ChemistryICSE Class 9 · Study of Gas Laws

Charles's Law, Absolute Zero, the Gas Equation and S.T.P. (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

Charles's law: at constant pressure the volume of a fixed mass of gas is directly proportional to its absolute (Kelvin) temperature, V₁/T₁ = V₂/T₂ with T = t + 273. Combined with Boyle's law it gives P₁V₁/T₁ = P₂V₂/T₂. Volumes are compared at S.T.P. (273 K and 760 mm Hg) using V = P₁V₁ × 273/(T₁ × 760).

Extend the V–t line backwards and it reaches zero volume at −273 °C, which is 0 K, so always convert to kelvin: T = t + 273.

Illustration drawn for QuestionBuddy.

Understand Charles's Law, Kelvin Scale, the Gas Equation and S.T.P.

Charles’s law. At constant pressure a gas expands by $\frac{1}{273}$ of its volume at 0 °C for every 1 °C rise: $V_{t} = V_{0} (1 + \frac{t}{273})$ . Extending the line backwards, V would be zero at −273 °C, called absolute zero. On the Kelvin scale 0 K = −273 °C and a kelvin equals a degree Celsius:

$T (K) = t (° C) + 273.$

Statement. At constant pressure the volume of a fixed mass of gas is directly proportional to its absolute temperature:

$V \propto T \Rightarrow \frac{V}{T} = k \Rightarrow \frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}} .$

Explanation. Heating makes molecules move faster and hit harder; to keep the pressure the same the gas must expand so that the hits per unit area become less frequent.

Graphs (P constant). V against T (K): straight line through the origin. V against t (°C): straight line that meets the temperature axis at −273 °C when extended.

The gas equation. For a fixed mass of gas, going from $(P_{1}, V_{1}, T_{1})$ to $(P_{2}, V_{2}, T_{2})$ by a Boyle step followed by a Charles step gives

$\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}} .$

Rules: temperature in kelvin; same units for P (and for V) on both sides; if T is constant it reduces to Boyle’s law; if P is constant, to Charles’s law; if V is constant, $\frac{P _{1}}{T _{1}} = \frac{P _{2}}{T _{2}}$ .

S.T.P. Standard temperature 0 °C (273 K) and standard pressure 1 atm = 760 mm Hg. To reduce a volume to S.T.P. take $P_{2} = 760$ mm Hg and $T_{2} = 273$ K:

$V_{S . T . P .} = \frac{P _{1} V _{1}}{T _{1}} \times \frac{273}{760} (P_{1} in mm Hg) .$

Typical checks. Why can the Celsius temperature not be used? Because V is proportional to T only on the Kelvin scale: doubling 27 °C to 54 °C does not double the volume.

Charles's Law, Kelvin Scale, the Gas Equation and S.T.P.: formulas and key facts

Kelvin scale: $T (K) = t (° C) + 273$; Absolute zero = −273 °C.
Charles's law: $\frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}}$; Constant P, T in kelvin.
Gas equation: $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$; Fixed mass of gas.
Reduction to S.T.P.: $V_{S . T . P .} = \frac{P _{1} V _{1}}{T _{1}} \times \frac{273}{760}$; P₁ in mm Hg.

How to solve Charles's Law, Kelvin Scale, the Gas Equation and S.T.P. questions

Convert temperatures first; it is the commonest error.

1Convert every temperature to kelvin (add 273).
2Write the two states (P, V, T) in a table.
3Cross out what is constant to see which law applies.
4Substitute in $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$ and solve.
5Quote the unit and convert the answer back to °C if the temperature is asked.

Charles's Law, Kelvin Scale, the Gas Equation and S.T.P. in real life

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.

A sealed can near a stove

Suppose a rigid sealed can of gas at 27 °C and 1 atm is left near a stove until it reaches 127 °C.

What is the pressure inside (volume constant)?

Try it first, then show the solutionSolution

$T_{1} = 300$ K, $T_{2} = 400$ K.
$P_{2} = P_{1} \times \frac{T _{2}}{T _{1}} = 1 \times \frac{400}{300} \approx 1.33$ atm.

Answer: About 1.33 atm.

Memory hook: Kelvin, then ratio.

Tricks to remember Charles's Law, Kelvin Scale, the Gas Equation and S.T.P.

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

Add 273
Celsius to kelvin: add 273. Always.
Use it when: Every gas problem.
PV over T is constant
PV/T has the same value for the same gas in any state.
Use it when: Gas equation.
STP: 273 and 760
S.T.P. is 273 K and 760 mm Hg.
Use it when: Standard conditions.

Charles's Law, Kelvin Scale, the Gas Equation and S.T.P.: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Absolute zero · 1 mark · MCQBoard level

The absolute zero of temperature on the Celsius scale is:
(a) 0 °C
(b) 273 °C
(c) −273 °C
(d) −100 °C

Try it first, then show the solutionSolution

Volume would become zero at −273 °C.

Answer: (c) −273 °C

Example 2: Celsius and kelvin · 2 marks · Short answerBoard level

Convert 127 °C to kelvin and 300 K to Celsius. What is the temperature of absolute zero in kelvin?

Try it first, then show the solutionSolution

$127 + 273 = 400$ K. (1)
$300 - 273 = 27$ °C; absolute zero is 0 K. (1)

Answer: 400 K; 27 °C; 0 K

Example 3: Heating and cooling at constant pressure · 3 marks · Short answerHarder board level

A gas occupies 600 cm³ at 27 °C at constant pressure. Find its volume (a) at 177 °C and (b) at −123 °C.

Try it first, then show the solutionSolution

$T_{1} = 300$ K. (½)
(a) $T_{2} = 450$ K: $V_{2} = 600 \times \frac{450}{300} = 900$ cm³. (1)
(b) $T_{2} = 150$ K: $V_{2} = 600 \times \frac{150}{300} = 300$ cm³. (1½)

Answer: 900 cm³; 300 cm³

Example 4: Both P and T change · 3 marks · Short answerHarder board level

3.0 L of a gas at 1.0 atm and 27 °C is changed to 2.0 atm and 127 °C. Find the final volume.

Try it first, then show the solutionSolution

$T_{1} = 300$ K, $T_{2} = 400$ K. (1)
$V_{2} = \frac{P _{1} V _{1} T _{2}}{T _{1} P _{2}} = \frac{1.0 \times 3.0 \times 400}{300 \times 2.0}$ (1)
$= 2.0$ L. (1)

Answer: 2.0 L

Example 5: Reducing to S.T.P. · 4 marks · Case-basedHOTS / competency

A gas occupies 560 cm³ at 27 °C and 700 mm Hg.
(a) Convert the temperature to kelvin.
(b) Find the volume at S.T.P.
(c) Find the volume if the same gas at 700 mm Hg is heated at constant pressure to 54 °C.
(d) State why the temperature must not be used in °C in these calculations.

Try it first, then show the solutionSolution

(a) $T = 300$ K. (½)
(b) $V = 560 \times \frac{700}{300} \times \frac{273}{760} \approx 469$ cm³. (1½)
(c) $V = 560 \times \frac{327}{300} \approx 610$ cm³. (1)
(d) Volume is proportional to the absolute temperature, not to the Celsius one; at 0 °C the volume is not zero. (1)

Answer: (a) 300 K (b) about 469 cm³ (c) about 610 cm³ (d) V ∝ T (kelvin)

Where students lose marks in Charles's Law, Kelvin Scale, the Gas Equation and S.T.P.

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

Mistake: Celsius used in the formula
Only kelvin is valid.
Wrong working
V₂ = 600 × 177/27
Right working
V₂ = 600 × 450/300
Do this: Add 273 first.
Usually costs: The numerical.
Mistake: S.T.P. pressure taken as 76 mm Hg
760 mm Hg (or 76 cm Hg).
Wrong working
P = 76 mm Hg.
Right working
P = 760 mm Hg.
Do this: Check the unit.
Usually costs: The answer.
Mistake: Pressure units different on two sides
Use the same unit.
Wrong working
P₁ = 0.9 atm, P₂ = 760 mm Hg.
Right working
Convert 0.9 atm to 684 mm Hg.
Do this: Convert first.
Usually costs: The answer.

Exam tips

Convert to kelvin before anything else.
State "constant pressure" or "constant volume" next to the law.
For S.T.P. problems write 273 K and 760 mm Hg explicitly.
Give the gas equation derivation in two steps when asked.

Practice questions on Charles's Law, Kelvin Scale, the Gas Equation and S.T.P.

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 · MCQBoard level
Standard temperature and pressure (S.T.P.) are:
(a) 0 K and 760 mm Hg
(b) 273 K and 760 mm Hg
(c) 273 °C and 760 mm Hg
(d) 25 °C and 1 atm
Hint
0 °C is 273 K.
Answer
(b)
Q2 · 1 mark · Very short answerBoard level
Convert −173 °C to kelvin.
Hint
Add 273.
Answer
100 K.
Q3 · 2 marks · Short answerHarder board level
400 cm³ of a gas at 17 °C is heated to 87 °C at constant pressure. Find the new volume.
Hint
290 K to 360 K.
Answer
$V_{2} = 400 \times \frac{360}{290} \approx 497$ cm³.
Q4 · 3 marks · Short answerHarder board level
A rigid flask of fixed volume holds a gas at 1.0 atm and 27 °C. It is heated to 127 °C. Find the final pressure.
Hint
Volume constant: P₁/T₁ = P₂/T₂.
Answer
$P_{2} = 1.0 \times \frac{400}{300} = 1.33$ atm.
Q5 · 2 marks · Assertion-reasonHarder board level
Assertion (A): In gas-law calculations the temperature must be in kelvin.
Reason (R): The volume of a gas at constant pressure is proportional to its Celsius temperature.
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Hint
Which temperature scale gives V ∝ T?
Answer
(c)
Q6 · 3 marks · Short answerHOTS / competency
At 0 °C a gas has volume 273 cm³ at constant pressure. Using $V_{t} = V_{0} (1 + \frac{t}{273})$ , find the volume at 100 °C and at −73 °C, and state the rate of change of volume per °C and the temperature at which the extended graph reaches V = 0.
Hint
V changes by 1 cm³ per °C here.
Answer
At 100 °C: $273 \times (1 + \frac{100}{273}) = 373$ cm³. At −73 °C: $273 \times (1 - \frac{73}{273}) = 200$ cm³. The volume changes by 1 cm³ per °C and reaches zero at −273 °C.

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.Charles's law?Show answer
At constant P, V ∝ T (kelvin) for a fixed mass of gas.
2.Absolute zero?Show answer
−273 °C or 0 K.
3.Gas equation?Show answer
P₁V₁/T₁ = P₂V₂/T₂.
4.Standard temperature and pressure?Show answer
273 K and 760 mm Hg (1 atm).
5.Conversion from °C to K?Show answer
T = t + 273.

Charles's Law, Kelvin Scale, the Gas Equation and S.T.P.: frequently asked questions

What is Charles's law?

At constant pressure the volume of a fixed mass of gas is directly proportional to its absolute (Kelvin) temperature.

What is absolute zero?

The lowest possible temperature, −273 °C (0 K), at which the volume of an ideal gas would be zero.

How do you reduce a gas volume to S.T.P.?

Use P₁V₁/T₁ = P₂V₂/T₂ with P₂ = 760 mm Hg and T₂ = 273 K.