Properties of Gases¶
Syllabus mapping
CH-11-02 — Quantitative Chemistry. Covers: using particle theory to describe ideal gases; solving problems with \(PV = nRT\) for ideal gases at standard temperature and pressure (STP); using data to analyse the pressure–volume relationship at constant temperature, the temperature–pressure relationship at constant volume, and the temperature–volume relationship at constant pressure; real-world problems combining stoichiometry with reacting masses, solutions and gases.
Wording above is paraphrased for teaching use — check the official syllabus before treating any line as verbatim NESA text.
You need to know: how particle theory explains gas pressure, why gas calculations need temperature in kelvin, how pressure, volume and temperature affect each other, and how to use \(PV = nRT\) and the molar volume at STP.
Assumed knowledge check¶
Before this page makes sense, you should be comfortable with:
- Moles and solution calculations
- The particle model — gases are made of particles that move freely and fill their container
- Reading a graph and spotting a straight-line relationship — see Trends and Relationships
Core¶
Particle theory and ideal gases¶
Gases are made of tiny particles moving fast in random directions, bouncing off each other and the walls of their container. When particles hit a wall, they push on it. All those pushes added up are the gas's pressure.
To make calculations simple, chemists use a model called an ideal gas. It assumes:
- The particles themselves take up no space — the gas is almost all empty space.
- There are no attractions between the particles.
- Collisions are perfectly bouncy — no energy is lost.
- The particles' average kinetic energy (how fast they move) depends only on temperature.
Real gases behave almost exactly like this at normal temperatures and pressures, so the ideal gas model works very well for this course.
Temperature in kelvin¶
For gas calculations, temperature must be in kelvin (K):
So \(25\ ^\circ\text{C} = 298.15\ \text{K}\) and \(0\ ^\circ\text{C} = 273.15\ \text{K}\).
Why? The kelvin scale starts at absolute zero (\(0\ \text{K} = -273.15\ ^\circ\text{C}\)), the temperature where particles would have no kinetic energy at all. Because it starts at a true zero, doubling the kelvin temperature really does double the particles' energy. That isn't true in °C — 20 °C is not "twice as hot" as 10 °C.
Pressure is measured in kilopascals (kPa). Normal air pressure at sea level is about \(101.3\ \text{kPa}\).
Pressure and volume (constant temperature)¶
Squash a gas into half the volume and its pressure doubles.
Why: the same particles, moving at the same speed, now hit the walls twice as often because they have less room.
| Volume (L) | 4.0 | 2.0 | 1.0 |
|---|---|---|---|
| Pressure (kPa) | 50 | 100 | 200 |
| \(P \times V\) | 200 | 200 | 200 |
Pressure and volume are inversely proportional: one goes up as the other goes down, and their product stays constant. A graph of \(P\) against \(V\) is a curve; a graph of \(P\) against \(\dfrac{1}{V}\) is a straight line through the origin.
Worked example
A \(2.0\ \text{L}\) sample of gas at \(100\ \text{kPa}\) is compressed to \(0.50\ \text{L}\) at the same temperature. What is the new pressure?
Check: volume went down by 4 times, so pressure went up by 4 times ✓
Temperature and pressure (constant volume)¶
Heat a gas in a sealed, rigid container and its pressure goes up.
Why: hotter particles move faster, so they hit the walls harder and more often.
Pressure is directly proportional to kelvin temperature: double \(T\), double \(P\). This is why aerosol cans warn "do not heat", and why tyre pressures go up on a hot day.
Worked example
A car tyre is at \(220\ \text{kPa}\) at \(15\ ^\circ\text{C}\). After a long drive the air inside is at \(45\ ^\circ\text{C}\). Assuming the volume doesn't change, what is the new pressure?
Convert to kelvin: \(T_1 = 288.15\ \text{K}\), \(T_2 = 318.15\ \text{K}\)
Temperature and volume (constant pressure)¶
Heat a gas in a container that can expand (like a balloon or a syringe with a free-moving plunger) and its volume goes up.
Why: faster particles push harder on the walls, so the container expands until the pressure inside matches the pressure outside again.
Volume is directly proportional to kelvin temperature. A balloon taken from a warm room into a freezer visibly shrinks.
Worked example
A balloon has a volume of \(3.0\ \text{L}\) at \(25\ ^\circ\text{C}\). It's put in a freezer at \(-18\ ^\circ\text{C}\). What is its new volume?
Convert to kelvin: \(T_1 = 298.15\ \text{K}\), \(T_2 = 255.15\ \text{K}\)
Summary of the three relationships¶
| Kept constant | Relationship | Equation | Graph (with \(T\) in kelvin) |
|---|---|---|---|
| Temperature | \(P\) up → \(V\) down | \(P_1V_1 = P_2V_2\) | \(P\) vs \(V\): curve. \(P\) vs \(\frac{1}{V}\): straight line |
| Volume | \(T\) up → \(P\) up | \(\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}\) | \(P\) vs \(T\): straight line through the origin |
| Pressure | \(T\) up → \(V\) up | \(\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}\) | \(V\) vs \(T\): straight line through the origin |
🖼️ Diagram needed — three small graphs side by side matching the table above (P vs V curve; P vs T straight line; V vs T straight line), each with a tiny particle-in-a-box sketch underneath showing the "why". TODO — placeholder until sourced/drawn.
The ideal gas law¶
All three relationships (plus the number of moles) combine into one equation:
| Symbol | Means | Unit to use |
|---|---|---|
| \(P\) | Pressure | kPa |
| \(V\) | Volume | L |
| \(n\) | Amount of gas | mol |
| \(R\) | Gas constant, \(8.314\) | J mol⁻¹ K⁻¹ (works with kPa and L) |
| \(T\) | Temperature | K |
With pressure in kPa and volume in litres, \(R = 8.314\) works directly — no extra conversions needed. Check your data sheet for the value of \(R\).
Worked example
How many moles of gas are in a \(5.00\ \text{L}\) container at \(100\ \text{kPa}\) and \(25\ ^\circ\text{C}\)?
Convert: \(T = 298.15\ \text{K}\)
Standard temperature and pressure (STP) and molar volume¶
STP means \(0\ ^\circ\text{C}\) (\(273.15\ \text{K}\)) and \(100\ \text{kPa}\). At STP, one mole of any gas takes up the same volume — the molar volume:
So at STP, a handy shortcut:
(Your data sheet lists molar volumes — always check which conditions a value is for.)
Worked example
What volume does \(0.50\ \text{mol}\) of oxygen gas occupy at STP?
Putting it together: stoichiometry with gases¶
Gas questions use the same four-step method as reacting-mass and solution questions. Just use \(V = n \times 22.71\) (at STP) or \(PV = nRT\) for the gas step.
Worked example — mass to gas volume
What volume of carbon dioxide, measured at STP, is produced when \(10.0\ \text{g}\) of calcium carbonate reacts with excess hydrochloric acid?
Step 1: \(n(\ce{CaCO3}) = \dfrac{10.0}{100.09} = 0.0999\ \text{mol}\)
Step 2: \(\ce{CaCO3} : \ce{CO2} = 1 : 1\), so \(n(\ce{CO2}) = 0.0999\ \text{mol}\)
Step 3: \(V = 0.0999 \times 22.71 = 2.27\ \text{L}\)
Worked example — solution to gas volume
Excess magnesium is added to \(50.0\ \text{mL}\) of \(1.00\ \text{mol L}^{-1}\) hydrochloric acid. What volume of hydrogen gas forms at STP?
Step 1: \(n(\ce{HCl}) = C \times V = 1.00 \times 0.0500 = 0.0500\ \text{mol}\)
Step 2: \(\ce{HCl} : \ce{H2} = 2 : 1\), so \(n(\ce{H2}) = \dfrac{0.0500}{2} = 0.0250\ \text{mol}\)
Step 3: \(V = 0.0250 \times 22.71 = 0.568\ \text{L}\) (568 mL)
Practical¶
The syllabus asks you to use data to analyse each relationship. A simple set-up for each:
- Pressure–volume (constant T): a sealed syringe connected to a pressure sensor (or a syringe pushed down with known masses on top). Record pressure at several volumes, then plot \(P\) against \(\dfrac{1}{V}\).
- Temperature–pressure (constant V): a sealed flask connected to a pressure sensor, sitting in water baths at different temperatures. Plot \(P\) against \(T\) (in kelvin).
- Temperature–volume (constant P): a syringe or a capillary tube with a trapped bubble of air, in water baths at different temperatures. Plot \(V\) against \(T\).
Give the gas time to reach the water's temperature before each reading. See Representing Data for why plotting \(P\) against \(\dfrac{1}{V}\) is better than \(P\) against \(V\).
Advanced
The combined gas law. When pressure, volume and temperature all change (for a fixed amount of gas):
Example: \(2.00\ \text{L}\) of gas at \(100\ \text{kPa}\) and \(27\ ^\circ\text{C}\) is changed to \(200\ \text{kPa}\) and \(127\ ^\circ\text{C}\). \(V_2 = \dfrac{P_1V_1T_2}{T_1P_2} = \dfrac{100 \times 2.00 \times 400.15}{300.15 \times 200} = 1.33\ \text{L}\).
Finding the molar mass of a gas. If \(1.00\ \text{g}\) of a gas occupies \(0.516\ \text{L}\) at STP: \(n = \dfrac{0.516}{22.71} = 0.0227\ \text{mol}\), so \(MM = \dfrac{m}{n} = \dfrac{1.00}{0.0227} = 44.0\ \text{g mol}^{-1}\) — consistent with \(\ce{CO2}\).
Gas volume ratios. Because equal volumes of gases (at the same temperature and pressure) contain equal numbers of moles, the mole ratio in an equation is also the volume ratio for gases. In \(\ce{N2 + 3H2 -> 2NH3}\), \(1\ \text{L}\) of nitrogen reacts with \(3\ \text{L}\) of hydrogen to make \(2\ \text{L}\) of ammonia — no molar masses needed.
Not at STP? Use \(PV = nRT\) for the gas step instead of 22.71. Don't use 22.71 L at any conditions other than \(0\ ^\circ\text{C}\) and \(100\ \text{kPa}\).
Extension
Beyond the syllabus for this course — won't appear in the HSC, included for interest / depth study inspiration.
When gases aren't ideal. At very high pressure, the particles are squashed so close that their own size matters. At very low temperature, they move slowly enough for attractions between them to matter — eventually the gas turns into a liquid. Real gases then don't follow \(PV = nRT\) exactly. The van der Waals equation adds two correction terms, one for particle size and one for attractions.
📎 Depth study idea
Finding absolute zero. Measure the pressure of a fixed volume of air at several temperatures (ice water up to near-boiling water), plot pressure against temperature in °C, and extrapolate the straight line back to where pressure would be zero. How close does your result get to \(-273.15\ ^\circ\text{C}\)? Evaluate what limits your accuracy.
Video/visual resources¶
- 🖥️ PhET Simulation — Gases Intro — a simpler starting point: pressure, volume and temperature with one variable held constant
- 🖥️ PhET Simulation — Gas Properties — hold one variable constant and watch the particles while you change the others
- 🎥 TODO: source a short demonstration video (e.g. a balloon in liquid nitrogen, or marshmallows in a syringe under reduced pressure).
Check yourself¶
-
Convert (a) \(30\ ^\circ\text{C}\) and (b) \(-20\ ^\circ\text{C}\) to kelvin.
Answer
(a) \(30 + 273.15 = 303.15\ \text{K}\). (b) \(-20 + 273.15 = 253.15\ \text{K}\).
-
Use particle theory to explain why the pressure of a gas increases when it's heated in a sealed container.
Answer
Heating makes the particles move faster. They hit the walls of the container more often and with more force, so the pressure increases.
-
\(6.0\ \text{L}\) of gas at \(120\ \text{kPa}\) is compressed at constant temperature until the pressure is \(300\ \text{kPa}\). What is the new volume?
Answer
\(V_2 = \dfrac{P_1V_1}{P_2} = \dfrac{120 \times 6.0}{300} = 2.4\ \text{L}\)
-
A syringe holds \(1.5\ \text{L}\) of air at \(20\ ^\circ\text{C}\). At constant pressure, what is the volume at \(80\ ^\circ\text{C}\)?
Answer
\(T_1 = 293.15\ \text{K}\), \(T_2 = 353.15\ \text{K}\). \(V_2 = \dfrac{1.5 \times 353.15}{293.15} = 1.8\ \text{L}\)
-
What pressure is exerted by \(0.100\ \text{mol}\) of gas in a \(2.00\ \text{L}\) container at \(300\ \text{K}\)?
Answer
\(P = \dfrac{nRT}{V} = \dfrac{0.100 \times 8.314 \times 300}{2.00} = 125\ \text{kPa}\)
-
What volume of hydrogen gas at STP is produced when \(0.486\ \text{g}\) of magnesium reacts with excess acid? (\(\ce{Mg + 2HCl -> MgCl2 + H2}\))
Answer
\(n(\ce{Mg}) = \dfrac{0.486}{24.31} = 0.0200\ \text{mol}\). Ratio \(1 : 1\), so \(n(\ce{H2}) = 0.0200\ \text{mol}\). \(V = 0.0200 \times 22.71 = 0.454\ \text{L}\).