure of

100

kPa and a temperature of

300

K. Determine the new volume of the gas at a pressure of

200

kPa and a temperature of

600

K. Show your work using the combined gas law.

Show worked answer →

A 3-point Part C calculation with the combined gas law from Table T.

Write the combined gas law: $\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$ . Solve for $V_2$ : $V_2 = \dfrac{P_1 V_1 T_2}{T_1 P_2} = \dfrac{(100)(4.0)(600)}{(300)(200)}$ .

Compute: $\dfrac{240000}{60000} = 4.0$ L. The doubling of pressure (which would halve the volume) is exactly offset by the doubling of Kelvin temperature (which would double the volume), so the volume is unchanged at $4.0$ L.

Markers reward the correct combined-gas-law setup (1 point), correct substitution with Kelvin temperatures (1 point), and the answer $4.0$ L (1 point).

Regents (Part A style)1 marksAt constant temperature, as the pressure on a fixed mass of gas increases, the volume of the gas (1) increases (2) decreases (3) remains the same (4) doubles

Show worked answer →

A 1-point Part A item on the pressure-volume relationship (Boyle's law). The answer is (2) decreases.

At constant temperature, pressure and volume are inversely related: squeezing the gas into a smaller space raises its pressure, and lowering the pressure lets it expand. So as pressure increases, volume decreases (their product $PV$ stays constant). This follows from the kinetic molecular theory, since the same particles hitting a smaller area more often raise the pressure.

Markers reward recognizing the inverse pressure-volume relationship at constant temperature.

Related dot points

Sources & how we know this

Physical Setting/Chemistry Core Curriculum — New York State Education Department (2002)
Reference Tables for Physical Setting/Chemistry, 2011 Edition — New York State Education Department (2011)