ChemistryICSE Class 9 · Chapter 7

Study of Gas Laws ICSE Class 9: Boyle's Law, Charles's Law and the Gas Equation

Updated 15 October 2026 · Fact-checked

Boyle's law says that at constant temperature the volume of a fixed mass of gas is inversely proportional to its pressure (P₁V₁ = P₂V₂). Charles's law says that at constant pressure volume is directly proportional to Kelvin temperature (V₁/T₁ = V₂/T₂). Together they give P₁V₁/T₁ = P₂V₂/T₂.

What Study of Gas Laws is about

Gases are easy to squeeze and expand a lot when heated. Their volume depends on pressure and temperature, so a volume of gas means nothing unless P and T are stated. The gas laws describe these changes for a fixed mass of gas.

The CISCE 2028 syllabus (Class IX, topic 7) covers: the behaviour of gases under changes of temperature and pressure, explained by molecular motion; Boyle's law and Charles's law with mathematical derivation, graphical representation, statements and simple calculations; absolute zero and the Kelvin scale; the gas equation $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$ with simple calculations; conversion between Celsius and Kelvin; and standard temperature and pressure (S.T.P.).

Weightage: CISCE publishes no chapter weightage for ICSE Chemistry. Study of Gas Laws is topic 7 of the Class IX part of the 2028 syllabus; the Class IX paper (80 marks, 2 hours) is set and marked by the school.Source: CISCE ICSE Chemistry (Science Paper 2) syllabus, examination year 2028 (Class IX)

Key concepts

Gases and molecular motion

The kinetic theory pictures a gas as tiny particles (atoms or molecules) that:

are far apart (most of a gas is empty space, so gases compress easily);
move randomly at high speed in straight lines, colliding with each other and with the walls;
move faster when heated (average kinetic energy rises with temperature);
attract each other only very weakly.

Pressure of a gas is caused by its molecules hitting the walls of the container. More hits per second, or harder hits, mean more pressure.

Units: pressure in atmospheres, mm or cm of mercury, or pascals: 1 atm = 76 cm Hg = 760 mm Hg = 1.013 × 10⁵ Pa. Volume in cm³, dm³ (litres) or m³: 1 dm³ = 1 L = 1000 cm³.

Worked example

Use molecular motion to explain why a balloon kept in sunlight becomes tighter.

Try it first, then show the solutionSolution

Heating makes the air molecules inside move faster.
They hit the inside of the balloon more often and harder, raising the pressure.
The rubber stretches until the pressures balance.

Answer: Faster molecules hit harder and more often.

Boyle's law: statement, derivation and graphs

Statement: at constant temperature, the volume of a fixed mass of a gas is inversely proportional to its pressure.

Derivation: $V \propto \frac{1}{P}$ (T constant) $\Rightarrow V = \frac{k}{P} \Rightarrow P V = k$ . So for two states of the same gas at the same temperature,
$P_{1} V_{1} = P_{2} V_{2}$

Molecular explanation: halving the volume packs the same molecules into half the space, so they hit each unit area of wall twice as often: the pressure doubles.

Graphs (T constant):

P against V: a smooth curve (rectangular hyperbola) falling as V rises.
P against 1/V (or V against 1/P): a straight line through the origin.
PV against P (or V): a horizontal straight line, since PV is constant.

Worked example

A gas occupies 500 cm³ at 760 mm Hg. Find its volume at 950 mm Hg, temperature constant.

Try it first, then show the solutionSolution

$P_{1} V_{1} = P_{2} V_{2}$
$760 \times 500 = 950 \times V_{2}$
$V_{2} = \frac{760 \times 500}{950} = 400$ cm³

Answer: 400 cm³

Charles's law, absolute zero and the Kelvin scale

Charles found that at constant pressure, a gas expands by 1/273 of its volume at 0 °C for every 1 °C rise in temperature (and contracts by the same amount for every 1 °C fall): $V_{t} = V_{0} (1 + \frac{t}{273})$ .

So at -273 °C the volume would become zero. This temperature is absolute zero, the lowest possible temperature, where molecular motion would (ideally) stop. In practice gases liquefy before this. (More precisely it is -273.15 °C.)

The Kelvin (absolute) scale starts at absolute zero: 0 K = -273 °C, and one kelvin is the same size as one degree Celsius.
$T (K) = t (° C) + 273 t (° C) = T (K) - 273$

Statement of Charles's law: at constant pressure, the volume of a fixed mass of a gas is directly proportional to its absolute (Kelvin) temperature.

Derivation: $V \propto T$ (P constant) $\Rightarrow V = k T \Rightarrow \frac{V}{T} = k$ , so
$\frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}}$

Molecular explanation: heating makes molecules move faster and hit harder; for the pressure to stay the same, the gas must expand so that hits per unit area become less frequent.

Graphs (P constant):

V against T (K): a straight line through the origin.
V against t (°C): a straight line that, extended backwards, meets the temperature axis at -273 °C.

Worked example

A gas occupies 300 cm³ at 27 °C. Find its volume at 127 °C, pressure constant.

Try it first, then show the solutionSolution

$T_{1} = 27 + 273 = 300$ K; $T_{2} = 127 + 273 = 400$ K
$\frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}} \Rightarrow V_{2} = \frac{300 \times 400}{300}$
$V_{2} = 400$ cm³

Answer: 400 cm³

The gas equation

When both pressure and temperature change, combine the two laws.

Derivation: take a gas from $(P_{1}, V_{1}, T_{1})$ to $(P_{2}, V_{2}, T_{2})$ in two steps.
1. Change pressure from $P_{1}$ to $P_{2}$ at constant $T_{1}$ (Boyle): $P_{1} V_{1} = P_{2} V^{'}$ , so $V^{'} = \frac{P _{1} V _{1}}{P _{2}}$ .
2. Change temperature from $T_{1}$ to $T_{2}$ at constant $P_{2}$ (Charles): $\frac{V ^{'}}{T _{1}} = \frac{V _{2}}{T _{2}}$ , so $V^{'} = \frac{V _{2} T _{1}}{T _{2}}$ .

Equating the two values of $V^{'}$ : $\frac{P _{1} V _{1}}{P _{2}} = \frac{V _{2} T _{1}}{T _{2}}$ , which rearranges to
$\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$

So $\frac{P V}{T}$ is constant for a fixed mass of gas.

Rules for calculations:

Temperatures must be in kelvin.
P and V may be in any units, as long as each is the same on both sides.
If T is constant it cancels (Boyle); if P is constant it cancels (Charles); if V is constant, $\frac{P _{1}}{T _{1}} = \frac{P _{2}}{T _{2}}$ .

Worked example

A gas occupies 500 cm³ at 15 °C and 750 mm Hg. Find its volume at 100 °C and 600 mm Hg.

Try it first, then show the solutionSolution

$T_{1} = 288$ K, $T_{2} = 373$ K
$V_{2} = \frac{P _{1} V _{1} T _{2}}{T _{1} P _{2}} = \frac{750 \times 500 \times 373}{288 \times 600}$
$V_{2} = 809.5$ cm³ (to 1 decimal place)

Answer: 809.5 cm³

Standard temperature and pressure (S.T.P.)

Because the volume of a gas depends on P and T, volumes are compared at an agreed set of conditions, standard temperature and pressure (S.T.P.):

Standard temperature = 0 °C = 273 K
Standard pressure = 1 atm = 76 cm Hg = 760 mm Hg (1.013 × 10⁵ Pa)

To reduce a volume to S.T.P., use the gas equation with $P_{2} = 760$ mm Hg and $T_{2} = 273$ K:
$V_{S T P} = \frac{P _{1} V _{1}}{T _{1}} \times \frac{273}{760}$ (P₁ in mm Hg)

Worked example

Reduce 380 cm³ of a gas at 27 °C and 700 mm Hg to S.T.P.

Try it first, then show the solutionSolution

$T_{1} = 300$ K; $P_{2} = 760$ mm, $T_{2} = 273$ K
$V_{2} = \frac{700 \times 380 \times 273}{300 \times 760}$
$V_{2} = 318.5$ cm³

Answer: 318.5 cm³

Chemistry in real life: Study of Gas Laws

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Tyre pressure on a hot day

Suppose a tyre is pumped to 2 atm at 17 °C in the morning, and by afternoon the air inside reaches 47 °C. Assume the volume of the tyre does not change.

Find the afternoon pressure.

Try it first, then show the solutionSolution

V constant, so $\frac{P _{1}}{T _{1}} = \frac{P _{2}}{T _{2}}$ .
$T_{1} = 290$ K, $T_{2} = 320$ K.
$P_{2} = 2 \times \frac{320}{290} = 2.21$ atm

Answer: About 2.21 atm.

Memory hook: Hotter gas, harder hits, higher pressure.

Weather balloon

Suppose a weather balloon holds 10 m³ of helium at 1 atm and 27 °C at launch and rises to where the pressure is 0.25 atm and the temperature -53 °C.

Find its new volume (assume it does not burst and the gas stays at the outside conditions).

Try it first, then show the solutionSolution

$T_{1} = 300$ K, $T_{2} = 220$ K
$V_{2} = \frac{P _{1} V _{1} T _{2}}{T _{1} P _{2}} = \frac{1 \times 10 \times 220}{300 \times 0.25}$
$V_{2} = 29.3$ m³

Answer: About 29.3 m³.

Memory hook: Low pressure wins over the cold: the balloon swells.

Bubble rising in a lake

Suppose a diver releases an air bubble of 2 cm³ at a depth where the pressure is 2.5 atm and the water is 7 °C, and it rises to the surface where the pressure is 1 atm and the water is 27 °C.

Find the bubble's volume at the surface.

Try it first, then show the solutionSolution

$T_{1} = 280$ K, $T_{2} = 300$ K
$V_{2} = \frac{2.5 \times 2 \times 300}{280 \times 1}$
$V_{2} = 5.36$ cm³

Answer: About 5.36 cm³.

Memory hook: Less pressure and more warmth: bubbles grow as they rise.

Topics in this chapter

Study of Gas Laws formula sheet

Boyle's law: $P_{1} V_{1} = P_{2} V_{2}$; T and mass constant.
Charles's law: $\frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}}$; P and mass constant; T in kelvin.
Gas equation: $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$; Fixed mass.
Kelvin and Celsius: $T = t + 273$; 0 K = -273 °C.
Charles's original form: $V_{t} = V_{0} (1 + \frac{t}{273})$; V₀ at 0 °C.
S.T.P.: $273 K, 760 mm Hg = 1 atm$; Standard conditions.

Study of Gas Laws: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Boyle's law graphs

Draw the P-V, P-1/V and PV-P graphs for a fixed mass of gas at constant temperature.

Draw

P-V: P on y-axis, V on x-axis; a smooth falling curve not touching either axis.
P-1/V: a straight line from the origin rising to the right.
PV-P: a horizontal straight line.

Label

P
V
1/V
PV
T constant

Watch out: Write 'temperature constant' on each graph.

Charles's law graphs

Draw V against T (K) and V against t (°C) at constant pressure.

Draw

V-T: a straight line from the origin (0 K) rising to the right.
V-t: a straight line rising to the right; draw the backward extension as a dotted line meeting the temperature axis at -273 °C.

Label

V
T (K)
t (°C)
-273 °C (absolute zero)
P constant

Watch out: Use a dotted extension: real gases liquefy before absolute zero.

How the school exam asks Study of Gas Laws

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Statements and derivations	State Boyle's or Charles's law; derive its mathematical form; derive the gas equation.	State with 'fixed mass' and the constant quantity; proportionality → constant → two-state form.
Graphs	Sketch P-V, P-1/V, PV-P, V-T and V-t graphs; read a value; identify the law.	Learn the shape and what is on each axis.
Molecular explanation	Explain each law in terms of molecular motion.	Collisions per unit area per second; speed of molecules.
Numericals	Boyle, Charles, gas-equation and S.T.P. problems; find V, P or T.	List data, convert to K, write the formula, substitute, units.
Temperature conversion	Convert °C to K and back; explain absolute zero.	Add or subtract 273.

Tricks to remember Study of Gas Laws

Short memory aids. Say each one aloud once, then use it on a question the same day.

Boyle is a see-saw
Pressure up, volume down: their product stays balanced (PV = k).
Use it when: Inverse proportion.
Charles likes it hot
Heat it, it grows: V/T = k (in kelvin).
Use it when: Direct proportion.
Always Kelvin
Before any T goes into a formula, add 273.
Use it when: Avoiding the commonest error.
S.T.P. = 'zero and seventy-six'
0 °C (273 K) and 76 cm Hg.
Use it when: Standard conditions.

Study of Gas Laws: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Graph identification · 1 mark · MCQExam level

For a fixed mass of gas at constant temperature, a straight line through the origin is obtained when one plots:
(a) P against V
(b) P against 1/V
(c) PV against P
(d) V against t (°C)

Try it first, then show the solutionSolution

P ∝ 1/V, so P against 1/V is a straight line through the origin.

Answer: (b)

Example 2: Temperature conversion · 2 marks · Short answerExam level

Convert (a) -73 °C to kelvin, (b) 546 K to Celsius. (c) What is absolute zero on the Celsius scale?

Try it first, then show the solutionSolution

(a) -73 + 273 = 200 K
(b) 546 - 273 = 273 °C
(c) -273 °C

Answer: 200 K; 273 °C; -273 °C.

Example 3: Boyle's law pressure · 2 marks · Short answerExam level

4 L of a gas at 1 atm is compressed to 1.5 L at constant temperature. Find the new pressure.

Try it first, then show the solutionSolution

$P_{2} = \frac{P _{1} V _{1}}{V _{2}} = \frac{1 \times 4}{1.5}$
$P_{2} = 2.67$ atm

Answer: 2.67 atm

Example 4: Doubling the volume · 2 marks · Short answerHarder exam level

To what temperature (°C) must a gas at 27 °C be heated at constant pressure so that its volume doubles?

Try it first, then show the solutionSolution

$\frac{V _{1}}{T _{1}} = \frac{2 V _{1}}{T _{2}} \Rightarrow T_{2} = 2 T_{1} = 600$ K
$t = 600 - 273 = 327$ °C

Answer: 327 °C (not 54 °C).

Example 5: Gas cylinder at S.T.P. · 3 marks · Short answerHarder exam level

A 10 L cylinder holds a gas at 15 atm and 27 °C. What volume would this gas occupy at S.T.P.?

Try it first, then show the solutionSolution

$P_{1} = 15$ atm, $V_{1} = 10$ L, $T_{1} = 300$ K; $P_{2} = 1$ atm, $T_{2} = 273$ K
$V_{2} = \frac{15 \times 10 \times 273}{300 \times 1}$
$V_{2} = 136.5$ L

Answer: 136.5 L

Example 6: Derive the gas equation · 4 marks · Long answerHarder exam level

Derive $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$ from Boyle's and Charles's laws.

Try it first, then show the solutionSolution

Step 1 (Boyle, T₁ constant): P₁ → P₂, V₁ → V': $V^{'} = \frac{P _{1} V _{1}}{P _{2}}$
Step 2 (Charles, P₂ constant): T₁ → T₂, V' → V₂: $V^{'} = \frac{V _{2} T _{1}}{T _{2}}$
Equate: $\frac{P _{1} V _{1}}{P _{2}} = \frac{V _{2} T _{1}}{T _{2}}$
Rearrange: $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$

Answer: As above.

Example 7: Find the temperature · 3 marks · Short answerHOTS / competency

100 cm³ of a gas is at 27 °C and 760 mm Hg. At what temperature (°C) will it occupy 120 cm³ at 836 mm Hg?

Try it first, then show the solutionSolution

$T_{2} = \frac{P _{2} V _{2} T _{1}}{P _{1} V _{1}} = \frac{836 \times 120 \times 300}{760 \times 100}$
$T_{2} = 396$ K
$t = 396 - 273 = 123$ °C

Answer: 123 °C

Example 8: Explain from data · 3 marks · Short answerHOTS / competency

A student records P (kPa) and V (cm³) for a gas at constant temperature: (100, 60), (150, 40), (200, 30), (300, 20). (a) Show the data obey Boyle's law. (b) Predict V at 250 kPa. (c) What would the PV against P graph look like?

Try it first, then show the solutionSolution

(a) PV = 6000 in every case: constant, so Boyle's law holds.
(b) V = 6000/250 = 24 cm³.
(c) A horizontal straight line at PV = 6000.

Answer: PV constant; 24 cm³; horizontal line.

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 2 + 2 + 1 marks.

Syringe and hot water

Suppose a student traps 40 cm³ of air in a sealed gas syringe at 27 °C and 1 atm. She first pushes the plunger until the volume is 30 cm³ (temperature unchanged), then puts the syringe in hot water at 77 °C and lets the plunger move freely, keeping the new pressure.

(i) · 2 marks
Find the pressure after compression.
Answer
Answer: $P_{2} = \frac{1 \times 40}{30} = 1.33$ atm.
(ii) · 2 marks
Find the volume in hot water at that pressure.
Answer
Answer: $V = 30 \times \frac{350}{300} = 35$ cm³.
(iii) · 1 mark
Which law applies to each step?
Answer
Answer: Compression: Boyle's law; heating: Charles's law.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

State Charles's law and write its mathematical form. (2 marks) (2 marks)

Step	Marks
At constant pressure, volume of a fixed mass of gas is directly proportional to absolute temperature	1
$V_{1} / T_{1} = V_{2} / T_{2}$ (or V/T = constant)	1

Reduce 380 cm³ of gas at 27 °C and 700 mm Hg to S.T.P. (3 marks) (3 marks)

Step	Marks
T₁ = 300 K, T₂ = 273 K, P₂ = 760 mm	1
$V_{2} = \frac{700 \times 380 \times 273}{300 \times 760}$	1
V₂ = 318.5 cm³	1

Explain Boyle's law on the basis of molecular motion and sketch the P-V graph. (4 marks) (4 marks)

Step	Marks
Pressure is due to molecules colliding with the walls	1
Smaller volume → more collisions per unit area per second → higher pressure (T constant, so speed unchanged)	1
Graph: P on y, V on x	1
Smooth curve falling as V increases (hyperbola), not touching the axes	1

Where students lose marks in Study of Gas Laws

The slips examiners see most often, and what to write instead.

Mistake: Using Celsius in the formula
The Celsius value is put straight into a proportion that only holds for kelvin.
Wrong working
$V_{2} = 300 \times \frac{127}{27} = 1411$ cm³
Right working
$V_{2} = 300 \times \frac{400}{300} = 400$ cm³
Do this: Convert to kelvin first, every time.
Usually costs: 1-2 marks.
Mistake: Final answer with no working
CISCE examiners' comments on the ICSE Class 10 paper, 2025: in a calculation of a gas volume at S.T.P., some candidates substituted values incorrectly and a few gave only the final answer without showing the steps.
Wrong working
V = 318.5 cm³
Right working
Data → formula → substitution → answer with unit.
Do this: Write every step; marks are given for each.
Usually costs: Most of the marks.
Mistake: Doubling the Celsius temperature
Students double 27 °C to 54 °C to double the volume.
Wrong working
Volume doubles at 54 °C.
Right working
Double the kelvin temperature: 300 K → 600 K = 327 °C.
Do this: Proportionality is with T in kelvin only.
Usually costs: 2 marks.
Mistake: Leaving out 'fixed mass' or the constant quantity
Statements are incomplete.
Wrong working
Volume is inversely proportional to pressure.
Right working
At constant temperature, the volume of a fixed mass of a gas is inversely proportional to its pressure.
Do this: Three parts: condition, fixed mass, relationship.
Usually costs: 1 mark.
Mistake: Wrong graph shape
P-V and V-T graphs are swapped.
Wrong working
Boyle's law: V against P is a straight line through the origin.
Right working
P against V is a curve; P against 1/V is a straight line through the origin.
Do this: Inverse → curve; direct → straight line.
Usually costs: 1 mark.

Practice questions on Study of Gas Laws

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The temperature at which the volume of a gas theoretically becomes zero is:
(a) 0 °C
(b) 273 K
(c) -273 °C
(d) 100 K
Hint
Absolute zero.
Answer
(c)
Q2 · 1 mark · MCQExam level
S.T.P. means:
(a) 0 K and 1 atm
(b) 273 K and 76 cm Hg
(c) 25 °C and 1 atm
(d) 273 °C and 760 mm Hg
Hint
0 °C.
Answer
(b)
Q3 · 2 marks · Short answerExam level
250 cm³ of a gas at 7 °C is heated to 77 °C at constant pressure. Find the new volume.
Hint
280 K and 350 K.
Answer
$V_{2} = 250 \times \frac{350}{280} = 312.5$ cm³
Q4 · 2 marks · Short answerHarder exam level
1000 cm³ of a gas at S.T.P. is heated to 819 °C at constant pressure. Find its volume.
Hint
1092 K = 4 × 273 K.
Answer
$V_{2} = 1000 \times \frac{1092}{273} = 4000$ cm³
Q5 · 2 marks · Short answerHarder exam level
A gas at 27 °C and 1 atm is cooled at constant pressure until its volume is halved. Find the final temperature in °C.
Hint
Half of 300 K.
Answer
T₂ = 150 K = -123 °C.
Q6 · 2 marks · Short answerHarder exam level
22.4 L of a gas at S.T.P. is taken to 27 °C and 2 atm. Find its new volume.
Hint
Gas equation.
Answer
$V_{2} = \frac{1 \times 22.4 \times 300}{273 \times 2} = 12.31$ L
Q7 · 3 marks · Short answerHarder exam level
Explain Charles's law using the motion of molecules. Why must the temperature be in kelvin for V ∝ T?
Hint
Speed; zero point.
Answer
Heating makes molecules faster, so they hit the walls harder and more often; to keep pressure constant the gas expands, spreading the hits over a larger area. V ∝ T only on the Kelvin scale because its zero is where the volume would be zero; on the Celsius scale V is not zero at 0 °C.
Q8 · 3 marks · Short answerExam level
Give the shape of the graph for each: (a) V against 1/P at constant T, (b) PV against V at constant T, (c) V against t (°C) at constant P.
Hint
Straight, horizontal, intercept.
Answer
(a) Straight line through the origin. (b) Horizontal straight line. (c) Straight line which, extended, cuts the temperature axis at -273 °C.
Q9 · 2 marks · Short answerHarder exam level
At what temperature (°C) will a gas at S.T.P. occupy 1.2 times its volume, the pressure being kept constant?
Hint
273 × 1.2.
Answer
T = 327.6 K = 54.6 °C.
Q10 · 3 marks · Short answerHOTS / competency
A gas cylinder can withstand 10 atm. It contains gas at 6 atm at 27 °C. Is it safe if a fire heats it to 327 °C? Show working.
Hint
V constant.
Answer
$P_{2} = 6 \times \frac{600}{300} = 12$ atm, more than 10 atm: not safe; it may burst.
Q11 · 3 marks · Short answerHOTS / competency
An air bubble triples in volume as it rises from the bottom of a lake to the surface. Assuming the temperature is the same throughout, and atmospheric pressure is 1 atm, find the pressure at the bottom. Which law did you use?
Hint
P₁V = 1 × 3V.
Answer
P₁V₁ = P₂V₂: P₁ × V = 1 × 3V, so P₁ = 3 atm. Boyle's law.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Absolute zero in °C?Show answer
-273 °C (0 K).
2.Standard temperature and pressure?Show answer
273 K and 760 mm Hg (1 atm).
3.1 atm in cm of mercury?Show answer
76 cm Hg.
4.What is constant in Boyle's law?Show answer
Temperature (and mass).
5.Graph of V against T (K) for Charles's law?Show answer
Straight line through the origin.
6.What causes gas pressure?Show answer
Molecules colliding with the container walls.
7.Gas equation?Show answer
P₁V₁/T₁ = P₂V₂/T₂.
8.By what fraction does a gas expand per °C at constant pressure?Show answer
1/273 of its volume at 0 °C.

Best order to study it

Learn the kinetic picture of a gas and what causes pressure.
Learn Boyle's law: statement, derivation, molecular explanation, graphs.
Learn Charles's law, absolute zero and the Kelvin scale.
Learn the derivation of the gas equation.
Learn S.T.P. and practise 15-20 numericals of all types.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you state both laws with all three parts?
Can you derive the gas equation in two steps?
Can you sketch all five graphs?
Can you convert any temperature between °C and K?
Can you reduce a volume to S.T.P.?
Can you explain each law by molecular motion?

Study of Gas Laws: frequently asked questions

Should I use 273 or 273.15?

Use 273 unless the question says otherwise; ICSE calculations use 273 K for 0 °C.

Do I have to convert mm Hg to atm?

No. Pressure and volume can be in any units as long as you use the same unit on both sides of the equation. Only temperature must be in kelvin.

Why can't a gas actually reach zero volume?

Real gases turn into liquids and then solids before reaching -273 °C, and their molecules have size of their own. Absolute zero comes from extending the graph.

Is the pressure law (P ∝ T) in the syllabus?

It is not named separately, but it follows from the gas equation with constant volume: \dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}, so such a calculation is a fair gas-equation question.

Based on: CISCE ICSE 2028 Science (52) Chemistry (Science Paper 2) syllabus, Classes IX and X; CISCE Analysis of Pupil Performance, ICSE (Class 10) Chemistry (Science Paper 2), 2025