Topic 9.1: Kinetic Theory of Temperature and Pressure

What Topic 9.1 requires

Topic 9.1 carries two learning objectives, and between them they define the two quantities the rest of Unit 9 moves around.

• 9.1.A Describe the pressure a gas exerts on its container in terms of atomic motion within that gas.
• 9.1.B Describe the temperature of a system in terms of the atomic motion within that system.

One essential knowledge statement sits under 9.1.A, with three sub-statements.

• 9.1.A.1 Atoms in a gas collide with and exert forces on other atoms in the gas and with the container in which the gas is contained.
• 9.1.A.1.i Collisions involving pairs of atoms or an atom and a fixed object can be described and analyzed using conservation of momentum principles.
• 9.1.A.1.ii The pressure exerted by a gas on a surface is the ratio of the sum of the magnitudes of the perpendicular components of the forces exerted by the gas's atoms on the surface to the area of the surface. The CED prints $P = \frac{F_{\perp}}{A}$ as the relevant equation.
• 9.1.A.1.iii Pressure exists throughout the gas itself, not just at the boundary between the gas and the container.

One statement sits under 9.1.B, with two sub-statements.

• 9.1.B.1 The temperature of a system is characterized by the average kinetic energy of the atoms within that system.
• 9.1.B.1.i The Maxwell-Boltzmann distribution provides a graphical representation of the energies and speeds of atoms at a given temperature.
• 9.1.B.1.ii The root-mean-square speed corresponding to the average kinetic energy for an ideal gas is related to the temperature of the gas by $K_{\text{avg}} = \frac{3}{2} k_B T = \frac{1}{2} m v_{\text{rms}}^2$.

The boundary statement is two sentences and both of them matter:

"AP Physics 2 only expects students to perform qualitative and quantitative analysis of collisions in one and two dimensions. Students are not expected to know the functional form of the Maxwell-Boltzmann distribution but are expected to be familiar with how features of the distribution are related to the temperature of the gas."

The second sentence gets quoted with its ending cut off, and the truncated version reads as though the distribution is optional. It is not. The exemption covers the functional form only, and the requirement to connect features of the curve to the temperature of the gas survives it.

The CED lists this topic's suggested skills twice, in the Unit at a Glance table and in the sidebar of the topic page, and here the two lists agree: 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

Unit 9 is weighted at 15 to 18 percent of the multiple-choice section, with a suggested 10 to 16 class periods.

Pressure is a running total of collisions (9.1.A.1)

Read 9.1.A.1.ii one clause at a time, because the sentence is built to be precise rather than readable.

• The forces exerted by the gas's atoms on the surface. Every atom that reaches the wall pushes on it during the brief moment of contact. One atom's push is a spike lasting a fraction of a nanosecond. Pressure is what those spikes average out to.
• The perpendicular components. An atom arriving at an angle carries momentum along the surface as well as into it. Only the component perpendicular to the surface contributes to the pressure. The sideways part is real, but it is not what $P$ measures.
• The sum of the magnitudes. You add sizes, not vectors. Every atom pushes outward on the container, so the contributions reinforce rather than cancel, and a vector sum would give the wrong answer for a curved wall.
• To the area of the surface. Double the wall area and roughly twice as many atoms hit it per second, so the force doubles and the pressure does not change. Pressure is an intensive property; force is not.

Two independent handles raise a gas's pressure, and keeping them apart is what a qualitative question turns on.

• Hit harder. Faster atoms deliver more momentum per bounce. That is the temperature handle, and it is where the second half of this topic goes.
• Hit more often. More atoms in the same space, or the same atoms in a smaller space, means more collisions per second on each square meter. That is the density handle, and it is where Topic 9.2 goes.

Pressure itself is a scalar. It has a size and no direction, even though the force it produces on a given surface points along that surface's normal. That is the definition AP Physics 1 Topic 8.2 sets up for fluids, and $P = F_{\perp}/A$ is printed twice on the AP Physics 2 sheet: once in the Thermal Physics block and once in Mechanics and Fluids.

One bounce, analysed with momentum (9.1.A.1.i)

Essential knowledge 9.1.A.1.i licenses you to treat a single atom hitting a wall as an ordinary collision problem analysed with conservation of momentum principles. Do that once and the pressure picture becomes arithmetic you already know from Topic 4.2.

Take one atom of mass $m$ travelling straight at a wall. Declare the sign convention before anything else: the direction the atom travels on its way in is positive, and that convention holds for the rest of this page. The atom arrives with velocity $+v$ and, in an elastic bounce off a wall too massive to move, leaves with velocity $-v$. Its momentum change is

The atom loses $2mv$ of momentum in the positive direction, so by Newton's third law the wall gains $2mv$ of momentum in that direction. The magnitude of the impulse delivered to the wall by one collision is $2mv$, not $mv$: the factor of two is the reversal, and dropping it halves every pressure you go on to calculate.

Nothing about that bounce changes the atom's kinetic energy: it arrives with $\frac{1}{2}mv^2$ and leaves with $\frac{1}{2}m(-v)^2$, the same number. Momentum is a vector and kinetic energy is not. Topic 4.4 is where that split is set up.

To get from one bounce to a pressure, add up the impulses over a stretch of time and divide:

then divide by the area. The second worked example below runs those two lines with numbers. The boundary statement caps how far this goes: AP Physics 2 expects qualitative and quantitative analysis of collisions in one and two dimensions, which covers the head-on bounce and the angled bounce, and stops short of the full kinetic-theory derivation of $PV = \frac{1}{3}Nmv_{\text{rms}}^2$. That result is not printed on the equation sheet and you will not be asked to produce it.

Pressure exists throughout the gas, not just at the walls (9.1.A.1.iii)

Essential knowledge 9.1.A.1.iii exists to kill one specific mental picture: that pressure is something a gas does to its container, and that the middle of the gas has no pressure because there is nothing there to push on.

Imagine a flat sheet of paper suspended inside a box of gas. Atoms strike both faces. Each face feels a pressure, and it is the same pressure the walls feel. Now make the sheet imaginary. The collisions are still happening across that surface; you have only removed the thing that was in the way. Pressure is a property of the gas at a point, defined by the momentum crossing any surface you care to draw there, and it does not need a wall to exist.

Two consequences worth carrying into problems.

• A balloon or a piston floating inside the gas feels pressure on every side. The reason it does not accelerate is that the pushes balance, not that they are absent.
• A gas sample in an AP Physics 2 problem has one pressure, not a pressure that varies with height. In a liquid the pressure grows with depth, $P = P_0 + \rho g h$, because the fluid above has weight (Topic 8.2 covers that). The ideal gas model rules the analogous gas effect out by assumption: essential knowledge 9.2.A.1 states that the only appreciable forces on the atoms are those that occur during collisions, which leaves no gravitational term inside the sample. So $P$ in $PV = nRT$ is a single number for the whole sample.

The CED opens Unit 9 by saying that students investigate what they cannot see by examining the properties of ideal gases. Pressure at an interior point is exactly that: a quantity with no visible surface attached to it.

Temperature is average kinetic energy (9.1.B.1)

Essential knowledge 9.1.B.1 says the temperature of a system is characterized by the average kinetic energy of the atoms within that system. The equation above is the quantitative version, printed on the AP Physics 2 equation sheet, with $k_B = 1.38 \times 10^{-23}$ J/K from the sheet's constants table.

Four things about that equation decide whether you use it correctly.

• It is per atom. $K_{\text{avg}}$ is the average kinetic energy of one atom, in joules. It is not the energy of a mole and it is not the energy of the sample. The energy of the whole sample is $N$ times this, which the sheet writes separately as $U = \frac{3}{2} N k_B T = \frac{3}{2} nRT$, and that belongs to Topic 9.4.
• $T$ is in kelvin, with no exceptions. Kinetic energy cannot be negative, and neither can an average of non-negative numbers, so a temperature scale that runs negative cannot go into this equation. Celsius is such a scale. Kelvin is not.
• Mass does not appear. Two gases at the same temperature have the same average kinetic energy per atom regardless of what those atoms are. That is the fact skill 2.C comparison questions are built on.
• The kinetic energy in question is translational. The AP Physics 2 Table of Information lists, among the conventions used on the exam unless otherwise stated, that ideal gases are monatomic. A single atom moving through space has translational kinetic energy and nothing else to store energy in, which is why the coefficient is exactly $\frac{3}{2}$ and why the CED says "atoms" rather than "molecules" throughout this unit.

The proportionality is the part worth internalising. $K_{\text{avg}} \propto T$ with $T$ in kelvin, so doubling the absolute temperature doubles the average kinetic energy, and doubling the Celsius reading does not. Going from 27 degrees Celsius to 54 degrees Celsius takes you from 300 K to 327 K, a nine percent rise, not one hundred percent. Convert first, then reason about ratios.

Root-mean-square speed, and what it is not (9.1.B.1.ii)

The sheet prints one equation with two equals signs in it:

The left half defines the average kinetic energy from the temperature. The right half says that if you want a single speed to represent the sample, the honest one is the speed an atom would need in order to have exactly the average kinetic energy. That speed is the root-mean-square speed. Setting the two outer expressions equal and solving gives

That rearranged form is not printed. The sheet gives you the equality and expects you to do the algebra, so practise the rearrangement rather than trying to recall a formula that will not be in front of you.

Read the symbols carefully, because two of them are traps.

• $m$ is the mass of one atom in kilograms, not the mass of the sample and not the molar mass. If a problem hands you a molar mass $M$ in grams per mole, convert to kilograms per mole and divide by Avogadro's number: $m = M/N_0$ with $N_0 = 6.02 \times 10^{23}$ per mole from the sheet. A factor of a thousand and a factor of $10^{23}$ both live in that one step.
• $v_{\text{rms}}$ is not the average speed. It is the square root of the average of the squares, which is a different operation and a different number. Squaring gives extra weight to the fast atoms, so the rms speed sits above the ordinary mean speed, which in turn sits above the most probable speed at the peak of the distribution. Three different speeds, one curve, and the AP equation uses only the rms one.

The two dependencies are what skill 2.D questions ask you to predict.

The last row surprises people. Compressing a gas at constant temperature raises its pressure, because the atoms hit the walls more often, but it does not speed them up at all.

Reading a Maxwell-Boltzmann distribution (9.1.B.1.i)

Essential knowledge 9.1.B.1.i says the Maxwell-Boltzmann distribution provides a graphical representation of the energies and speeds of atoms at a given temperature. The boundary statement then removes the functional form from the syllabus and keeps the graph reading. So what you owe is a set of statements about the shape of the curve.

What the axes are. Speed runs along the horizontal axis, starting at zero, with no upper bound drawn. The vertical axis is the relative number of atoms per unit speed. A single point on the curve is not a count of atoms; an area under a stretch of the curve is.

What is fixed. The total area under the whole curve is the total number of atoms. If the sample is sealed and nothing leaks, that area is the same at every temperature. This is what makes the curves at two temperatures cross: one cannot sit above the other everywhere.

What raising the temperature does.

• The peak moves to the right, because the typical speed increases.
• The peak drops in height, because the fixed area has to spread over a wider range of speeds.
• The curve broadens and the high-speed tail thickens. A modest temperature rise makes a large proportional difference to the number of very fast atoms.
• The curve still starts at zero and still approaches zero at large speeds. It never becomes symmetric.

What changing the gas does. At the same temperature, a lighter gas has its peak further right and a longer tail, because $v_{\text{rms}} \propto 1/\sqrt{m}$. The average kinetic energy is identical for both; only the speeds differ.

Skill 1.A asks you to create diagrams, tables, charts, or schematics to represent physical situations, and the CED backs it with a sample instructional activity: students are shown a thermodynamic process on a PV, PT, or VT diagram together with a speed distribution for the initial state, and are asked to draw the distribution for the final state or explain why it is the same graph. That second option is the interesting one. An isothermal compression changes pressure and volume and leaves the distribution untouched, because temperature is the only thing the curve depends on.

What the AP Physics 2 equation sheet gives you here

The AP Physics 2 Table of Information carries a Thermal Physics block of exactly eight equations. Two of them belong to Topic 9.1:

• $P = \dfrac{F_{\perp}}{A}$
• $K_{\text{avg}} = \dfrac{3}{2} k_B T = \dfrac{1}{2} m v_{\text{rms}}^2$

The other six are $\frac{Q}{\Delta t} = \frac{kA\Delta T}{L}$ and $Q = mc\Delta T$ (Topic 9.5), $PV = nRT = Nk_BT$ (Topic 9.2), and $U = \frac{3}{2}nRT = \frac{3}{2}Nk_BT$, $W = -P\Delta V$ and $\Delta U = Q + W$ (Topic 9.4). Eight is the whole block; the full sheet is at /formulas/ap-physics-2. The AP Physics 1 sheet has no thermal physics section, so seven of those eight are new to you here. The one exception is $P = F_{\perp}/A$, which AP Physics 1 also prints, in its fluids equations.

The constants you need are printed too: $k_B = 1.38 \times 10^{-23}$ J/K, $N_0 = 6.02 \times 10^{23}$ per mole, $R = 8.31$ J/(mol K), and 1 atm $= 1.0 \times 10^5$ Pa.

One symbol warning. The letter $k$ does three jobs on this sheet. In the Thermal Physics block it is the thermal conductivity in $\frac{Q}{\Delta t} = \frac{kA\Delta T}{L}$. In the Electricity block it is the Coulomb constant, $9.0 \times 10^9$ N m squared per C squared. Boltzmann's constant is always written with its subscript, $k_B$, and the sheet's own symbol list for Thermal Physics defines $k$ as thermal conductivity and nothing else. Write the subscript every time.

Three things Topic 9.1 uses that are not printed anywhere, and that you therefore have to supply yourself:

• $v_{\text{rms}} = \sqrt{3k_BT/m}$, the rearrangement of the sheet equation.
• $m = M/N_0$, the step from molar mass to the mass of one atom.
• The conversion from degrees Celsius to kelvin. The unit-symbol table on the Table of Information lists both "degree Celsius" and "kelvin" and prints no relation between them, so carry the offset in your head.

How Topic 9.1 is tested, and where it leads

The four suggested skills tell you the shape of the questions.

• 2.C, compare physical quantities between two or more scenarios. This is the skill behind an item that gives you two gases, or one gas at two temperatures, and asks which quantity is larger. The reliable move is to ask what each quantity depends on: $K_{\text{avg}}$ on $T$ alone, $v_{\text{rms}}$ on $T$ and $m$, pressure on both plus how crowded the gas is.
• 3.B and 3.C, apply a law to make a claim and justify a claim with evidence. A claim like "the argon atoms are slower" earns its marks from the sentence after it, and that sentence should name the relationship: equal temperature means equal average kinetic energy, and equal kinetic energy with more mass means less speed.
• 1.A, create diagrams, tables, charts, or schematics. Sketching a speed distribution before and after a change answers a comparison question and shows the reasoning at the same time.

The unit opener names the fourth free-response question, the Qualitative/Quantitative Translation, as where this unit's skills land: it asks students to make a claim with supporting evidence and reasoning without reference to equations, then derive an equation or set of equations for the same scenario, then connect the two. Topic 9.1 suits that question because it has a verbal model and a printed equation for identical physics. The exam itself is 3 hours long: 42 multiple-choice questions worth 50 percent in 85 minutes, then 4 free-response questions worth 50 percent in 95 minutes, calculator allowed on both sections.

Where the topic goes next:

• Topic 9.2, The Ideal Gas Law packages the same collisions as one macroscopic state equation, $PV = nRT = Nk_BT$. Everything on this page is the reason that equation has the form it does.
• Topic 9.3 reuses the collision picture for atoms of two systems in contact, to explain why energy flows from hot to cold.
• Topic 9.4 sums $K_{\text{avg}}$ over all $N$ atoms to get internal energy, then tracks how heating and work change it. For the PV-diagram routine and the sign conventions, work through the thermodynamics and PV diagrams guide, which handles that procedure end to end.