Explain gas behavior with kinetic-molecular theory and solve pressure, volume, temperature and amount problems

Learning goal: Use the kinetic-molecular theory to explain the pressure, volume and temperature of a gas through the motion and collisions of its particles; identify and convert units of pressure, volume and temperature, using the Kelvin scale; describe and apply the relationships between any two of pressure, volume, temperature and amount when the others are held constant; and solve problems with the ideal gas law PV = nRT, including the molar volume of a gas at standard temperature and pressure.

The particle picture. The kinetic-molecular theory describes a gas as a very large number of tiny particles in constant, rapid, random motion, far apart compared with their size, moving in straight lines until they collide with one another or the container, with no attractions between them and no energy lost in collisions. Every gas property follows. Pressure is the force of the particles' collisions on each unit area of the container wall: more collisions per second, or harder collisions, mean higher pressure. Temperature on the Kelvin scale is directly proportional to the average kinetic energy of the particles: double the Kelvin temperature and the average kinetic energy doubles. Volume is simply the space the particles fly around in. Because the particles are so far apart, a gas is compressible and any two gases mix completely. Real gases deviate from this picture at high pressure and low temperature, when the particles are close enough for their attractions to matter; those deviations are reserved for later study, and the gases here are treated as ideal.

Units, and why Kelvin. Pressure is measured in kilopascals (kPa), atmospheres (atm) and millimeters of mercury (mmHg, also called torr): 1 atm = 101.325 kPa = 760 mmHg exactly. Volume is in liters or milliliters; 1 L = 1000 mL. Temperature in gas calculations must be in kelvin, K = °C + 273.15, usually rounded to °C + 273. The reason is not convention. Only on the Kelvin scale is temperature proportional to average kinetic energy, with zero at absolute zero, the temperature at which, in the ideal-gas picture, the particles' kinetic energy would extrapolate to zero (real substances are no longer gases long before, and quantum physics leaves a tiny residual motion). On the Celsius scale doubling 10 °C to 20 °C does not double anything physical, and a temperature of 0 °C would make every gas law predict zero volume or zero pressure. To convert 745 mmHg to atmospheres: 745 / 760 = 0.980 atm; to kilopascals: 0.980 × 101.325 = 99.3 kPa.

Two variables at a time. With four quantities, pressure P, volume V, temperature T and amount of gas n, the laws describe what happens when two are held constant. Pressure and volume, with T and n fixed: they are inversely proportional, P₁V₁ = P₂V₂. Squeeze a gas into half the volume and the particles hit the walls twice as often, so the pressure doubles. Volume and temperature, with P and n fixed: directly proportional, V₁/T₁ = V₂/T₂. Warm a gas in a container that can expand, such as a balloon, and the faster particles push the walls outward until the collision rate per area is back to matching the outside pressure. Pressure and temperature, with V and n fixed: directly proportional, P₁/T₁ = P₂/T₂. Warm a gas in a rigid container, such as a tire or a sealed can, and the harder, more frequent collisions raise the pressure; this is why tire pressure rises on a hot road and why a sealed can must never be heated. Volume and amount, with P and T fixed: directly proportional, because equal volumes of any gases at the same temperature and pressure contain equal numbers of particles, which is Avogadro's law; blowing more air into a balloon makes it larger. The three two-variable laws combine into P₁V₁/T₁ = P₂V₂/T₂ for a fixed amount of gas. The names attached to these laws matter less than being able to state which pair is related, in which direction, and why the particles behave that way.

The ideal gas law. For a single unchanging state of a gas the four quantities are tied together by PV = nRT, where R is the ideal gas constant. Its value depends on the units: 8.31 L·kPa/(mol·K) when pressure is in kilopascals and volume in liters, 0.0821 L·atm/(mol·K) when pressure is in atmospheres; the energy form 8.31 J/(mol·K) is the same constant. Match the units of R to the units of the data, and always use kelvin. At standard temperature and pressure, 0 °C (273 K) and 1 atm (101.3 kPa), one mole of an ideal gas occupies V = nRT/P = 1 × 8.31 × 273 / 101.3 = 22.4 L, the molar volume, which allows gas volumes at STP to be converted to moles directly; at other conditions use the full equation. Some references define STP with 100 kPa, which gives 22.7 L, so state the conditions you are using.

A worked model: a model airbag. A real airbag fills with gas made by a fast chemical reaction; a safe classroom model uses baking soda and vinegar in a sealable plastic bag: NaHCO₃ + CH₃COOH → NaCH₃COO + H₂O + CO₂. How much baking soda is needed to fill a 5.00 L bag at room pressure, 101 kPa, and 22 °C, when the bag also holds 250 mL of vinegar? The liquid takes up room, so the gas has 5.00 − 0.25 = 4.75 L to fill. Convert temperature: 22 + 273 = 295 K. Moles of CO₂ needed: n = PV/RT = (101 × 4.75) / (8.31 × 295) = 479.75 / 2451 = 0.1957 mol. Each mole of NaHCO₃ makes one mole of CO₂, so 0.1957 mol of NaHCO₃ is needed, and with molar mass 84.01 g/mol that is 0.1957 × 84.01 = 16.4 g. The vinegar must be in excess: 250 mL of 5 percent vinegar holds about 0.21 mol of acetic acid, a little more than the 0.196 mol required. (Ignoring the liquid and using the whole 5.00 L would give 0.206 mol and 17.3 g, an overestimate.) When the experiment is done, the bag usually looks less than full: some carbon dioxide dissolves in the liquid, some escapes while sealing, the bag stretches, and the reaction may not finish. The gap between prediction and result is where the thinking happens.

Everyday applications. A hot-air balloon rises because heating the air inside makes it expand at constant pressure; the same volume of hot air holds fewer particles than cold air, so it weighs less than the cold air it displaces. A tire blowout on a hot day is the pressure-temperature relation in a rigid container. A sealed bag of snacks puffs up on a mountain because the outside pressure falls and the gas inside expands. In each case name the two changing variables, the fixed ones and the particle explanation.

Investigation: the airbag model and a syringe test. Materials: sealable freezer bags, baking soda, white vinegar, a measuring cup, a balance, a thermometer, a tissue, a large capped plastic syringe with no needle, goggles. Part A: measure a bag's capacity with water and dry it; calculate the baking soda needed as in the model, using the room temperature you measure and the gas space left for the gas, which is the bag's capacity minus the volume of vinegar; pour in the vinegar (5 percent acetic acid; about 250 mL for 16 g of baking soda, which is why a smaller bag and less baking soda are often used), wrap the baking soda in a tissue and tuck it into an upper corner held shut by pinching, seal the bag, then release the tissue into the vinegar and watch; record whether the bag fills fully, partially or bursts. Part B: draw 30 mL of air into the syringe, cap it, and push the plunger to 20 mL and 15 mL, noting how the resistance grows; release and watch the plunger return; then hold the barrel in warm hands and note any movement. For numbers, not only impressions: if the school has a gas pressure sensor or a syringe fitted with a gauge, record the pressure at 30, 25, 20 and 15 mL and check that pressure times volume stays nearly constant; otherwise use a gas-law computer simulation to collect at least five data pairs each for pressure against volume, pressure against temperature and volume against temperature, holding the other quantities fixed, and graph each relationship. Record volumes and observations. A result that supports the claim: the bag inflates roughly as predicted and more baking soda gives more inflation; the plunger resists more and more as the volume shrinks, faster than in simple proportion to the push (at 20 mL the trapped air is at about 1.5 times its starting pressure, at 15 mL about 2 times), and springs back, and warming the trapped air nudges it outward. What the model cannot show: you cannot see particles or count collisions; the bag's volume is not fixed, so the pressure inside is barely above room pressure; and a syringe gives no pressure reading unless a gauge is attached. Do the bag experiment outdoors or over a sink, wear goggles and stand back; do not use a glass container.

Worked example 1 — pressure and volume. A gas occupies 2.50 L at 101 kPa. It is allowed to expand at constant temperature to 3.00 L. New pressure: P₂ = P₁V₁ / V₂ = 101 × 2.50 / 3.00 = 84.2 kPa. The volume rose, so the pressure fell, as inverse proportion requires; the particles have more room and strike each unit of wall less often. Check the direction before trusting the arithmetic.

Worked example 2 — volume and temperature. A balloon holds 1.20 L at 20 °C and is warmed to 80 °C at constant pressure. Convert: 293 K and 353 K. V₂ = V₁ × T₂ / T₁ = 1.20 × 353 / 293 = 1.45 L. Using Celsius by mistake, 1.20 × 80 / 20 = 4.8 L, quadruples the volume, which is absurd; the kelvin ratio 353/293 is only 1.20, matching the modest expansion you would actually see.

Worked example 3 — pressure and temperature in a tire. A tire holds air at an absolute pressure of 220 kPa at 15 °C. (Gas laws need absolute pressure; a tire gauge reads the pressure above atmospheric, so a gauge reading must have about 101 kPa added before calculating and removed again afterwards.) After driving, the air inside reaches 45 °C and the volume is essentially unchanged. Convert: 288 K and 318 K. P₂ = P₁ × T₂ / T₁ = 220 × 318 / 288 = 243 kPa. The pressure rose by about 10 percent, which is why tire pressures are set cold. The particles hit the rigid walls harder and more often; the number of particles did not change.

Worked example 4 — ideal gas law for an unknown. What is the pressure of 0.500 mol of nitrogen in a rigid 10.0 L cylinder at 27 °C? T = 300 K. P = nRT/V = 0.500 × 8.31 × 300 / 10.0 = 125 kPa. Then: what volume does that same 0.500 mol occupy at STP? Either apply the law, V = 0.500 × 8.31 × 273 / 101.3 = 11.2 L, or use the molar volume, 0.500 mol × 22.4 L/mol = 11.2 L; the two agree, as they must.

Representation task: three particle pictures and a units card. Draw a box of gas particles three times: the original, the same box at half the volume and same temperature (more collisions per area, higher pressure), and the original box at double the Kelvin temperature (longer arrows, harder collisions, higher pressure). Under each write which variables changed, which were fixed and the law that applies. Then make a units card: the three pressure units with their conversions, the Kelvin conversion, the two forms of R, and the molar volume at STP with its conditions stated.

Independent performance: airbag prediction and gas-law problem set. Carry out the airbag model with a calculated mass of baking soda and write a report. Success criteria: the bag capacity and the vinegar volume measured and the gas space stated; the room temperature and pressure used, with the temperature in kelvin; the ideal gas law calculation of moles of carbon dioxide and the mass of baking soda, with units canceling; the observed result; at least two explanations for any difference between prediction and observation; and a problem set of five gas-law calculations from your teacher, each stating which quantities are constant, whether the unknown should rise or fall before calculating, and the answer with units to correct significant figures.