Every topic so far has been about bodies with a definite size and shape: blocks, trolleys, charged spheres, coils of wire. A gas has neither. It spreads out to fill whatever holds it, and three quantities describe its state: its pressure, its volume and its temperature. Change one of them and at least one of the others changes too. This first page of the topic works out how, in the simplest case: a fixed mass of gas, sealed so that none gets in or out, with one of the three quantities held constant while the other two change. That gives three processes and three short laws. The rest of the topic builds on them. The next page joins the three laws into one equation and reads them from graphs, the third explains them with molecules, and the last brings in heat and work.
The gas laws are a small but steady part of the physics paper. Every paper so far has had one or two gas-law questions, all between Q25 and Q33. One of them, about a gas whose pressure is reduced at constant temperature, was printed in the same words in December 2025 (Q28), January 2026 (Q28) and March 2026 (Q25). December 2025, Q33 asked for a volume after heating at constant pressure, and April 2026, Q33 for a pressure after heating at constant volume. All three are worked through below; March 2026, Q32, which reads a path on a graph, is on The Ideal Gas Equation & Gas Graphs page. None of them needs more than one law and one step. What they test is whether you read the process word correctly and put the temperatures in kelvin, and their wrong options come from a short list of slips, which this page names as they come up.
The state of a gas
The pressure p of a gas is the force it exerts on each square metre of the walls that hold it, p = F/S, where S is the area. Its unit is the pascal: 1 Pa = 1 N/m². The air around you is at about 1.0 × 10⁵ Pa. The volume V is the space the gas fills, which is the volume of its container. Its SI unit is the cubic metre; the litre and the millilitre are common too, with 1 L = 10⁻³ m³ and 1 mL = 10⁻⁶ m³. The temperature T says how hot the gas is. Later on this page you will see that the gas laws need it measured from a particular zero, and not from the freezing point of water.
The laws on this page are about a fixed mass of gas. The papers say "a certain mass of ideal gas" or "a fixed amount of gas"; in Chinese, 一定质量的气体. The gas is sealed in, so none escapes and none is added, and each law compares two states of the same gas. An ideal gas (理想气体) is the model gas the laws describe exactly. Real gases such as air follow it closely at ordinary temperatures and pressures, and the Molecules, Temperature & Internal Energy page explains why.
When the state of a fixed mass of gas changes, all three quantities can change at once. The simplest experiments hold one of them constant and look at how the other two are linked. That gives three processes. The papers name the process, not the law, so these three words carry the questions:
The three processes.Isothermal (等温, изотермический): the temperature is held constant. Isobaric (等压, изобарный): the pressure is held constant. Isochoric (等容, изохорный): the volume is held constant. The Greek roots help: iso- means equal, therm- is heat, bar- is weight, as in barometer, and chor- is space. In each process one quantity is fixed and the other two change.
Three ways to change the state of a fixed mass of gas, each named for the quantity held fixed. Isothermal: the cylinder stands in a water bath and the piston is pushed in slowly, so T stays the same while p rises and V falls. Isobaric: a freely sliding piston with a load on it keeps the pressure the same while the heated gas expands. Isochoric: a rigid, closed container keeps the volume the same while the pressure of the heated gas rises.
Questions often describe the setup instead of naming the process, so it helps to know how each quantity is held fixed. The temperature stays constant when the gas is in contact with something at a steady temperature, such as a large water bath or the room, and the change is made slowly, so that the gas has time to stay at that temperature. The pressure stays constant under a piston that slides freely. The piston is in equilibrium, with the gas pushing it up and its weight and the atmosphere pushing it down, and as long as the downward forces do not change, the gas must push up with the same pressure; The Ideal Gas Equation & Gas Graphs page calculates that pressure. The volume stays constant in a rigid, closed container, such as a steel cylinder or a glass flask.
Try it! · Check-in question
A steel gas cylinder is sealed and left in strong sunshine, and the gas inside warms up. Which kind of change does the gas undergo?
A steel cylinder keeps its shape, so the volume of the gas cannot change: the process is isochoric, at constant volume, so the answer is C. As the gas warms, its pressure rises. A confuses a fixed mass with a fixed temperature. No gas gets in or out, but the sunshine warms it, so its temperature changes. B confuses closed with constant pressure. A constant pressure needs a volume that is free to change, such as a gas under a sliding piston; in a closed steel cylinder the pressure rises as the gas warms. D misreads the process words. In every one of the three, one quantity is fixed and the other two change. Here V is fixed while p and T change together, which is exactly an isochoric change.
Isothermal: pressure and volume
Start with the temperature fixed. Seal some air in a syringe whose nozzle is connected to a pressure gauge, and push the piston in. The air pushes back harder and harder: the gauge reading rises as the volume falls. Push slowly, so that the air stays at room temperature. A quick push warms the air, for a reason the First Law of Thermodynamics page explains, and then the temperature is no longer fixed.
The figure shows readings from such an experiment. Pulled out to 80 mL, the air is at 50 kPa. Halve the volume to 40 mL and the pressure doubles to 100 kPa; halve it again to 20 mL and the pressure doubles again, to 200 kPa. At 10 mL it would be 400 kPa. Multiply each pair of readings and the product is the same every time: pV = 4000 kPa·mL. So at constant temperature the pressure is inversely proportional to the volume. Squeeze a gas into half the space and it pushes twice as hard on the walls.
The same air in a sealed syringe, kept at room temperature, with the volume drawn to scale. Each time the volume halves, the pressure doubles, and the product pV stays at 4000 kPa·mL.
Isothermal change. For a fixed mass of gas at constant temperature, pV = constant, so p₁V₁ = p₂V₂. The pressure and the volume are inversely proportional: multiply one by a factor and the other is divided by the same factor. Any units will do, as long as both states use the same ones. This is Boyle's law.
Worked example · a new volume from a new pressure
A fixed mass of gas occupies 2.4 L at 1.0 × 10⁵ Pa. Its pressure is raised to 1.6 × 10⁵ Pa while its temperature is kept constant. What is its volume now?
Step 1: the temperature is constant, so p₁V₁ = p₂V₂.
Step 2: V₂ = p₁V₁/p₂ = (1.0 × 10⁵ × 2.4) ÷ (1.6 × 10⁵) = 1.5 L. The powers of ten cancel, and the litres never needed converting to m³.
Check the direction: the pressure went up, so the volume should go down, and it did. The factor gives the same result in one line: the pressure is multiplied by 1.6, so the volume is divided by 1.6, and 2.4 ÷ 1.6 = 1.5 L.
Worked example · pressure down at constant temperature (December 2025, Q28 · January 2026, Q28 · March 2026, Q25)
A certain mass of ideal gas, with its temperature kept constant and pressure reduced, its volume would ( ).
(A) remains unchanged (B) decrease (C) all options above are possible (D) increase
Step 1: "a certain mass" means a fixed mass, and "temperature kept constant" makes the change isothermal, so pV = constant.
Step 2: the pressure falls. For the product pV to stay the same, the volume must rise, by the same factor: halve the pressure and the volume doubles. The answer is D.
B treats pressure and volume as directly proportional, so that both fall together. The syringe shows the opposite: a gas given more space pushes less hard on its walls. A, an unchanged volume, would make pV fall along with the pressure. C avoids choosing an answer. The stem fixes the mass and the temperature, which is everything pV = constant needs, so only one of the options can happen.
This question was printed word for word in three papers, and it is worked through once, here. Only the order of the options changed: "increase" is B in January 2026 and C in March 2026. January also printed "all options above are possible" as option A, before the options it refers to; it is wrong there too. Learn the working, not the letter.
The hedge. "All options above are possible" has been one of the four options every time this question was printed. It is wrong whenever the question fixes the mass of gas and holds one quantity constant, because then the law decides the answer. It could be right only if something the law needs were missing: if gas could leak in or out, for example, or if the temperature were allowed to change as well.
Try it! · Check-in question
The outlet of a bicycle pump is blocked, and the pump holds 300 mL of air at 1.0 × 10⁵ Pa. The handle is pushed in slowly, so that the air stays at room temperature, until the air takes up 200 mL. What is its pressure now?
The handle moves slowly and the air stays at room temperature, so the change is isothermal: p₁V₁ = p₂V₂, and p₂ = 1.0 × 10⁵ × 300 ÷ 200 = 1.5 × 10⁵ Pa, so the answer is A. The volume is divided by 1.5, so the pressure is multiplied by 1.5. B multiplies the pressure by 200/300, as if the pressure fell with the volume; squeezing a gas raises its pressure, as the harder push needed on the handle shows. C divides by the change in volume, 100 mL, instead of by the new volume: 300/100 = 3. Boyle's law compares the two volumes themselves. D treats "isothermal" as "nothing changes", but only the temperature is held fixed; p and V change together.
Temperature from absolute zero
Now hold the volume fixed instead, and change the temperature. Seal some air in a rigid glass flask connected to a pressure gauge, and stand the flask in water baths at different temperatures, waiting each time for the air to reach the temperature of the water. The warmer the water, the higher the reading, and equal steps of temperature give equal steps of pressure: plotted against the Celsius temperature t, the readings lie on a straight line. But the line does not pass through the origin. At 0 °C the air still has most of its pressure, and warming it from 50 °C to 100 °C raises the pressure by only about 15%, not by a factor of 2. The pressure is not proportional to the Celsius temperature.
Extend the line backwards, to temperatures colder than any water bath. It reaches zero pressure at about −273 °C. Repeat the experiment with a different amount of air in the flask, or with a different gas, and the line has a different slope, but it reaches zero at the same temperature, −273 °C. No real gas gets there, because every gas turns to liquid first, so this part of the line exists only on paper. But it marks something real. A gas cannot push with less than no pressure, so −273 °C is the lowest temperature there can be. It is called absolute zero.
The pressure of the air in two sealed flasks, A and B, which hold different amounts of air, measured between 0 °C and 100 °C (solid). Neither line passes through 0 °C at zero pressure, so p is not proportional to t. Extended backwards (dashed), both reach zero pressure at the same temperature, −273 °C. Measured from there, in kelvin, each line starts at zero.
That gives temperature a better zero. Measure it from absolute zero, in steps the same size as Celsius degrees, and you have the absolute temperature T, measured in kelvin (K). Since absolute zero is at −273 °C, T = t + 273: 0 °C is 273 K, 27 °C is 300 K and 100 °C is 373 K. On this scale each line in the figure starts at zero, so the pressure is proportional to T. The exact value is 273.15, but the papers use 273, and so does this page.
Absolute temperature. T = t + 273, where t is in °C and T is in kelvin (K); back again, t = T − 273. Absolute zero, 0 K, is −273 °C. A change of 1 °C is a change of 1 K, so a temperature difference is the same number on both scales: from 27 °C to 54 °C is a rise of 27 K. A ratio of temperatures is not the same on both scales, and the gas laws need T in kelvin. In Chinese, T is 热力学温度; in Russian, абсолютная температура.
Try it! · Check-in question
Which temperature has twice the absolute temperature of 27 °C?
27 °C is 27 + 273 = 300 K. Twice that is 600 K, and back in Celsius 600 − 273 = 327 °C, so the answer is D. A doubles the Celsius number. 54 °C is 327 K, only 1.09 times 300 K. B finds 600 K but writes it down as °C, without converting back. C adds 273 to 600 instead of subtracting it: going from kelvin to Celsius, the number gets smaller, since 0 K is −273 °C.
Isochoric: pressure and temperature
The flask experiment of the last section was already an isochoric change: the glass held the volume fixed while the temperature changed. Measured in kelvin, the pressure is proportional to the temperature, p ∝ T, so p/T has the same value at every temperature. Double the absolute temperature of a gas in a rigid container and its pressure doubles.
Isochoric change. For a fixed mass of gas at constant volume, p/T = constant, so p₁/T₁ = p₂/T₂, with T in kelvin. The pressure is proportional to the absolute temperature: heat the gas and its pressure rises by the same factor as T.
Why does the temperature have to be converted, when the pressure can be in Pa, kPa or any other unit? Changing the unit of pressure multiplies both pressures by the same number, and that number cancels in the ratio p₂/p₁. Changing from kelvin to Celsius does something different: it subtracts 273 from both temperatures, and a subtraction does not cancel in a ratio. 327/300 is 1.09, but (327 − 273)/(300 − 273) = 54/27 = 2. A ratio says how many times larger one quantity is than another, and that only makes sense when both are measured from a true zero. For a pressure or a volume, zero means none at all. For temperature, the true zero is absolute zero, not the freezing point of water.
Kelvin in every ratio. Pressures and volumes can be in any unit, as long as both states use the same one. Temperatures must be in kelvin, T = t + 273, before they go into p₁/T₁ = p₂/T₂ or V₁/T₁ = V₂/T₂.
April 2026 Q33 drawn to scale. Measured from 0 °C, the end temperature looks twice the start (orange). But the pressure of a gas is proportional to its temperature measured from absolute zero, the whole bar, and that grows only from 300 K to 327 K, a factor of 1.09.
Worked example · the Celsius-ratio trap (April 2026, Q33)
A fixed amount of gas undergoes isochoric change. Its temperature increases from 27°C to 54°C. The pressure becomes ( ) times its original value.
A. 1.09 B. 1 C. 2 D. 1.5
Step 1: "isochoric" means at constant volume. The English paper uses the word without explaining it; the Chinese paper says 等容变化. For a fixed amount of gas at constant volume, p/T is constant, so p₂/p₁ = T₂/T₁.
Step 2: kelvin. T₁ = 27 + 273 = 300 K and T₂ = 54 + 273 = 327 K.
Step 3: p₂/p₁ = 327/300 = 1.09. The gas was heated at constant volume, so its pressure rose, and a factor a little above 1 fits a warming of 27 K on 300 K. The answer is A.
C, 2, is the Celsius ratio, 54/27, and it is among the options for that reason. B, 1, reads "isochoric" as "nothing changes": the volume is fixed, but the pressure is free to change with the temperature. D, 1.5, does not follow from any common slip.
The Celsius ratio. To a gas, 54 °C is not twice as hot as 27 °C. A ratio of Celsius temperatures can even come out negative, or with a zero underneath: cooling from 27 °C to −3 °C would give a factor of −3/27, and warming from 0 °C to anything would mean dividing by zero. If a ratio of two temperatures comes out far from 1 for a change of a few tens of degrees, or below zero, Celsius has crept in. Convert first: T = t + 273.
Try it! · Check-in question
Before a long drive, the air in a car tyre is at 7 °C and its pressure is 2.8 × 10⁵ Pa. After the drive the air is at 47 °C. Taking the tyre's volume as constant, what is the pressure of the air now?
The volume is taken as constant, so this is an isochoric change and p/T is constant. In kelvin the air goes from 7 + 273 = 280 K to 47 + 273 = 320 K. It is heated, so the pressure rises by the factor 320/280: p₂ = 2.8 × 10⁵ × 320 ÷ 280 = 3.2 × 10⁵ Pa, so the answer is B. A treats a constant volume as a constant pressure; only V is fixed. C turns the fraction upside down, 280/320, and gets a pressure that falls as the air warms. D uses the Celsius ratio, 47/7 ≈ 6.7, which predicts an almost sevenfold rise for a warming of 40 K, about one seventh of the absolute temperature.
Isobaric: volume and temperature
The third law, for constant pressure, follows from the first two. Take 3.0 L of gas at 1.0 × 10⁵ Pa and 300 K, and bring it to 450 K, with its pressure at the end the same as at the start, in two steps.
Step 1, at constant volume: heat the gas from 300 K to 450 K. The absolute temperature is multiplied by 1.5, so by p/T = constant the pressure is too. It rises to 1.5 × 10⁵ Pa, while the volume stays at 3.0 L.
Step 2, at constant temperature: keep the gas at 450 K and let it expand until its pressure is back to 1.0 × 10⁵ Pa. The pressure is divided by 1.5, so by pV = constant the volume is multiplied by 1.5. It grows to 4.5 L.
The gas now has its starting pressure, a temperature 1.5 times higher and a volume 1.5 times larger. A fixed mass of gas at a given pressure and temperature always has the same volume, whatever route it took to get there, so heating it directly at constant pressure from 300 K to 450 K must also end at 4.5 L. Any other factor works in the same way, so at constant pressure V ∝ T. Experiment agrees. Heat a gas under a freely sliding piston and the piston rises; the volumes measured at different Celsius temperatures lie on a straight line, and that line, extended backwards, reaches zero volume at the same −273 °C.
Isobaric change. For a fixed mass of gas at constant pressure, V/T = constant, so V₁/T₁ = V₂/T₂, with T in kelvin. The volume is proportional to the absolute temperature: heat the gas and it expands by the same factor as T.
Worked example · a new volume at constant pressure (December 2025, Q33)
An ideal gas of a certain mass undergoes an isobaric process (pressure is kept constant) from state A to state B. Suppose that at state A, the volume is V1 = 1.5 m³ and the temperature T1 = 300 K. At state B, the temperature is T2 = 400 K. The volume at state B is ( ).
(A) 6 m³ (B) 4.5 m³ (C) 1.5 m³ (D) 2 m³
Step 1: a fixed mass at constant pressure, so V/T is constant and V₂ = V₁ × T₂/T₁.
Step 2: the temperatures are already in kelvin, so nothing needs converting.
Step 3: the gas is heated, so its volume grows, by the factor 400/300: V₂ = 1.5 × 400/300 = 2 m³. The answer is D.
A, 6 m³, is 1.5 × 400/100, and B, 4.5 m³, is 1.5 × 300/100: both divide by the temperature change, 100 K, instead of by the starting temperature. C, 1.5 m³, reads "isobaric" as "nothing changes": the pressure is held fixed, but the volume follows the temperature.
Dividing by the temperature change. 400 − 300 = 100 K is the size of the step, and it belongs nowhere in V₁/T₁ = V₂/T₂. The law compares the two temperatures themselves, each measured from absolute zero: the volume grows by the factor 400/300 because the temperature does. 1.5 × 400/100 = 6 m³ and 1.5 × 300/100 = 4.5 m³ are the two ways of making this slip, and December's paper offered both.
Try it! · Check-in question
A balloon holds 3.0 L of air at 27 °C. It is put in a fridge, and the air inside cools to −3 °C while its pressure stays the same. What is the volume of the air now?
The pressure stays the same, so the change is isobaric and V/T is constant. In kelvin the air goes from 300 K to 270 K, so the volume shrinks by the factor 270/300 = 0.9: V₂ = 3.0 × 0.9 = 2.7 L, so the answer is B. A turns the fraction upside down, 300/270, and has the cooled balloon growing. C reads "the pressure stays the same" as "nothing changes"; only p is fixed, and V follows T. D uses the Celsius ratio, −3/27, and gets a negative volume. That is impossible, and it shows why Celsius temperatures cannot go into a ratio: 0 °C is not the zero of anything the gas has.
Reading a gas-law question
The three laws have the same shape. In each, one quantity is held fixed, and the other two are either inversely proportional, as p and V are, or proportional, as p and V each are to T in kelvin.
Three processes, three laws, for a fixed mass of gas. Isothermal, T fixed: pV = constant. Isochoric, V fixed: p/T = constant. Isobaric, p fixed: V/T = constant. T is always in kelvin.
Exam wording. The papers name the process, never the law: "with its temperature kept constant" (December 2025, Q28 and its reprints), "an isobaric process (pressure is kept constant)" (December 2025, Q33) and "undergoes isochoric change" (April 2026, Q33). Learning the laws by their names is a trap, because the names do not match across languages. English books call V/T = constant Charles's law, and p/T = constant the pressure law or Gay-Lussac's law. Chinese and Russian books call V/T = constant Gay-Lussac's law (盖-吕萨克定律, закон Гей-Люссака) and p/T = constant Charles's law (查理定律, закон Шарля). Only pV = constant is named after Boyle everywhere (玻意耳定律; in Russian, закон Бойля — Мариотта). Go by the process word and you need none of the names.
Every question on this page has been answered with the same four steps. The fourth catches the commonest slip of all, a fraction turned upside down, which gives an answer that moves the wrong way.
1
Check that the gas is a fixed mass. "A certain mass", "a fixed amount", a sealed container: no gas gets in or out, so the laws apply.
2
Find what is held fixed. Read the process word, or the setup. A water bath or "temperature kept constant" fixes T, so pV = constant. A freely sliding piston or "pressure kept constant" fixes p, so V/T = constant. A rigid, closed container fixes V, so p/T = constant.
3
Put every temperature in kelvin. T = t + 273. If the answer is asked for in °C, convert back at the end: t = T − 273.
4
Decide the direction, then use the factor. Say first whether the unknown should rise or fall: heating raises p at constant volume and V at constant pressure, and raising p lowers V at constant temperature. Then multiply by the ratio of the two given values, the way up that moves the answer in that direction.
Worked example · reading the setup
A glass bottle of air is closed with a cork at 27 °C, when the air inside is at 1.0 × 10⁵ Pa. The bottle is then stood in hot water until the air inside is at 87 °C. The cork will pop out if the pressure inside reaches 1.3 × 10⁵ Pa. Does it?
Step 1: while the cork holds, no air gets in or out, so the air is a fixed mass.
Step 2: nothing in the question names the process, but a glass bottle does not stretch, so the volume is fixed. The change is isochoric, with p/T constant.
Step 3: T₁ = 27 + 273 = 300 K and T₂ = 87 + 273 = 360 K.
Step 4: the air is heated at constant volume, so its pressure rises, by the factor 360/300 = 1.2. p₂ = 1.2 × 1.0 × 10⁵ = 1.2 × 10⁵ Pa. That is less than 1.3 × 10⁵ Pa, so the cork stays in.
The same law gives the temperature at which the cork would pop: T = 300 × 1.3 = 390 K, which is 117 °C, hotter than boiling water. The Celsius ratio would have predicted a pressure of 1.0 × 10⁵ × 87/27 ≈ 3.2 × 10⁵ Pa, and a cork flying out of the bottle.
The iso- word fixes one quantity, not three. Isobaric means the pressure stays the same; the volume and the temperature still change. Isochoric means the volume stays the same; the pressure and the temperature still change. Isothermal means the temperature stays the same; the pressure and the volume still change. Both real calculations offered a "nothing changes" option: 1.5 m³ in December 2025, Q33 and a factor of 1 in April 2026, Q33.
Try it! · Check-in question
A certain mass of ideal gas is cooled while its volume is kept constant. Its pressure would
A constant volume makes this an isochoric change, so p/T is constant: when T falls, p falls by the same factor, and the answer is D. A has the direction backwards, as if pressure and temperature were inversely linked. That is the pattern for p and V at constant temperature, not for p and T. B fixes the wrong quantity: the volume is held constant, and the pressure is free to change. C avoids choosing an answer. A fixed mass of gas with a fixed volume and a falling temperature is everything p/T = constant needs, so only one result is possible.
A check on the size of an answer helps too. Near room temperature the absolute temperature is about 300 K, so a change of 30 K changes it by about a tenth, and changes the pressure at constant volume, or the volume at constant pressure, by about a tenth as well. An answer two or three times the starting value, for a change of a few tens of degrees, means Celsius has crept in.
Try it! · Check-in question
A fixed mass of gas is heated from 20 °C to 40 °C while its pressure is kept constant. What happens to its volume?
The change is isobaric, so V/T is constant, with T in kelvin: 20 °C is 293 K and 40 °C is 313 K. The volume is multiplied by 313/293 ≈ 1.07, an increase of about 7%, so the answer is A. B doubles the volume because 40 is twice 20, the Celsius ratio again; measured from absolute zero, the temperature has risen by only 20 K on 293 K. C reads "constant pressure" as "nothing changes". D has the right size but the wrong direction, as if V and T were inversely proportional; a gas heated at constant pressure expands.
Your turn! · Practice questions
Practice · Question 1
An air bubble with a volume of 2.0 cm³ is released at the bottom of a lake, where the pressure is 3.0 × 10⁵ Pa. It rises to the surface, where the pressure is 1.0 × 10⁵ Pa. The water is at the same temperature all the way up. What is the bubble's volume at the surface?
No air gets in or out of the bubble, and its temperature stays the same, so the change is isothermal: p₁V₁ = p₂V₂. The pressure falls by a factor of 3, from 3.0 × 10⁵ Pa to 1.0 × 10⁵ Pa, so the volume grows by a factor of 3: V₂ = 2.0 × 3.0 ÷ 1.0 = 6.0 cm³, so the answer is C. A divides the volume by 3, as if pressure and volume were directly proportional; bubbles grow as they rise. B keeps the volume as it was, but pV must stay the same, and p has fallen. D divides by the change in pressure, 2.0 × 10⁵ Pa, instead of by the new pressure: 2.0 × 3.0 ÷ 2.0 = 3.0. Boyle's law compares the two pressures themselves.
Practice · Question 2
A fixed mass of gas at 127 °C is cooled at constant pressure until its volume is half of what it was. What is its temperature now?
At constant pressure V/T is constant, so halving V halves T, measured in kelvin. 127 °C is 400 K; half of it is 200 K, which is 200 − 273 = −73 °C, so the answer is A. B halves the Celsius number, 127 ÷ 2 = 63.5. That is 336.5 K, only about 16% colder than 400 K and nowhere near half. C finds 200 K but writes it down as °C; the question asks in °C, so convert back. D doubles the temperature, 800 K = 527 °C, as if volume and temperature were inversely proportional. A gas cooled at constant pressure shrinks, so a smaller volume needs a lower temperature.
Practice · Question 3
A glass jar is closed with a tight lid while the air inside is at 87 °C and 1.0 × 10⁵ Pa. The jar then cools to 27 °C, and its volume does not change. What is the pressure of the air inside now?
The closed jar keeps its volume, so the change is isochoric and p/T is constant. In kelvin, 87 °C is 360 K and 27 °C is 300 K. The air cools, so its pressure falls: p₂ = 1.0 × 10⁵ × 300 ÷ 360 ≈ 8.3 × 10⁴ Pa, so the answer is D. That is less than the pressure of the air outside, which is why a jar closed while hot is hard to open once it has cooled. A uses the Celsius ratio, 27/87, and cuts the pressure to less than a third for a cooling of one sixth in kelvin. B turns the fraction upside down, 360/300, and has the pressure rising as the air cools. C divides by the temperature change, 60 K: 1.0 × 10⁵ × 300 ÷ 60 = 5.0 × 10⁵ Pa. The law compares the two temperatures, not the step between them.
Practice · Question 4
The volume of a fixed mass of gas is reduced by 20% while its temperature is kept constant. How does its pressure change?
At constant temperature pV is constant. Reducing the volume by 20% leaves 0.80 of it, so the pressure is divided by 0.80, which multiplies it by 1.25: an increase of 25%, so the answer is B. A assumes that inverse proportion means equal percentages, but 1 ÷ 0.80 is 1.25, not 1.20. Inverse proportion works with factors, not with percentages. C treats p and V as directly proportional. D divides by the 20% that was taken away instead of by the 80% that is left: 1 ÷ 0.20 = 5, a rise of 400%.
Practice · Question 5
A gas is sealed in a vertical cylinder of uniform cross-section by a piston that slides freely without friction. At 27 °C the piston is 30 cm above the base of the cylinder. The gas is heated to 87 °C. How high above the base is the piston now?
The piston slides freely, and the forces pushing it down, its weight and the atmosphere, do not change, so the gas pressure does not change either: the process is isobaric, and V/T is constant. The cylinder is uniform, so the volume is the base area times the height, V = Sh, and the height is proportional to the volume. In kelvin the gas goes from 300 K to 360 K, so h₂ = 30 × 360 ÷ 300 = 36 cm, and the answer is C. A uses the Celsius ratio, 30 × 87/27 ≈ 96.7 cm, more than three times the height for a warming of one fifth in kelvin. B turns the fraction upside down, 300/360, and has the heated gas shrinking. D takes "constant pressure" to mean that the piston stays still, but a constant pressure is exactly what lets the piston rise as the gas expands.
Things to remember
Concept
What to know
A fixed mass of gas
Sealed: none gets in or out. Its state is p (in Pa, 1 Pa = 1 N/m²), V (in m³, 1 L = 10⁻³ m³) and T. Each law compares two states of the same gas.
The process words
Isothermal (等温, изотермический): T fixed. Isobaric (等压, изобарный): p fixed. Isochoric (等容, изохорный): V fixed. One quantity is fixed and the other two change.
Isothermal
pV = constant, p₁V₁ = p₂V₂: p and V are inversely proportional. December 2025 Q28, reprinted as January 2026 Q28 and March 2026 Q25: pressure down, volume up.
Absolute temperature
T = t + 273, in kelvin; absolute zero is 0 K = −273 °C. A step of 1 °C is a step of 1 K, but a ratio of Celsius temperatures means nothing.
Isochoric
p/T = constant, p₁/T₁ = p₂/T₂. April 2026 Q33: 327 K ÷ 300 K = 1.09, not 54 ÷ 27 = 2.
Water bath or a slow change: T fixed. Freely sliding piston: p fixed. Rigid closed container: V fixed.
The method
Fixed mass? What is held fixed? Kelvin. Decide the direction first, then multiply by the ratio that moves the answer that way. Near 300 K, a few tens of degrees change p or V by a few per cent to a few tens of per cent, never by a factor of 2.
Law names
Go by the process word. English Charles's law is V/T = constant; Chinese and Russian Charles's law (查理定律, закон Шарля) is p/T = constant, and their Gay-Lussac's law is V/T = constant.
The wrong options
The Celsius ratio; dividing by ΔT instead of by T₁; "isobaric" or "isochoric" read as "nothing changes"; p and V treated as directly proportional; the option "all options above are possible", which avoids choosing.
