Lesson C · 5.15–5.22
By the end you should explain gas pressure with collisions, convert °C to kelvin, and choose $p_1V_1=p_2V_2$ or $p_1/T_1=p_2/T_2$.
How I start gases
A gas is a crowd of particles flying in random directions. Every time a particle hits a wall it delivers a tiny force. Billions of hits each second on each square metre are the pressure. If your explain answer never says “collide” or “collision”, I cannot give the first mark. We are not doing specific heat capacity or melting and boiling graphs on this Double Award course — those are 5.2P and 5.8P–5.14P.
5.15
Gas particles move randomly and rapidly. They travel in straight lines between collisions. When a particle hits the wall of the container it changes direction, so the wall exerts a force on the particle and, by Newton’s third law, the particle exerts an equal force on the wall. Pressure is that force spread over the area of the wall:
$$p = \frac{F}{A}$$The pressure is the same on every wall of a rectangular box of still gas (apart from a tiny gravitational effect we ignore). It acts at right angles to each surface.
3-mark “explain how a gas exerts a pressure”
Particles move randomly / in all directions [1]. They collide with the walls of the container [1]. Each collision exerts a force; force per unit area is pressure [1].
5.16–5.19
Cool a gas and the particles slow down. There is a lowest possible temperature: the particles have (almost) zero kinetic energy, they stop colliding with the walls, and the pressure would fall to zero. That temperature is absolute zero.
$$\text{absolute zero} = -273^\circ\text{C} = 0\,\text{K}$$The kelvin scale starts there. A kelvin interval is the same size as a Celsius degree, so you convert by adding or subtracting $273$ — never by a factor of $273$.
$$T(\text{K}) = \theta(^\circ\text{C}) + 273 \qquad \theta(^\circ\text{C}) = T(\text{K}) - 273$$| Celsius | Kelvin | Note |
|---|---|---|
| $-273^\circ\text{C}$ | $0\,\text{K}$ | Absolute zero. No negative kelvin |
| $0^\circ\text{C}$ | $273\,\text{K}$ | Ice point. Not “zero kelvin” |
| $20^\circ\text{C}$ | $293\,\text{K}$ | A typical room |
| $100^\circ\text{C}$ | $373\,\text{K}$ | Steam point |
Specification 5.18: if you raise the temperature, the average speed of the molecules increases. Specification 5.19: the kelvin temperature is proportional to the average kinetic energy of the particles. Double $T$ in kelvin and you double the average KE. That is why $0^\circ\text{C}$ is not “zero energy” — $0^\circ\text{C}=273\,\text{K}$, so the particles still have plenty of KE.
“The particles stop at $0^\circ\text{C}$” is wrong
$0^\circ\text{C}$ is the melting point of ice, not absolute zero. Particles in a gas at $0^\circ\text{C}$ are still moving quickly. Absolute zero is $-273^\circ\text{C}$.
5.20 · Qualitative
Always state what is being held constant. Both stories below assume a fixed mass of gas (the same number of particles, no leaks).
Boyle — smaller $V$, same $T$
The particles have less space. They hit the walls more frequently. Same average speed (temperature unchanged), so each hit is about as hard as before, but there are more hits per second. Pressure rises.
Pressure law — higher $T$, same $V$
Average speed increases, so average KE increases. Particles hit the walls more often and harder (larger force per collision). Pressure rises.
4-mark “explain why $p$ increases when $T$ increases at constant $V$”
Temperature in kelvin is proportional to average KE / particles move faster [1]. They collide with the walls more frequently [1]. They collide with a greater force [1]. Force on a given area increases, so pressure increases [1].
3-mark “explain why $p$ increases when $V$ decreases at constant $T$”
Particles are in a smaller volume / closer together [1]. They collide with the walls more frequently [1]. Greater force per unit area, so pressure increases [1]. Do not say they move faster — $T$ is constant, so average speed is unchanged.
5.21–5.22
| Law | Held constant | Equation | Units warning |
|---|---|---|---|
| Boyle (5.22) | Temperature (and mass) | $p_1 V_1 = p_2 V_2$ | $p$ and $V$ can stay in kPa and $\text{cm}^3$ if both sides match |
| Pressure law (5.21) | Volume (and mass) | $p_1/T_1 = p_2/T_2$ | $T$ must be kelvin |
You may also write $pV=\text{constant}$ and $p/T=\text{constant}$. Same physics. I still want you to label “1” as before and “2” as after.
Board example — Boyle
A trapped volume of air is $40\,\text{cm}^3$ at $100\,\text{kPa}$. It is compressed slowly to $20\,\text{cm}^3$. The temperature stays constant. Find the new pressure.
$p_1=100\,\text{kPa}$, $V_1=40\,\text{cm}^3$, $V_2=20\,\text{cm}^3$.
$$p_2 = \frac{p_1 V_1}{V_2} = \frac{100\times 40}{20} = 200\,\text{kPa}$$Volume halved, pressure doubled. That is the sense-check. I did not convert $\text{cm}^3$ because both volumes are in the same unit and they cancel.
Board example — $p/T$
A sealed can of fixed volume contains gas at $120\,\text{kPa}$ and $20^\circ\text{C}$. It is heated to $100^\circ\text{C}$. Calculate the new pressure.
Convert first: $T_1=20+273=293\,\text{K}$, $T_2=100+273=373\,\text{K}$.
$$p_2 = p_1 \times \frac{T_2}{T_1} = 120 \times \frac{373}{293} = 153\,\text{kPa}$$If you used $20$ and $100$, you would get $600\,\text{kPa}$ — five times too large, and you should smell a rat.
The mistake I mark most often on 5.21
Leaving the temperatures in °C. $p\propto T$ is only true on the kelvin scale, because only then does $p=0$ at $T=0$. A graph of $p$ against °C does not go through the origin; it hits zero at $-273^\circ\text{C}$.
Choosing the equation in 10 seconds
1.
Convert $27^\circ\text{C}$ to kelvin and $310\,\text{K}$ to °C.
2.
A gas at $250\,\text{kPa}$ occupies $8.0\times 10^{-4}\,\text{m}^3$. It expands at constant temperature to $1.0\times 10^{-3}\,\text{m}^3$. Calculate the new pressure.
3.
Why must temperatures be in kelvin in $p_1/T_1=p_2/T_2$?