Physics · 2 Thermal physics

2.1.3 Gases and the absolute scale of temperature

Before you start

The particles of a gas are far apart and move randomly at high speed in all directions, colliding with each other and with the walls of their container (2.1.2).
Pressure is force per unit area, $p = \frac{F}{A}$ , measured in pascals: $1$ Pa $= 1$ N/m² (1.8).
When the temperature of a gas rises, its particles gain kinetic energy, so on average they move faster (2.1.2).
There is a lowest possible temperature, $- 273$ °C, called absolute zero, where the particles have least kinetic energy (2.1.2).
Rearranging an equation such as $a \times b = c \times d$ to make one letter the subject.

Syllabus objectives covered (3)

Describe qualitatively, in terms of particles, the effect on the pressure of a fixed mass of gas of: (a) a change of temperature at constant volume (b) a change of volume at constant temperature
Convert temperatures between kelvin and degrees Celsius; recall and use the equation T (in K) = θ (in °C) + 273
Recall and use the equation pV = constant for a fixed mass of gas at constant temperature, including a graphical representation of this relationship

Part 1 of 7

Where the pressure of a gas comes from

Every time a gas particle hits a wall of its container and bounces off, it pushes on the wall. One collision is a tiny force for a tiny moment. Billions of collisions every second add up to a steady force on the wall.
Pressure is that force spread over the area of the wall: the force per unit area. So the pressure of a gas is the total force of its particles' collisions on each square metre of wall.
This gives you exactly two ways to raise the pressure. Either the particles hit each unit area of wall more often, or each collision pushes harder (with a greater force). Every pressure explanation in this lesson is one or both of these.
Notice that it is collisions with the walls that make the pressure. Particles also collide with each other, but those collisions do not push on the container. 'The particles collide more' does not explain anything until you say they collide with the walls.
Fixed mass means no gas gets in or out. The number of particles stays the same. Every situation in this subtopic is about a fixed mass of gas.

Part 2 of 7

Heating a gas whose volume cannot change

Picture a fixed mass of gas in a strong, sealed metal can. The can is rigid, so the volume of the gas cannot change. Now heat it.
Step 1. The temperature rises, so the particles gain kinetic energy. On average they move faster.
Step 2. Faster particles cross the can more quickly, so each particle hits the walls more often. Collisions are more frequent.
Step 3. Faster particles also hit the walls harder, so each collision exerts a greater force.
Step 4. More frequent collisions and a greater force per collision both give a greater force on each unit area of wall. The pressure increases.
apparatus diagram — not drawn yet
Two identical sealed rigid boxes side by side, each containing the same twelve dots for gas particles. Left box labelled 'cool': each dot has a short arrow in a random direction. Right box labelled 'hot': the same number of dots, each with a longer arrow in a random direction, and small star marks where several arrows meet the walls. A pressure gauge on top of each box reads lower on the left and higher on the right.
Same number of particles, same volume. Hotter particles move faster, so they hit the walls more often and harder.
The reverse argument works too. Cool the can and the particles slow down, so they hit the walls less often and with less force. The pressure falls.
This is why an aerosol can carries a warning not to heat it. The volume is fixed, so heating raises the pressure, and the can could burst.
The trap here is to say the particles get bigger or expand. They do not. The particles stay the same size; only their speed changes.

Part 3 of 7

Squeezing a gas at constant temperature

Now picture a fixed mass of gas in a sealed syringe. You push the plunger in slowly, so that the gas stays at room temperature. The volume decreases but the temperature does not change.
Step 1. The temperature is constant, so the average kinetic energy of the particles is unchanged. They move at the same average speed as before. Each collision with a wall exerts the same force as before.
Step 2. The same number of particles is now in a smaller volume. They are closer together and have less distance to travel between walls.
Step 3. So the particles hit each unit area of the walls more often: there are more collisions per second on each square metre.
Step 4. More collisions per second on each unit area means a greater force per unit area. The pressure increases.
apparatus diagram — not drawn yet
Two sealed syringes side by side, each containing the same ten dots for gas particles with arrows of the same length. Left: the plunger pulled well out, large volume, dots spread out. Right: the plunger pushed in to half the volume, the same ten dots closer together, arrows the same length as on the left. Both labelled 'same temperature'.
Same number of particles at the same speed in half the space, so more collisions per second with each unit area of wall.
Compare this with heating. Here only one of the two causes is at work: collisions are more frequent, but each one is no harder, because the speed has not changed.
The trap here is to say the particles move faster when the gas is squeezed. At constant temperature they do not. If your answer says they move faster, it contradicts 'constant temperature'.
The reverse argument: pull the plunger out at constant temperature, the volume increases, the particles hit each unit area of wall less often, and the pressure falls.

Part 4 of 7

Absolute zero and the kelvin scale

The Celsius scale puts its zero at the melting point of ice. That is a convenient choice, but nothing special happens to the particles there. Temperatures below $0$ °C are common, so Celsius temperatures can be negative.
From 2.1.2 you know there is a lowest possible temperature, absolute zero, at $- 273$ °C, where the particles have least kinetic energy. Nothing can be colder.
The kelvin scale starts at absolute zero. So $0$ K is absolute zero, and there are no negative temperatures in kelvin. It is called the absolute scale of temperature because its zero is absolute zero.
One kelvin is the same size as one degree Celsius. The two scales are the same ruler with the zero in a different place: the kelvin zero is $273$ steps below the Celsius zero.
apparatus diagram — not drawn yet
Two vertical temperature scales side by side, Celsius on the left and kelvin on the right, with horizontal lines joining equal temperatures: 100 °C and 373 K labelled 'water boils'; 0 °C and 273 K labelled 'ice melts'; −273 °C and 0 K labelled 'absolute zero'.
Each step on one scale is the same size as a step on the other. Only the zero is in a different place.
So to turn a Celsius temperature $θ$ into a kelvin temperature $T$ , add $273$ . You must recall this equation; Paper 4 does not give it to you.
$T (in K) = θ (in °C) + 273$
The unit is the kelvin, symbol K. There is no degree sign: write $300$ K, not $300$ °K.

Part 5 of 7

Converting between kelvin and degrees Celsius

Celsius to kelvin: add $273$ . For example, $27$ °C is $27 + 273 = 300$ K, and $- 40$ °C is $- 40 + 273 = 233$ K.
Kelvin to Celsius: rearrange the same equation by subtracting $273$ from both sides.
$θ (in °C) = T (in K) - 273$
For example, $373$ K is $373 - 273 = 100$ °C, and $50$ K is $50 - 273 = - 223$ °C. A kelvin temperature below $273$ K is always below $0$ °C, so its Celsius value is negative.
A quick check that you went the right way: the number in kelvin is always $273$ bigger than the number in °C. If your kelvin answer is the smaller number, you subtracted when you should have added.
A change in temperature is the same number on both scales, because the steps are the same size. A rise from $20$ °C to $50$ °C is a rise of $30$ °C, and also a rise of $30$ K (from $293$ K to $323$ K). Do not add $273$ to a temperature change.
Do not treat Celsius numbers as ratios. Going from $20$ °C to $40$ °C is not doubling the temperature: in kelvin it is $293$ K to $313$ K. Only the kelvin scale is measured from the true zero.

Part 6 of 7

pV = constant, and how to use it

You have seen that squeezing a gas at constant temperature raises its pressure. Measurements show something more exact: for a fixed mass of gas at constant temperature, the pressure multiplied by the volume always gives the same number.
$p V = constant$
This relationship is known as Boyle's law. You must recall the equation on Paper 4, and it holds only when both conditions are met: the mass of gas is fixed and the temperature is constant.
It means $p$ and $V$ are inversely proportional. Halve the volume and the pressure doubles. Make the volume three times as big and the pressure becomes a third. The product $p V$ never changes.
Because $p V$ is the same before and after a change, you can write the two states equal to each other. This is the form you use in calculations. The subscript 1 means before the change and 2 means after it.
$p_{1} V_{1} = p_{2} V_{2}$
Units: use any pressure unit and any volume unit, as long as you use the same units on both sides. If $p_{1}$ is in kPa, then $p_{2}$ comes out in kPa. If $V_{1}$ is in cm³, then $V_{2}$ comes out in cm³. You do not need to convert to Pa and m³.
Method: write down the four quantities, mark the one you want, write $p_{1} V_{1} = p_{2} V_{2}$ , rearrange by dividing both sides by whatever multiplies the unknown, then substitute. For an unknown $p_{2}$ , divide both sides by $V_{2}$ :
$p_{2} = \frac{p_{1} V_{1}}{V_{2}}$
Then sense-check the direction. A smaller volume must give a larger pressure, and a larger volume a smaller pressure. If your answer goes the other way, you have multiplied where you should have divided.

Part 7 of 7

Graphs of pV = constant

Plot pressure $p$ on the vertical axis against volume $V$ on the horizontal axis. Because $p$ and $V$ are inversely proportional, the graph is a curve, not a straight line.
The curve falls steeply at small volumes and gets less steep as the volume increases. It gets closer and closer to each axis without ever touching it, because $p V$ is never zero: neither $p$ nor $V$ can reach zero.
To test a set of results for this relationship, do not judge the shape by eye. Multiply $p$ by $V$ for each pair of readings. If the products are equal (or equal within experimental error), the gas obeys $p V =$ constant.
Now plot $p$ against $\frac{1}{V}$ instead. Since $p = constant \times \frac{1}{V}$ , this graph is a straight line through the origin. A straight line through the origin is the clearest graphical evidence that $p$ is inversely proportional to $V$ .
$p = constant \times \frac{1}{V}$
graph diagram — not drawn yet
Two graphs side by side. Left: pressure p / kPa on the vertical axis against volume V / cm³ on the horizontal axis, a smooth curve through (20, 200), (40, 100) and (80, 50), falling steeply then levelling out, approaching both axes but touching neither. Right: p / kPa on the vertical axis against 1/V / cm⁻³ on the horizontal axis, a straight line through the origin and the points (0.0125, 50), (0.025, 100) and (0.050, 200).
The same results plotted two ways. p against V curves; p against 1/V is a straight line through the origin.
The trap here is to sketch $p$ against $V$ as a straight line sloping down. A straight line that falls to the horizontal axis would mean the pressure reaches zero at some finite volume, which $p V =$ constant does not allow.

Worked examples

Example 1. A fixed mass of air is trapped in a sealed syringe. (a) The plunger is clamped so it cannot move, and the syringe is put into hot water. Explain, in terms of particles, why the pressure of the air increases. (b) The syringe is taken out and left to cool back to room temperature. The plunger is then unclamped and pushed in slowly, so the temperature of the air stays constant. Explain, in terms of particles, why the pressure of the air increases.

Show the working

Before writing, identify what is constant in each part. In (a) the volume is constant and the temperature rises. In (b) the temperature is constant and the volume falls. That decides which causes you can use.
(a) Temperature rises, so the air particles gain kinetic energy and move faster on average.
Faster particles hit the walls more often, and each collision exerts a greater force.
So the force per unit area on the walls is greater: the pressure increases. Notice you used both causes, because the speed changed.
(b) The temperature is constant, so the particles move at the same average speed as before. Say this: it shows you know the speed has not changed.
The same number of particles is in a smaller volume, so they hit each unit area of the walls more often.
More collisions per second on each unit area gives a greater force per unit area: the pressure increases. Only one cause is at work here.

Answer: (a) The particles move faster, so they collide with the walls more often and with greater force; the force per unit area (pressure) increases. (b) The particles move at the same average speed, but in a smaller volume they collide with each unit area of wall more often; the force per unit area (pressure) increases.

Example 2. (a) Normal human body temperature is $37$ °C. Convert this to kelvin. (b) Liquid nitrogen boils at $77$ K. Convert this to degrees Celsius. (c) The air in a room warms from $20$ °C to $80$ °C. State the rise in temperature in kelvin, and the final temperature in kelvin.

Show the working

(a) Celsius to kelvin: add $273$ .
$T = 37 + 273 = 310 K$
Check: the kelvin number is the bigger one. Good.
(b) Kelvin to Celsius: subtract $273$ . The answer is negative because $77$ K is below $273$ K, which is the melting point of ice.
$θ = 77 - 273 = - 196 °C$
(c) The rise is $80 - 20 = 60$ °C. A kelvin is the same size as a degree Celsius, so the rise is also $60$ K. Do not add $273$ to a change.
The final temperature is a temperature, not a change, so you do add $273$ .
$T = 80 + 273 = 353 K$

Answer: (a) $310$ K. (b) $- 196$ °C. (c) A rise of $60$ K; final temperature $353$ K.

Example 3. A cylinder closed by a piston contains $240$ cm³ of gas at a pressure of $100$ kPa. The mass of gas is fixed. (a) The piston is pushed in slowly until the volume is $80$ cm³. The temperature stays constant. Calculate the new pressure. (b) Starting again from $240$ cm³ and $100$ kPa, calculate the volume of the gas when its pressure is $150$ kPa at the same temperature.

Show the working

Fixed mass and constant temperature: both conditions hold, so $p V =$ constant applies. Recall it as $p_{1} V_{1} = p_{2} V_{2}$ .
(a) Write the four quantities: $p_{1} = 100$ kPa, $V_{1} = 240$ cm³, $V_{2} = 80$ cm³, $p_{2} = ?$ . Divide both sides by $V_{2}$ to get $p_{2}$ on its own.
$p_{2} = \frac{p_{1} V_{1}}{V_{2}} = \frac{100 \times 240}{80} = 300 kPa$
Sense-check: the volume became a third of its value, so the pressure should be three times as big. $3 \times 100 = 300$ kPa. It matches.
The pressure is in kPa because $p_{1}$ was in kPa. The cm³ cancel, so you never needed to convert them.
(b) Now the unknown is $V_{2}$ : $p_{1} = 100$ kPa, $V_{1} = 240$ cm³, $p_{2} = 150$ kPa. Divide both sides by $p_{2}$ .
$V_{2} = \frac{p_{1} V_{1}}{p_{2}} = \frac{100 \times 240}{150} = 160 {cm}^{3}$
Sense-check: the pressure went up, so the volume must go down. $160$ cm³ is less than $240$ cm³. It matches.

Answer: (a) $300$ kPa. (b) $160$ cm³.

Example 4. A student traps a fixed mass of air and changes its volume slowly at constant temperature. Her results are: $V = 20$ cm³, $p = 200$ kPa; $V = 25$ cm³, $p = 160$ kPa; $V = 40$ cm³, $p = 100$ kPa; $V = 50$ cm³, $p = 80$ kPa; $V = 80$ cm³, $p = 50$ kPa. (a) Determine whether her results obey $p V =$ constant. (b) Predict the pressure when the volume is $32$ cm³. (c) Describe the shape of the graph of $p$ against $V$ , and of the graph of $p$ against $\frac{1}{V}$ .

Show the working

(a) Do not judge by eye. Multiply each pair.
$20 \times 200 = 4000, 25 \times 160 = 4000, 40 \times 100 = 4000, 50 \times 80 = 4000, 80 \times 50 = 4000$
Every product is $4000$ kPa cm³. The product is constant, so the results obey $p V =$ constant.
(b) Use the constant you just found. At $32$ cm³, $p \times 32$ must also be $4000$ .
$p = \frac{4000}{32} = 125 kPa$
Sense-check: $32$ cm³ lies between $25$ cm³ and $40$ cm³, so the pressure should lie between $160$ kPa and $100$ kPa. $125$ kPa does.
(c) $p$ against $V$ : a curve that falls steeply at small volumes and gets less steep as the volume increases, never touching either axis. $p$ against $\frac{1}{V}$ : a straight line through the origin.

Answer: (a) $p V = 4000$ kPa cm³ for every reading, so yes. (b) $125$ kPa. (c) $p$ against $V$ is a curve, falling and getting less steep, touching neither axis; $p$ against $\frac{1}{V}$ is a straight line through the origin.

Where people go wrong

Mistake: Saying that when a gas is heated, the particles expand or get bigger.

Instead: The particles stay the same size. They gain kinetic energy and move faster.

Pressure comes from collisions with the walls. Faster particles hit the walls more often and harder; their size has nothing to do with it.

Mistake: Saying that the particles move faster when a gas is compressed at constant temperature.

Instead: At constant temperature the average speed of the particles does not change. The pressure rises only because the particles hit each unit area of wall more often.

Temperature is linked to the average kinetic energy of the particles. If the temperature is constant, so is their average speed.

Mistake: Explaining pressure with 'the particles collide more' without saying what they collide with.

Instead: Say they collide with the walls (of the container) more often.

Pressure is the force per unit area on the walls. Collisions between particles do not push on the walls.

Mistake: Subtracting $273$ when converting °C to K, or adding $273$ to a temperature change.

Instead: Celsius to kelvin: add $273$ . Kelvin to Celsius: subtract $273$ . A change of temperature is the same number in °C and in K.

The kelvin number is always $273$ bigger than the Celsius number for the same temperature, and the steps on the two scales are the same size.

Mistake: Sketching $p$ against $V$ as a straight line sloping down to the axis.

Instead: It is a curve that gets less steep and never touches either axis. The straight line through the origin is $p$ against $\frac{1}{V}$ .

$p$ is inversely proportional to $V$ , so doubling $V$ halves $p$ . A straight falling line would give zero pressure at a finite volume, which $p V =$ constant does not allow.

Check yourself

A sealed, rigid can of gas is heated. Explain, in terms of particles, why the pressure inside the can increases.
Answer
The particles gain kinetic energy and move faster, so they hit the walls more often and with greater force. The force per unit area on the walls (the pressure) increases.
A gas is compressed slowly at constant temperature. Does the average speed of its particles increase, decrease or stay the same? What does change?
Answer
It stays the same, because the temperature is constant. What changes is how often the particles hit each unit area of the walls: more often, so the pressure increases.
Convert $25$ °C to kelvin.
Answer
$25 + 273 = 298$ K.
Convert $250$ K to degrees Celsius.
Answer
$250 - 273 = - 23$ °C.
The temperature of a gas falls by $15$ °C. State the fall in kelvin.
Answer
$15$ K. A change in temperature is the same number on both scales.
A fixed mass of gas at $120$ kPa has a volume of $0.50$ m³. At constant temperature it is compressed to $0.20$ m³. Calculate its new pressure.
Answer
$p_{2} = \frac{120 \times 0.50}{0.20} = 300$ kPa.
The pressure of a fixed mass of gas is made three times as big at constant temperature. What happens to its volume?
Answer
It becomes a third of its original value, because $p V$ stays constant.
For a fixed mass of gas at constant temperature, which graph is a straight line through the origin?
Answer
Pressure $p$ plotted against $\frac{1}{V}$ (one divided by the volume).

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A strong steel cylinder contains a fixed mass of gas. The volume of the cylinder does not change. At first the gas is at a temperature of $17$ °C.

(a) Calculate the temperature of the gas in kelvin. 1 mark

Mark scheme

B1 290 K; unit required

(b) The cylinder is left in sunlight and the temperature of the gas rises to $47$ °C. State the rise in temperature of the gas in kelvin. 1 mark

Mark scheme

B1 30 K; do not accept 303 K or 320 K given as the rise

(c) Explain, in terms of particles, why the pressure of the gas increases as its temperature rises. 3 marks No calculator

Mark scheme

B1 particles move faster / have more (average) kinetic energy; do not credit 'particles expand' or 'particles get bigger'
B1 particles collide with the walls (of the cylinder) more often / more frequently; insufficient: 'more collisions' with no reference to the walls
B1 (each collision exerts) a greater force / particles hit the walls harder / greater change in momentum per collision, so the force per unit area / pressure increases; owtte

Question 2. The outlet of a bicycle pump is sealed so that no air can escape. The pump contains $150$ cm³ of air at a pressure of $1.0 \times 10^{5}$ Pa. The piston is pushed in slowly, and the temperature of the air stays constant.

(a) Explain, in terms of particles, why the pressure of the air increases as the piston is pushed in. 2 marks No calculator

Mark scheme

B1 the same number of particles in a smaller volume / particles closer together, owtte
B1 particles collide with (each unit area of) the walls more often, so the force per unit area increases; this mark cannot be scored if the answer says the particles move faster (CON)

(b) Calculate the pressure of the air when its volume is $60$ cm³. 3 marks

Mark scheme

C1 $p_{1} V_{1} = p_{2} V_{2}$ or $p V =$ constant, in any form
C1 $1.0 \times 10^{5} \times 150 = p \times 60$ or $p = \frac{1.0 \times 10^{5} \times 150}{60}$
A1 $2.5 \times 10^{5}$ Pa (accept 250 000 Pa or 250 kPa); unit required

(c) Apart from the temperature, state one quantity that must stay constant for your calculation in (b) to be valid. 1 mark No calculator

Mark scheme

B1 the mass of air / the number of particles / the amount of air (accept: no air leaks out); do not accept 'the volume' (it changes)

Question 3. A student investigates a fixed mass of gas at constant temperature. She measures its pressure $p$ at six different volumes $V$ . Her results are: $V = 10$ cm³, $p = 240$ kPa; $V = 15$ cm³, $p = 160$ kPa; $V = 20$ cm³, $p = 120$ kPa; $V = 30$ cm³, $p = 80$ kPa; $V = 40$ cm³, $p = 60$ kPa; $V = 60$ cm³, $p = 40$ kPa.

(a) Determine whether her results obey the relationship $p V =$ constant. Show your working. 2 marks

Mark scheme

M1 $p \times V$ calculated for at least three of the readings, or shown for at least two pairs of readings that doubling $V$ halves $p$ , owtte
A1 all products equal 2400 (kPa cm³) / $p$ halves each time $V$ doubles, so the results obey $p V =$ constant

(b) Sketch a graph of $p$ against $V$ for this gas. You do not need to plot the points accurately. 2 marks No calculator

Mark scheme

B1 axes labelled $p$ (vertical) and $V$ (horizontal) with a curve along which $p$ decreases as $V$ increases
B1 the curve gets less steep as $V$ increases and does not touch either axis; a straight line scores 0 for this mark

(c) Calculate the pressure of the gas when its volume is $48$ cm³. 2 marks

Mark scheme

C1 $p \times 48 = 2400$ or $p = \frac{2400}{48}$ , or any correct $p_{1} V_{1} = p_{2} V_{2}$ using a pair from the results
A1 50 kPa (accept $5.0 \times 10^{4}$ Pa); unit required

(d) State what quantity she should plot against $p$ to obtain a straight line through the origin. 1 mark No calculator

Mark scheme

B1 $\frac{1}{V}$ / one divided by the volume

How this is examined

Calculating a new pressure or volume with pV = constant: a two- or three-mark Calculate about a gas compressed or expanded in a cylinder, syringe or pump, with the temperature stated as constant Writing $p_{1} V_{1} = p_{2} V_{2}$ (or $p V =$ constant) earns the first mark; correct numbers substituted earns the next; the final value with its unit (the same pressure or volume unit as the data) earns the last. A correct final answer with its unit scores all the marks even without working, but working can still earn marks if the answer is wrong. The equation is not given on Paper 4, so you must recall it.
Why heating a gas at constant volume raises its pressure: a two- or three-mark Explain about a sealed rigid container that is heated (or cooled), asking for an answer in terms of particles Three separate points: 'the particles move faster (have more kinetic energy)'; 'they collide with the walls more often'; 'each collision exerts a greater force, so the force per unit area (pressure) increases'. The word walls must be there.
Why squeezing a gas at constant temperature raises its pressure: a two-mark Explain about a piston or plunger pushed in at constant temperature, asking for an answer in terms of particles 'The same number of particles is in a smaller volume' and 'so they collide with (each unit area of) the walls more often, giving a greater force per unit area'. Do not say the particles move faster: that contradicts the constant temperature.
Testing data for pV = constant and sketching the graph: a table of pressure and volume readings followed by a Determine, a Sketch of p against V, or a question about which graph is a straight line For the test: $p \times V$ worked out for several readings and seen to be the same. For the sketch: a falling curve that gets less steep and touches neither axis. For the straight line: ' $p$ against $\frac{1}{V}$ , through the origin'.
Converting between kelvin and degrees Celsius: a one-mark conversion, on its own or as the first step of a longer gas question The correct number with the correct unit: add $273$ for °C to K, subtract $273$ for K to °C, and write K with no degree sign. A temperature change is not converted.

Words to know

Gas pressure: The force per unit area on a surface caused by gas particles colliding with it.
Fixed mass of gas: A quantity of gas that none escapes from and none is added to, so the number of particles stays the same.
Absolute zero: The lowest possible temperature, $0$ K or $- 273$ °C, where the particles have least kinetic energy.
Kelvin (K): The unit of temperature on the absolute scale. One kelvin is the same size as one degree Celsius. $T$ (in K) $= θ$ (in °C) $+ 273$ .
Absolute scale of temperature: The kelvin scale, which starts at absolute zero, so it has no negative values.
Inversely proportional: Two quantities whose product is constant: when one is doubled, the other is halved. For a fixed mass of gas at constant temperature, $p$ is inversely proportional to $V$ .
Boyle's law: For a fixed mass of gas at constant temperature, $p V =$ constant.

All Physics lessons Next: 2.2.1 Thermal expansion of solids, liquids and gases