Kinetic theory of pressure: atoms colliding with surfaces

Highlights only0 highlighted

Pump up a basketball and it goes from floppy to hard, yet nothing solid has been added inside it.

What you added is more air: more atoms and molecules, each moving at hundreds of meters per second, each bouncing off the inside of the rubber.

The ball is hard because of an enormous number of tiny collisions every second.

This section builds the idea of pressure from that picture.

It is the start of the kinetic theory of gases, the model that links what atoms do (move, collide, carry momentum and kinetic energy) to what you measure in the lab (pressure, temperature, volume).

Every later idea in this unit, from the ideal gas law to the first law of thermodynamics, rests on it.

On the exam. Thermodynamics is 15-18% of the multiple-choice section (Section I), the same share as Electric Force, Field, and Potential and as Electric Circuits.

In Section II any of the four free-response questions can be set on it: recent exams have used thermodynamics for a Translation Between Representations question, a Mathematical Routines question and an experimental design question.

So every section of these notes ends with practice in the free-response shapes as well as multiple choice.

One more note if you are using an older textbook: fluids (buoyancy, pressure with depth, Bernoulli's equation) used to be taught in this course but now belongs to AP Physics 1, and it is not part of this unit.

How the points work

This box applies to every section of the unit.

Each scoring row is worth one point, labeled Point A1, A2, B1 ... by part letter. No points are printed against the parts on the question paper.

The four free-response questions sit in a fixed order with fixed totals: Mathematical Routines (MR) 10, Translation Between Representations (TBR) 12, Experimental Design and Analysis (LAB) 10, Qualitative/Quantitative Translation (QQT) 8. A multiple-choice question is one point with one key.

Derive: the first point is for a multistep derivation that starts from an equation on the Table of Information or a named principle (Newton's second law, conservation of energy). A correct final expression written on its own earns the later points but never that first one.

Justify: points go to the links of your reasoning (what changes, why, so what), not to naming a law. Your justification may use equations but must include reasoning beyond the algebra. A selection ("increases", "Incorrect") earns its point only with an attempt at a relevant justification.

Consistency credit: a later answer that follows correctly from your own earlier wrong answer still earns its points.

Magnitudes and directions: a magnitude point usually ignores a sign; a direction (up, toward the wall, into the gas) is scored strictly.

Units are required on every final numerical answer. Give 2 or 3 significant figures; units are scored, significant figures have not been.

Graphs: separate points for linearizing axes, labeled linear scales, plotting, a best-fit line, the slope relation and a value in range.

TBR Part D and QQT Part C: open by naming the earlier part you use ("In Part B, ...") and state the functional dependence.

Atoms in a gas: always moving, always colliding

You need to know that 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.

In air at room temperature a typical molecule travels at about 500 m/s, faster than a passenger jet, but it goes only about a ten-millionth of a meter before it hits another molecule.

Between collisions it moves in a straight line at constant velocity, because (in the model we use) the only appreciable forces act during collisions.

The result is a chaotic, never-ending rain of impacts on every surface the gas touches.

The diagram below shows the picture to keep in your head: atoms moving in random directions at a range of speeds, colliding with each other and with the walls.

One atom is shown bouncing off the right-hand wall.

While it is in contact, the wall pushes it back into the gas, and by Newton's third law it pushes the wall outward with an equal and opposite force.

Atoms of a gas drawn as small gray circles inside a rectangular container, each with a short arrow showing its velocity in a random direction. One atom near the right wall is shown arriving at an angle and leaving at the same angle on the other side of the perpendicular, and an arrow outside the wall shows the force that atom exerts on the wall, pointing outward and perpendicular to the wall.

Physics 1 recap: momentum change in a collision with a wall

You met momentum and impulse in AP Physics 1, and both are on the Mechanics page of the Table of Information:

\vec{p} = m\vec{v} \qquad \vec{F}_{\text{net}} = \frac{\Delta\vec{p}}{\Delta t} \qquad \vec{J} = \vec{F}_{\text{avg}}\Delta t = \Delta\vec{p}

Momentum is a vector, so a change in direction is a change in momentum even when the speed stays the same.

Take a ball of mass $m$ moving at speed $v$ straight toward a wall (call that direction $+x$ ).

It bounces straight back elastically, with the same speed.

Its momentum goes from $+mv$ to $-mv$ , so

\Delta p = p_{\text{after}} - p_{\text{before}} = (-mv) - (+mv) = -2mv

The change is twice the momentum, not zero and not $mv$ .

The negative sign says the change points away from the wall: the wall pushed the ball back.

The impulse the wall gave the ball is $-2mv$ , so by Newton's third law the ball gave the wall an impulse of $+2mv$ , toward the wall.

You need to know that collisions involving pairs of atoms or an atom and a fixed object can be described and analyzed using conservation of momentum principles.

For an atom and a fixed wall, the wall is attached to the Earth, so the momentum the atom loses goes into the wall and the Earth, which barely move.

For two atoms colliding with each other, the total momentum of the pair is the same before and after.

When two atoms of the same mass collide head-on elastically, they simply swap velocities, which is one way energy passes from fast atoms to slow ones (you will use this in the section on thermal equilibrium).

Collisions at an angle (two dimensions). Atoms rarely hit a wall head-on.

Split the velocity into a component perpendicular to the wall and a component parallel to it.

In an elastic collision with a smooth fixed wall, the wall can push only perpendicular to its surface, so:

the parallel component is unchanged;

the perpendicular component reverses, so the change in momentum has magnitude $2mv_{\perp} = 2mv\cos\theta$ , where $\theta$ is measured from the perpendicular (the normal) to the wall.

The diagram below shows a helium atom striking a wall at $30°$ from the perpendicular.

It leaves at $30°$ on the other side, like a ball banked off a cushion on a pool table.

A helium atom approaches a fixed wall along a line 30 degrees from the dashed perpendicular to the wall and rebounds along a line 30 degrees on the other side of the perpendicular with the same speed v. The component of velocity parallel to the wall is unchanged; the perpendicular component reverses.

From many collisions to pressure

One atom gives the wall a tiny impulse, about $10^{-23}$ kg·m/s.

But there are about $2.5 \times 10^{22}$ molecules in a liter of air at room conditions, and every square meter of a container wall is struck around $10^{27}$ times a second (Worked example 2 below estimates this).

All those impulses add up to a steady outward force on every part of the wall, the way a hard rain on a car roof produces a steady roar instead of separate taps.

The force on a larger patch of wall is larger, simply because more atoms hit it.

So the useful quantity is the force per unit area.

You need to know that 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:

P = \frac{F_{\perp}}{A}

Symbol

Quantity

Unit

$P$

pressure

pascal, Pa (1 Pa = 1 N/m²)

$F_{\perp}$

sum of the magnitudes of the perpendicular force components on the surface

newton, N

$A$

area of the surface

square meter, m²

Three details of that definition matter:

Perpendicular components only. Each atom pushes on the surface only perpendicular to it (the parallel part of its velocity is unchanged), so only perpendicular force components count.

Magnitudes. Pressure has no direction of its own; it is a scalar. A gas at pressure $P$ pushes outward on a floor, a ceiling and a side wall alike, always perpendicular to whichever surface is there. Force is the vector: $F = PA$ , directed perpendicular to the surface, from the gas toward the surface.

Area in square meters. 1 cm² = $10^{-4}$ m² and 1 mm² = $10^{-6}$ m². Squaring the conversion factor is where most numerical slips happen.

Atmospheric pressure at sea level is about $1.0 \times 10^5$ Pa, the value of 1 atm printed on the Table of Information.

That is a force of $1.0 \times 10^5$ N on every square meter, the weight of about 10 metric tons.

You do not feel it because the air inside your body and on all sides of you pushes back equally.

Pressure is not just a wall effect

You need to know that pressure exists throughout the gas itself, not just at the boundary between the gas and the container.

Put a small pressure sensor anywhere in the middle of a tank of gas and it reads the same as at the walls (ignoring tiny effects of gravity).

That is because the atoms are everywhere: any surface you place inside the gas, facing any direction, is struck by atoms from that side.

Even with no sensor there, pick any flat patch inside the gas: the gas on one side of it pushes on the gas on the other side through the same collisions.

"Pressure is where the walls are" is a tempting idea because walls are where we measure it, but the walls only reveal a pressure that is already there.

Net force: when collisions on one side outnumber the other

An object in a gas feels a net force from the gas only when the collisions on one side outweigh those on the other.

Pour a little water in an empty soda can, heat it until steam drives the air out, then turn it upside down into cold water.

The steam inside condenses, the number of molecules striking the inside walls each second drops sharply, and the atmosphere's collisions on the outside are no longer balanced.

The can is crushed inward in a fraction of a second.

Nothing "sucked" the can in: the outside air pushed it, because more momentum per second was delivered to the outside surfaces than to the inside surfaces.

The photograph below shows such a can a moment after it was turned upside down into ice water: its sides have been pushed in by the air outside.

Photograph of an empty aluminum can held upside down by tongs with its open end in a bowl of ice water. The can's sides are crushed inward. A hot plate stands beside the bowl.

Is the answer a sensible size?

Situation

Typical pressure

Atmosphere at sea level

$1.0 \times 10^5$ Pa (1 atm)

Atmosphere in Denver, Colorado (about 1.6 km up)

about $0.83 \times 10^5$ Pa

Air inside a car tire (absolute)

about $3 \times 10^5$ Pa

A full scuba tank (absolute)

about $2 \times 10^7$ Pa

If a calculation gives a room's air pressure as a few pascals or a tire's as $10^{12}$ Pa, look for an area left in cm² or mm².

On the Table of Information

$P = \dfrac{F_{\perp}}{A}$ , on the Thermal Physics page (also on the Mechanics and Fluids page).

$\vec{p} = m\vec{v}$ , $\vec{F}_{\text{net}} = \dfrac{\Delta\vec{p}}{\Delta t}$ and $\vec{J} = \vec{F}_{\text{avg}}\Delta t = \Delta\vec{p}$ , on the Mechanics and Fluids page.

1 atm $= 1.0 \times 10^5$ Pa, $g = 9.8$ m/s², and the unified atomic mass unit, in the constants.

You need to know and use $P = F_{\perp}/A$ in all three directions (finding $P$ , $F_{\perp}$ or $A$ ); it is also printed on the Table of Information.

Derive or recall: not on the Table of Information

The momentum change of an atom that rebounds elastically from a fixed wall: magnitude $2mv_{\perp}$ , with the parallel component unchanged.

The pressure of a gas under a piston in equilibrium, from Newton's second law with each pressure entered as pressure times area (below).

Area conversions: 1 cm² = $10^{-4}$ m², 1 mm² = $10^{-6}$ m².

Physics 1 recap: free-body diagrams and Newton's second law for a piston

Recent exams have asked for a free-body diagram of a piston that sits on a trapped gas, followed by a derivation of the gas pressure.

The diagram below shows a vertical cylinder of gas closed by a piston of mass $M$ and area $A$ that slides without friction.

The piston is at rest.

A vertical cylinder, closed at the bottom and open at the top, holds a gas below a horizontal piston of mass M and area A. The piston can slide without friction. Air at atmospheric pressure is above the piston.

Three objects exert forces on the piston, so its free-body diagram has three arrows, each starting on the dot that represents the piston and pointing away from it:

$F_g$ , the gravitational force exerted by the Earth, downward, with magnitude $Mg$ ;

$F_{\text{atm}}$ , the force exerted by the air above, downward, with magnitude $P_{\text{atm}}A$ ;

$F_{\text{gas}}$ , the force exerted by the trapped gas below, upward, with magnitude $P_{\text{gas}}A$ .

Free-body diagram on a dot for the piston at rest. One long arrow points straight up from the dot, labeled F gas. Two shorter arrows point straight down, side by side, labeled F atm for the force from the atmosphere and F g for the gravitational force; the two downward arrows together are as long as the upward arrow.

Each arrow names the object that exerts the force.

A pressure is never an arrow on a free-body diagram: write $F_{\text{atm}}$ , not $P_{\text{atm}}$ .

And there is no " $F_{\text{piston}}$ ": the piston cannot exert a force on itself.

Newton's second law (AP Physics 1, on the Mechanics page as $\vec{a}_{\text{sys}} = \sum\vec{F}/m_{\text{sys}}$ ) for the piston at rest, taking up as positive:

\sum F = F_{\text{gas}} - F_{\text{atm}} - F_g = 0

Enter each pressure as pressure times area, using $F = PA$ from $P = F_{\perp}/A$ :

P_{\text{gas}}A - P_{\text{atm}}A - Mg = 0 \qquad\Longrightarrow\qquad P_{\text{gas}} = P_{\text{atm}} + \frac{Mg}{A}

This is the absolute pressure of the gas: the atmosphere's contribution plus the extra needed to hold up the piston.

A heavier piston or a smaller area means a larger gas pressure, and if the piston had no mass the gas pressure would equal atmospheric pressure.

Notice what the equation does not contain: the volume or the temperature of the gas.

For a piston free to slide, the forces fix the pressure, and the gas adjusts its volume to suit.

You will use this again in the section on the first law, where heating such a gas makes it expand at constant pressure.

Why the atom picture predicts the gas law (an algebra-based route)

You can get a feel for how atomic motion sets the pressure with one atom of mass $m$ in a cubical box of side $L$ , moving back and forth along $x$ with speed $v_x$ and bouncing elastically off the two end walls.

Each hit on the right wall gives the wall an impulse $2mv_x$ .

The atom returns to that wall after traveling $2L$ , so the time between hits is $\Delta t = 2L/v_x$ .

The average force on the wall from this one atom is

F = \frac{\Delta p}{\Delta t} = \frac{2mv_x}{2L/v_x} = \frac{mv_x^2}{L}

and the pressure it produces on the wall of area $L^2$ is

P = \frac{F}{L^2} = \frac{mv_x^2}{L^3} = \frac{mv_x^2}{V}

Two lessons come straight out of this.

A faster atom hits harder (a bigger $2mv_x$ ) and more often (a shorter $\Delta t$ ), so the pressure goes as the speed squared, which is the kinetic energy.

And spreading the same atom over a bigger volume lowers the pressure.

With $N$ atoms moving in all directions, the same reasoning gives a pressure proportional to $N$ times the average kinetic energy, divided by $V$ .

In the next section you will see that temperature measures that average kinetic energy, and the result becomes the ideal gas law $PV = Nk_BT$ .

You also need: momentum, impulse and Newton's third law from AP Physics 1, including momentum conservation in two dimensions for collisions at an angle.

Used again in the next section (temperature as average kinetic energy), in the section on thermal equilibrium (energy passed in atom-atom collisions), in the section on the first law (the piston's force balance), and in Unit 15, Modern Physics, where Compton scattering treats a photon striking an electron as a collision with momentum conserved in two dimensions, the same bookkeeping as an atom striking a wall here.

Key terms

Atoms in a gas: 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.

Collisions: Collisions involving pairs of atoms or an atom and a fixed object can be described and analyzed using conservation of momentum principles.

Pressure: 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. Relevant equation: $P = F_{\perp}/A$

Pressure in the gas: Pressure exists throughout the gas itself, not just at the boundary between the gas and the container.

Worked example 1: a direct use of $P = F_\perp / A$, with an area conversion

A square pressure sensor 2.0 cm on each side is mounted in the wall of a gas tank.

The gas exerts a total perpendicular force of 48 N on it.

Calculate the pressure of the gas, in pascals and in atmospheres.

Convert the area first: $A = (0.020 \text{ m})(0.020 \text{ m}) = 4.0 \times 10^{-4} \text{ m}^2$ .

$P = \dfrac{F_\perp}{A}$

$P = \dfrac{48 \text{ N}}{4.0 \times 10^{-4} \text{ m}^2} = 1.2 \times 10^{5} \text{ Pa}$

In atmospheres: $\dfrac{1.2 \times 10^5 \text{ Pa}}{1.0 \times 10^5 \text{ Pa/atm}} = 1.2$ atm.

Answer: $1.2 \times 10^5$ Pa, or 1.2 atm.

That is a little above atmospheric pressure, a sensible value for a lightly pressurized tank.

Using 4.0 cm² as if it were 4.0 m² would give 12 Pa, far too small for any gas you could fill a tank with.

The atom in the next example is the one shown in the wall-collision diagram above.

Worked example 2: the momentum change of one atom, and how many hits make a pressure (two dimensions)

A helium atom of mass $6.64 \times 10^{-27}$ kg moving at 1200 m/s strikes a fixed wall at $30°$ from the perpendicular and rebounds elastically at $30°$ on the other side.

Part A Calculate the magnitude of the atom's change in momentum.

Part B A gas at $1.0 \times 10^5$ Pa is made of atoms like this one, each delivering this impulse.

Estimate the number of such collisions per second on each square meter of wall.

Part A. Only the perpendicular component reverses.

With $\cos 30° = \sqrt{3}/2$ from the trigonometry table:

$v_\perp = v\cos 30° = (1200 \text{ m/s})(0.866) = 1.04 \times 10^3 \text{ m/s}$

$|\Delta p| = 2mv_\perp = 2(6.64 \times 10^{-27} \text{ kg})(1.04 \times 10^3 \text{ m/s}) = 1.38 \times 10^{-23} \text{ kg·m/s}$

The parallel component, $v\sin 30° = 600$ m/s, is the same before and after, so it contributes nothing.

Part B. On 1 m² the force is $F_\perp = PA = (1.0 \times 10^5 \text{ Pa})(1 \text{ m}^2) = 1.0 \times 10^5$ N. Force is impulse per unit time, so the number of impulses per second is

$\dfrac{1.0 \times 10^5 \text{ N}}{1.38 \times 10^{-23} \text{ kg·m/s}} \approx 7 \times 10^{27}$ collisions per second on each square meter.

Answer: A: $1.4 \times 10^{-23}$ kg·m/s.

B: about $7 \times 10^{27}$ per second per square meter.

A number that large is why the pressure on a wall feels perfectly steady, with no flicker from individual hits.

Using $2mv$ with the full 1200 m/s would overstate the impulse by about 15%.

Worked example 3: a symbolic derivation and a factor-of-change reading

A single atom of mass $m$ moves back and forth along one edge direction of a cubical box of side $L$ with speed $v$ , bouncing elastically off the two opposite walls.

Derive an expression for the average pressure the atom exerts on one of those walls, in terms of $m$ , $v$ and $L$ .

Then use your expression to describe what happens to that pressure if the atom's speed is doubled and the box is replaced by one with sides twice as long.

Start from a relationship on the Table of Information:

$\vec{F}_{\text{net}} = \dfrac{\Delta\vec{p}}{\Delta t}$

Each collision with the wall changes the atom's momentum by magnitude $\Delta p = 2mv$ .

The atom returns after traveling a distance $2L$ , so $\Delta t = \dfrac{2L}{v}$ :

$F = \dfrac{2mv}{2L/v} = \dfrac{mv^2}{L}$

$P = \dfrac{F_\perp}{A} = \dfrac{mv^2/L}{L^2}$

$P = \dfrac{mv^2}{L^3}$

Factor of change. $P$ is proportional to $v^2$ and inversely proportional to $L^3$ (the volume).

Doubling $v$ multiplies $P$ by 4.

Doubling $L$ multiplies the volume by 8, which divides $P$ by 8.

Together: $P_{\text{new}} = \dfrac{4}{8}P = \dfrac{1}{2}P$ .

How it scores: the first point is for starting from $F = \Delta p/\Delta t$ (or the impulse-momentum theorem) and $P = F_\perp/A$ ; separate points go to $\Delta p = 2mv$ and $\Delta t = 2L/v$ ; the final point is for $P = mv^2/L^3$ in the listed symbols.

A correct final expression written alone would earn the later points but not the first.

Worked example 4: the absolute pressure of a gas under a piston (a short exam-style chain)

A vertical cylinder of gas is closed by a frictionless piston of mass 2.0 kg and area 10 cm², as in the cylinder diagram above.

The air above the piston is at $P_{\text{atm}} = 1.0 \times 10^5$ Pa.

The piston is at rest.

Part A On a dot representing the piston, draw and label the forces exerted on the piston.

Part B Derive an expression for the pressure $P_{\text{gas}}$ of the trapped gas in terms of $M$ , $A$ , $P_{\text{atm}}$ and physical constants.

Part C Calculate $P_{\text{gas}}$ .

Part A. The free-body diagram above: $F_{\text{gas}}$ up; $F_{\text{atm}}$ and $F_g$ down; the up arrow as long as the two down arrows together, because the piston is at rest.

Part B. Newton's second law for the piston, with $a = 0$ and up positive:

$\sum F = F_{\text{gas}} - F_{\text{atm}} - F_g = 0$

$P_{\text{gas}}A - P_{\text{atm}}A - Mg = 0$

$P_{\text{gas}} = P_{\text{atm}} + \dfrac{Mg}{A}$

Part C. Convert the area: $10 \text{ cm}^2 = 10 \times 10^{-4} \text{ m}^2 = 1.0 \times 10^{-3} \text{ m}^2$ .

$\dfrac{Mg}{A} = \dfrac{(2.0 \text{ kg})(9.8 \text{ m/s}^2)}{1.0 \times 10^{-3} \text{ m}^2} = 1.96 \times 10^4 \text{ Pa}$

$P_{\text{gas}} = 1.0 \times 10^5 \text{ Pa} + 0.196 \times 10^5 \text{ Pa} = 1.2 \times 10^5 \text{ Pa}$

Points: Part A: one point for each correct, labeled force ( $F_g$ down, $F_{\text{atm}}$ down, $F_{\text{gas}}$ up).

Part B: Point B1 for starting from Newton's second law with the forces from Part A; Point B2 for writing each pressure force as pressure times area; Point B3 for the final expression.

Part C: one point for the value with units.

Answer: $P_{\text{gas}} = P_{\text{atm}} + Mg/A = 1.2 \times 10^5$ Pa.

The piston adds only about 20% to the atmosphere's pressure, a sensible result for a 2 kg piston.

A diagram with an arrow labeled " $P_{\text{gas}}$ " or "pressure" would lose that force's point, and leaving out $P_{\text{atm}}$ gives only the extra $2.0 \times 10^4$ Pa, which is not the gas pressure.

Worked example 5: where pressure exists (full and weaker answers)

A large tank holds gas at pressure $P$ .

A student claims: "The gas pushes only on the walls of the tank.

A small pressure sensor held in the middle of the tank, away from every wall, would read zero."

Indicate whether the claim is correct or incorrect.

Justify your answer.

Full-credit answer. Incorrect.

The atoms of the gas are moving in random directions throughout the whole tank, not only near the walls.

Any surface placed in the middle of the gas, including the face of the sensor, is struck by atoms on that side, and each collision exerts a perpendicular force on it.

So the sensor reads the same pressure $P$ as a sensor in the wall: the pressure exists throughout the gas, not just at its boundary.

Point A1: For indicating "incorrect" with an attempt at a relevant justification.

Point A2: For reasoning that atoms move throughout the gas (in all directions), so they collide with any surface placed in it.

Point A3: For linking those collisions to a force per unit area on the sensor, so it reads the gas pressure.

Weaker answer: "Incorrect, because pressure is the same everywhere in a gas."

This earns the selection point, but it restates the conclusion without saying why.

It never mentions atoms or collisions, so it earns neither reasoning point.

A justification has to supply the mechanism.

Worked example 6: a net force from unbalanced collisions (full and weaker answers)

A sealed, empty plastic water bottle is taken on a flight with the cap screwed on tightly at cruising altitude, where the cabin pressure is lower than at sea level.

After landing, the bottle is found partly crushed.

A student claims: "The air inside the bottle pulled the walls inward."

Indicate whether the claim is correct or incorrect.

Justify your answer.

Full-credit answer. Incorrect.

A gas can only push on a surface, through atoms colliding with it.

When the bottle was sealed, the air inside and outside were at the same pressure, so collisions on the two sides of the wall balanced.

At sea level the outside air is at a higher pressure, so outside atoms deliver more momentum per second to each square meter of the outer surface than inside atoms deliver to the inner surface.

The net force on the wall points inward, and the wall bends in until the inside pressure rises enough to balance it.

Point A1: For indicating "incorrect" with an attempt at a relevant justification.

Point A2: For reasoning that forces from a gas come from collisions that push on a surface (a gas cannot pull).

Point A3: For reasoning that the outside pressure (collisions per area per time, or momentum delivered) is larger than the inside pressure, so the net force on the walls is inward.

Weaker answer: "Incorrect, because the pressure outside was higher, so the bottle was crushed."

This earns the selection and the third link, but it never says why a gas cannot pull, so it misses Point A2.

Name the mechanism (collisions push) and then compare the two sides.

Exam tip

A free-body diagram shows forces, never pressures.

Label each arrow $F$ with a subscript naming the object that exerts it ( $F_g$ , $F_{\text{atm}}$ , $F_{\text{gas}}$ ), start every arrow on the dot, and in Newton's second law enter each pressure as pressure times area.

Absolute gas pressure from these forces was one of the hardest points on a recent paper, mostly because answers left out the atmosphere or wrote a pressure where a force belonged.

Common mistakes

Drawing pressure as a force. An arrow labeled $P$ or "pressure" scores nothing; write $F_{\text{gas}} = P_{\text{gas}}A$ . The mix-up is tempting because pressure "pushes", but it is force per area, a scalar.

A generic or self-made force. No "applied force", no " $F_{\text{piston}}$ " on the piston's own diagram. Every force has an object exerting it: the Earth, the air above, the gas below.

Leaving out the atmosphere. $P_{\text{gas}} = Mg/A$ is only the extra pressure due to the piston. It is tempting because the air above seems "empty", but it is pressing down at about $10^5$ Pa.

Using the whole velocity at an angle. Only the perpendicular component reverses, so $|\Delta p| = 2mv\cos\theta$ with $\theta$ from the perpendicular. Using $2mv$ treats every hit as head-on.

Writing $\Delta p = 0$ for an elastic bounce. The speed is unchanged, but the direction reverses, so the momentum changes by $2mv_\perp$ . The kinetic energy is what stays the same.

Thinking a vacuum or low pressure "sucks". A gas can only push. A can or bottle collapses because the outside gas pushes harder than the inside gas.

Forgetting to square area conversions. 1 cm² is $10^{-4}$ m², not $10^{-2}$ m².

Check your understanding

1. A gas exerts a total perpendicular force of 15 N on a flat surface of area 25 cm². What is the pressure of the gas on the surface?

(A) 0.60 Pa

(B) 60 Pa

(D) $6.0 \times 10^{5}$ Pa

Total [1]

2. An atom moving toward a smooth, fixed wall at an angle to the perpendicular rebounds elastically. A student claims that during the collision the wall exerts a force on the atom with a component parallel to the wall. Which of the following best evaluates the student's claim?

(A) Correct, because the atom's velocity has a component parallel to the wall, so the force must too.

(B) Correct, because the atom's direction of motion changes, which requires a force along every direction of motion.

(C) Incorrect, because only the perpendicular component of the atom's momentum changes, so the force on the atom is perpendicular to the wall.

(D) Incorrect, because an elastic collision involves no force, only a change of direction.

Total [1]

3. A container holds $N$ identical atoms. The atoms are replaced by $N/2$ atoms of the same kind, each moving twice as fast on average, and the volume of the container is unchanged. Using the kinetic model of pressure, by what factor does the pressure on the walls change?

(A) $\dfrac{1}{2}$

(B) $1$

(D) $4$

Total [1]

4. Qualitative/Quantitative Translation

Two identical rigid containers hold the same number of atoms at the same absolute temperature $T$ .

Container X holds helium, with atoms of mass $m$ .

Container Y holds argon, with atoms of mass $10m$ .

Each container is a cube with edges of length $L$ .

Part A A student claims: "The argon exerts the greater pressure on the walls, because each argon atom has more momentum, so each of its collisions with a wall delivers a larger impulse."

Indicate whether the claim is correct or incorrect.

Justify your answer without manipulating equations.

Part B Model one atom of each gas as moving back and forth along a line perpendicular to two opposite walls of its container, with the average kinetic energy of an atom of that gas.

Derive an expression for the ratio $\dfrac{F_Y}{F_X}$ of the average force exerted on one wall by the argon atom to the average force exerted on one wall by the helium atom.

Express your answer in terms of $m$ , $L$ , $T$ and physical constants, as appropriate.

Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

Part C Indicate whether your expression in Part B is consistent with your answer to Part A. Briefly justify your answer.

Total [8]

5. Two identical sealed boxes contain the same number of identical atoms with the same speeds. In Box 1 the atoms strike the walls at random angles. A student claims that if, in Box 2, every atom struck the walls at a glancing angle (moving almost parallel to the wall it hits), the pressure on the walls of Box 2 would be greater, because each atom would slide a long way along the wall during each hit. Indicate whether the claim is correct or incorrect. Justify your answer. Your justification may include expressions/equations but must include conceptual reasoning beyond algebraic solutions.

Total [3]

6. An argon atom of mass $6.6 \times 10^{-26}$ kg moving at 400 m/s strikes a wall head-on and rebounds elastically.

Part A Calculate the magnitude of the change in the atom's momentum.

Part B The gas in the container is at $1.0 \times 10^5$ Pa.

Calculate the number of such collisions per second that would be needed on 1.0 cm² of the wall to produce that pressure.

Total [3]

7. A student wants to determine atmospheric pressure in the lab using a vertical cylinder of gas closed by a light, frictionless piston of known area $A$ . The student can place slotted masses on the piston and has a pressure sensor connected to the gas. The gas temperature is kept at room temperature.

Part A Indicate what quantities could be graphed on each axis to produce a linear graph from which atmospheric pressure can be determined.

Part B Describe the feature of the graph that would give atmospheric pressure.

Total [2]

Answers

1. The answer is (C). $A = 25 \times 10^{-4} \text{ m}^2 = 2.5 \times 10^{-3} \text{ m}^2$ , so $P = \dfrac{15 \text{ N}}{2.5 \times 10^{-3} \text{ m}^2} = 6.0 \times 10^3$ Pa.

(A) divides by 25 with the area left in cm².
(B) converts cm² to m² by dividing by 100 instead of $10^4$ .
(D) treats the area as mm² ( $25 \times 10^{-6}$ m²).

2. The answer is (C). A smooth fixed wall can push only perpendicular to its surface; the parallel component of the atom's velocity, and so of its momentum, is unchanged.

(A) confuses the direction of the velocity with the direction of the force.
(B) tempts because the direction of motion changes, but only the perpendicular component changes.
(D) is wrong: any change of momentum needs a force, elastic or not.

3. The answer is (C). In the kinetic model the pressure is proportional to the number of atoms times the average of $v^2$ : $P \propto Nv^2/V$ . Halving $N$ and doubling the speed gives $\dfrac{1}{2} \times 4 = 2$ .

(A) uses only the change in the number of atoms.
(B) cancels the halving against a doubling, treating pressure as proportional to $v$ , not $v^2$ .
(D) uses only the change in speed.

4. Part A (3 points)

Point A1: For indicating "incorrect" with an attempt at a relevant justification.
Point A2: For reasoning that at the same temperature the atoms of both gases have the same average kinetic energy, so the more massive argon atoms move more slowly and hit a given wall less often, even though each hit delivers a larger impulse.
Point A3: For reasoning that the two effects cancel: the pressure depends on the number of atoms per unit volume and their average kinetic energy, which are the same in both containers, so the pressures are equal.

Example Response: Incorrect.

Both gases are at the same temperature, so their atoms have the same average kinetic energy.

An argon atom, ten times as massive, must therefore move more slowly than a helium atom.

Each argon collision does deliver a larger impulse, but a slower atom takes longer to cross the container and comes back to the wall less often.

The larger impulse per hit and the fewer hits each second cancel, so the same number of atoms with the same average kinetic energy in the same volume give the same pressure.

Part B (3 points)

Point B1: For a multistep derivation that starts from $F\Delta t = \Delta p$ (or $\Delta p = 2m_av$ for one collision) together with $K = \frac{1}{2}mv^2$ or $K_{\text{avg}} = \frac{3}{2}k_BT$ .

Point B2: For the average force of one atom of mass $m_a$ and speed $v$ on one wall: momentum change $2m_av$ per hit and time $\dfrac{2L}{v}$ between hits on that wall, so $F = \dfrac{2m_av}{2L/v} = \dfrac{m_av^2}{L} = \dfrac{2K}{L}$ .

Point B3: For both atoms having the same $K = \frac{3}{2}k_BT$ , so $\dfrac{F_Y}{F_X} = \dfrac{2K/L}{2K/L} = 1$ .

Scoring Note: A correct final expression written on its own earns B2 and B3 but not B1.

An answer that keeps the masses, such as $\sqrt{10}$ or 10, does not earn B3.

Part C (2 points)

Point C1: For indicating "consistent" (or a selection consistent with the student's own Parts A and B).

Point C2: For a functional-dependence reason: in Part B the force from one atom is $2K/L$ , which depends on the atom's kinetic energy and not on its mass; equal temperatures give equal $K$ , so each argon atom pushes on the wall as hard on average as each helium atom, and equal numbers of atoms give equal pressures, so the claim is incorrect, as in Part A.

5. (3 points)

Point A1: For indicating "incorrect" with an attempt at a relevant justification.
Point A2: For reasoning that in a collision with a wall only the perpendicular component of an atom's momentum changes (the parallel component is unchanged), so a glancing hit changes the momentum by only a small amount, $2mv_\perp$ with $v_\perp$ small.
Point A3: For linking the smaller momentum change per hit (and the force perpendicular to the wall) to a smaller sum of perpendicular force components per area, so the pressure in Box 2 would be smaller, not greater.

Example Response: Incorrect.

Pressure is the sum of the perpendicular force components on a surface divided by its area.

A smooth wall can push an atom only perpendicular to itself, so in each collision only the perpendicular component of momentum reverses, giving a change of $2mv_\perp$ .

In a glancing hit $v_\perp$ is very small, so each collision delivers only a small impulse, and sliding along the wall delivers none.

Box 2 would have a lower pressure.

(Scoring Note: the claim's "slides a long way" describes the unchanged parallel component, which contributes nothing to pressure.)

6. Part A (1 point)

Point A1: For $|\Delta p| = 2mv = 2(6.6 \times 10^{-26} \text{ kg})(400 \text{ m/s}) = 5.3 \times 10^{-23}$ kg·m/s, with units.

Part B (2 points)

Point B1: For $F_\perp = PA$ with the area converted: $(1.0 \times 10^5 \text{ Pa})(1.0 \times 10^{-4} \text{ m}^2) = 10$ N.

Point B2: For the number per second $= \dfrac{10 \text{ N}}{5.3 \times 10^{-23} \text{ kg·m/s}} = 1.9 \times 10^{23}$ per second.

Scoring Note: Part B earns full credit if it uses the student's own Part A value consistently.

7. (2 points)

Point A1: For indicating the gas pressure $P_{\text{gas}}$ on the vertical axis and the total mass $M$ placed on the piston on the horizontal axis (or the reverse).
Point A2: For describing that atmospheric pressure is the vertical intercept (the pressure when $M = 0$ ), because $P_{\text{gas}} = P_{\text{atm}} + (g/A)M$ is a straight line with slope $g/A$ and intercept $P_{\text{atm}}$ . If the axes are reversed, $P_{\text{atm}}$ is where the line crosses the pressure axis (the horizontal intercept).

Scoring Note: Because the gas stays at room temperature, its volume falls as masses are added, but that does not affect the force balance on the piston, which is all the graph uses.

Kinetic theory of pressure: atoms colliding with surfaces

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