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Study of Gas Laws — ICSE Class 9 Chemistry Important Questions

13 ICSE Class 9 Chemistry practice questions on Study of Gas Laws, each with a full model answer, covering 5 topics from the chapter in the question formats used in the exam.

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13
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Key concepts
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Board-favourite ICSE Gas Laws questions are Boyle's law (PV=constantPV=\text{constant}PV=constant at constant TTT), Charles' law (VT=constant\frac{V}{T}=\text{constant}V/T=constant at constant PPP), converting Celsius to absolute (Kelvin) temperature using T(K)=t(∘C)+273T(K)=t(^\circ C)+273T(K)=t(^ C)+273, and numericals on the combined gas equation P1V1T1=P2V2T2\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}P_1V_1/T_1=P_2V_2/T_2, often reducing a volume to S.T.P.

Syllabus reference (2026–27): CISCE (Council for the Indian School Certificate Examinations)

About Study of Gas Laws

In this ICSE Class 9 Chemistry chapter Study of Gas Laws you learn how the pressure, volume and temperature of a fixed mass of gas are related. Boyle's law links pressure and volume at constant temperature, Charles' law links volume and absolute temperature at constant pressure, and the combined gas equation ties all three together. You also learn the Kelvin (absolute) temperature scale and solve numericals, including reduction of gas volumes to standard temperature and pressure (S.T.P.).

Boyle's lawCharles' lawAbsolute (Kelvin) temperature scaleCombined gas equationNumericals and reduction to S.T.P.

Key concepts & formulas

Boyle's law

At constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure: V∝1PV\propto\dfrac{1}{P}V1/P, so PV=constantPV=\text{constant}PV=constant and P1V1=P2V2P_1V_1=P_2V_2P_1V_1=P_2V_2.

Charles' law

At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature: V∝TV\propto TV T, so VT=constant\dfrac{V}{T}=\text{constant}V/T=constant and V1T1=V2T2\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}V_1/T_1=V_2/T_2.

Absolute temperature

The Kelvin (absolute) scale starts at absolute zero, −273∘C-273^\circ C-273^ C. Convert with T(K)=t(∘C)+273T(K)=t(^\circ C)+273T(K)=t(^ C)+273. Temperatures in gas-law formulae must always be in kelvin.

Combined gas equation and S.T.P.

P1V1T1=P2V2T2\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}P_1V_1/T_1=P_2V_2/T_2. Standard temperature and pressure (S.T.P.) are 273 K (0∘C)273\text{ K}\,(0^\circ C)273 K\,(0^ C) and 760 mm Hg (1 atm)760\text{ mm Hg}\,(1\text{ atm})760 mm Hg\,(1 atm).

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Important questions with answers

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Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

Boyle's law is obeyed at constant temperature and states that:

  1. (a)

    V∝TV\propto TV T

  2. (b)

    PV=constantPV=\text{constant}PV=constant

  3. (c)

    VT=constant\dfrac{V}{T}=\text{constant}V/T=constant

  4. (d)

    P∝TP\propto TP T

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Answer: (b) PV=constantPV=\text{constant}PV=constant.

At constant temperature, V∝1PV\propto\dfrac{1}{P}V1/P, so the product PVPVPV is constant for a fixed mass of gas.

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Q2MCQEasy1 mark

The value of absolute zero on the Celsius scale is:

  1. (a)

    0∘C0^\circ C0^ C

  2. (b)

    100∘C100^\circ C100^ C

  3. (c)

    −273∘C-273^\circ C-273^ C

  4. (d)

    273∘C273^\circ C273^ C

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Answer: (c) −273∘C-273^\circ C-273^ C.

Absolute zero, the zero of the Kelvin scale, corresponds to −273∘C-273^\circ C-273^ C; T(K)=t(∘C)+273T(K)=t(^\circ C)+273T(K)=t(^ C)+273.

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Q3MCQModerate1 mark

A gas at constant temperature is compressed to half its original volume. Its pressure becomes:

  1. (a)

    Half

  2. (b)

    Double

  3. (c)

    Four times

  4. (d)

    Unchanged

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Answer: (b) Double.

By Boyle's law P1V1=P2V2P_1V_1=P_2V_2P_1V_1=P_2V_2. If V2=V12V_2=\dfrac{V_1}{2}V_2=V_1/2, then P2=P1V1V1/2=2P1P_2=\dfrac{P_1V_1}{V_1/2}=2P_1P_2=P_1V_1/V_1/2=2P_1.

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Q4MCQHOTS1 mark

A fixed mass of gas at 27∘C27^\circ C27^ C is heated at constant pressure until its volume doubles. The new temperature is:

  1. (a)

    54∘C54^\circ C54^ C

  2. (b)

    300∘C300^\circ C300^ C

  3. (c)

    327∘C327^\circ C327^ C

  4. (d)

    600∘C600^\circ C600^ C

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Answer: (c) 327∘C327^\circ C327^ C.

T1=27+273=300 KT_1=27+273=300\text{ K}T_1=27+273=300 K. By Charles' law V1T1=V2T2\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}V_1/T_1=V_2/T_2 with V2=2V1V_2=2V_1V_2=2V_1: T2=2×300=600 K=600−273=327∘CT_2=2\times300=600\text{ K}=600-273=327^\circ CT_2=2×300=600 K=600-273=327^ C.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): In all gas-law calculations the temperature must be taken in kelvin.

Reason (R): Volume is directly proportional to the Celsius temperature of the gas.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

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Answer: (c) A is true - kelvin must be used. R is false: volume is proportional to the absolute (kelvin) temperature, not the Celsius temperature; at 0∘C0^\circ C0^ C the volume is not zero.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Convert (i) 25∘C25^\circ C25^ C to kelvin and (ii) 200 K200\text{ K}200 K to degrees Celsius.

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(i) T(K)=25+273=298 KT(K)=25+273=\mathbf{298\text{ K}}T(K)=25+273=298 K.

(ii) t(∘C)=200−273=−73∘Ct(^\circ C)=200-273=\mathbf{-73^\circ C}t(^ C)=200-273=-73^ C.

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Q7Very ShortModerate2 marks

A gas occupies 250 cm3250\text{ cm}^3250 cm^3 at 760 mm Hg760\text{ mm Hg}760 mm Hg. What volume will it occupy at 1000 mm Hg1000\text{ mm Hg}1000 mm Hg, temperature remaining constant?

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By Boyle's law P1V1=P2V2P_1V_1=P_2V_2P_1V_1=P_2V_2.

V2=P1V1P2=760×2501000=1900001000=190 cm3V_2=\dfrac{P_1V_1}{P_2}=\dfrac{760\times250}{1000}=\dfrac{190000}{1000}=\mathbf{190\text{ cm}^3}V_2=P_1V_1/P_2=760×250/1000=190000/1000=190 cm^3.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

State Charles' law. A given mass of gas has a volume of 300 cm3300\text{ cm}^3300 cm^3 at 27∘C27^\circ C27^ C. Find its volume at 57∘C57^\circ C57^ C if the pressure is kept constant.

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Charles' law: At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature, i.e. V1T1=V2T2\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}V_1/T_1=V_2/T_2.

T1=27+273=300 KT_1=27+273=300\text{ K}T_1=27+273=300 K, T2=57+273=330 KT_2=57+273=330\text{ K}T_2=57+273=330 K, V1=300 cm3V_1=300\text{ cm}^3V_1=300 cm^3.

V2=V1 T2T1=300×330300=330 cm3V_2=\dfrac{V_1\,T_2}{T_1}=\dfrac{300\times330}{300}=\mathbf{330\text{ cm}^3}V_2=V_1\,T_2/T_1=300×330/300=330 cm^3.

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Q9Short AnswerModerate3 marks

A certain mass of gas occupies 2 litres2\text{ litres}2 litres at 27∘C27^\circ C27^ C and 740 mm Hg740\text{ mm Hg}740 mm Hg pressure. Find its volume at S.T.P.

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S.T.P.: P2=760 mm HgP_2=760\text{ mm Hg}P_2=760 mm Hg, T2=273 KT_2=273\text{ K}T_2=273 K. Given P1=740 mm HgP_1=740\text{ mm Hg}P_1=740 mm Hg, V1=2 LV_1=2\text{ L}V_1=2 L, T1=27+273=300 KT_1=27+273=300\text{ K}T_1=27+273=300 K.

Using P1V1T1=P2V2T2\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}P_1V_1/T_1=P_2V_2/T_2:

V2=P1V1T2T1P2=740×2×273300×760V_2=\dfrac{P_1V_1T_2}{T_1P_2}=\dfrac{740\times2\times273}{300\times760}V_2=P_1V_1T_2/T_1P_2=740×2×273/300×760

V2=404040228000=1.77 LV_2=\dfrac{404040}{228000}=\mathbf{1.77\text{ L}}V_2=404040/228000=1.77 L (approx).

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Q10Short AnswerHOTS3 marks

Why must a gas syllabus use the Kelvin scale rather than the Celsius scale in Charles' law? Support your answer with the idea of absolute zero.

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Charles' law says volume is directly proportional to temperature, so at zero temperature the volume should be zero. On the Celsius scale a gas at 0∘C0^\circ C0^ C still has a large volume, and negative Celsius temperatures would give impossible negative volumes - so proportionality fails.

On the Kelvin scale, temperature is measured from absolute zero (−273∘C)(-273^\circ C)(-273^ C), the lowest possible temperature at which the volume of an ideal gas would theoretically become zero. Only when TTT is in kelvin is VT\dfrac{V}{T}V/T truly constant, so the Kelvin scale must be used.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

(a) State Boyle's law and represent it graphically (PPP against VVV). (b) A gas cylinder holds 30 litres30\text{ litres}30 litres of gas at 12 atm12\text{ atm}12 atm and 27∘C27^\circ C27^ C. What volume would this gas occupy at S.T.P.?

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(a) Boyle's law: At constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure, V∝1PV\propto\dfrac{1}{P}V1/P, so PV=constantPV=\text{constant}PV=constant. A graph of PPP (y-axis) against VVV (x-axis) is a rectangular hyperbola.

ICSE Class 9 Chemistry — Study of Gas Laws: (a) State Boyle's law and represent it graphically (P against V). (b) A gas cylinder holds 30\text{ litres} of gas at 12\text{ atm} and

(b) Given V1=30 LV_1=30\text{ L}V_1=30 L, P1=12 atmP_1=12\text{ atm}P_1=12 atm, T1=27+273=300 KT_1=27+273=300\text{ K}T_1=27+273=300 K; at S.T.P. P2=1 atmP_2=1\text{ atm}P_2=1 atm, T2=273 KT_2=273\text{ K}T_2=273 K.

P1V1T1=P2V2T2⇒V2=P1V1T2T1P2=12×30×273300×1\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\Rightarrow V_2=\dfrac{P_1V_1T_2}{T_1P_2}=\dfrac{12\times30\times273}{300\times1}P_1V_1/T_1=P_2V_2/T_2 V_2=P_1V_1T_2/T_1P_2=12×30×273/300×1

V2=98280300=327.6 LV_2=\dfrac{98280}{300}=\mathbf{327.6\text{ L}}V_2=98280/300=327.6 L.

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Q12Long AnswerHOTS5 marks

A fixed mass of gas has a volume of 760 cm3760\text{ cm}^3760 cm^3 at 27∘C27^\circ C27^ C and 700 mm Hg700\text{ mm Hg}700 mm Hg. (i) Reduce this volume to S.T.P. (ii) The same gas is then heated to 77∘C77^\circ C77^ C at S.T.P. pressure; find the new volume. Show all working.

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(i) Reduction to S.T.P.

Given V1=760 cm3V_1=760\text{ cm}^3V_1=760 cm^3, P1=700 mm HgP_1=700\text{ mm Hg}P_1=700 mm Hg, T1=27+273=300 KT_1=27+273=300\text{ K}T_1=27+273=300 K; S.T.P.: P2=760 mm HgP_2=760\text{ mm Hg}P_2=760 mm Hg, T2=273 KT_2=273\text{ K}T_2=273 K.

V2=P1V1T2T1P2=700×760×273300×760=700×273300=191100300=637 cm3V_2=\dfrac{P_1V_1T_2}{T_1P_2}=\dfrac{700\times760\times273}{300\times760}=\dfrac{700\times273}{300}=\dfrac{191100}{300}=\mathbf{637\text{ cm}^3}V_2=P_1V_1T_2/T_1P_2=700×760×273/300×760=700×273/300=191100/300=637 cm^3 at S.T.P.

(ii) Heating to 77∘C77^\circ C77^ C at constant (S.T.P.) pressure.

Now V2=637 cm3V_2=637\text{ cm}^3V_2=637 cm^3 at T2=273 KT_2=273\text{ K}T_2=273 K; heat to T3=77+273=350 KT_3=77+273=350\text{ K}T_3=77+273=350 K at constant pressure, so use Charles' law:

V3=V2 T3T2=637×350273=222950273=816.7 cm3V_3=\dfrac{V_2\,T_3}{T_2}=\dfrac{637\times350}{273}=\dfrac{222950}{273}=\mathbf{816.7\text{ cm}^3}V_3=V_2\,T_3/T_2=637×350/273=222950/273=816.7 cm^3 (approx).

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

A student traps a fixed mass of dry air in a graduated tube and records its volume VVV at pressure PPP, keeping the temperature constant at 27∘C27^\circ C27^ C:

| PPP (mm Hg) | 400 | 500 | 800 | | VVV (cm3^3^3) | 60 | 48 | ? |

(i) Which gas law is being verified?
(ii) Show that the first two readings obey this law.
(iii) Calculate the missing volume at 800 mm Hg800\text{ mm Hg}800 mm Hg.
(iv) What would a graph of PPP against 1V\dfrac{1}{V}1/V look like?

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(i) The experiment verifies Boyle's law (constant temperature, varying PPP and VVV).

(ii) P1V1=400×60=24000P_1V_1=400\times60=24000P_1V_1=400×60=24000; P2V2=500×48=24000P_2V_2=500\times48=24000P_2V_2=500×48=24000. The product PVPVPV is constant, so the readings obey Boyle's law.

(iii) P3V3=24000⇒V3=24000800=30 cm3P_3V_3=24000\Rightarrow V_3=\dfrac{24000}{800}=\mathbf{30\text{ cm}^3}P_3V_3=24000 V_3=24000/800=30 cm^3.

(iv) Since P=constant×1VP=\text{constant}\times\dfrac{1}{V}P=constant×1/V, a graph of PPP against 1V\dfrac{1}{V}1/V is a straight line passing through the origin.

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All ICSE Class 9 Chemistry Chapters

The Language of ChemistryChemical Changes and ReactionsWaterAtomic Structure and Chemical BondingThe Periodic TableStudy of the First Element: HydrogenStudy of Gas LawsAtmospheric Pollution

Frequently asked questions

  • Do these Study of Gas Laws questions follow the latest ICSE syllabus?
    Yes — they are aligned to the CISCE 2026–27 syllabus for ICSE Class 9 Chemistry, so nothing here is outside the current course.

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