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ICSE Class 9 Chemistry Question 5 of 29

Study of Gas Laws — Question 13

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Question 13

At 0°C and 760 mm Hg pressure, a gas occupies volume of 100 cm3. Kelvin temperature of the gas is increased by one-fifth and the pressure is increased one and a half times. Calculate the final volume of the gas.

Answer

Initial conditions [S.T.P.] :

P1 = Initial pressure of the gas = 760 mm Hg
V1 = Initial volume of the gas = 100 cm3
T1 = Initial temperature of the gas = 0°C = 273 K

Final conditions :

P2 (Final pressure) = increased one and a half times of P1

= (1 + 12\dfrac{1}{2}) of 760

= 32\dfrac{3}{2} x 760

= 3 x 380

= 1140 mm Hg

V2 (Final volume) = ?

T2 (Final temperature) = increased by one-fifth of 273 K

= (1+15)\Big(1 + \dfrac{1}{5}\Big) of 273

= 65\dfrac{6}{5} x 273 = 16385\dfrac{1638}{5}

By Gas Law:

P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}

Substituting the values :

760×100273=1140×V216385760×100273=5×1140×V21638V2=760×100×1638273×5×1140V2=80 cm3\dfrac{760 \times 100 }{273} = \dfrac{1140 \times \text{V}_2}{\dfrac{1638}{5}} \\[1em] \dfrac{760 \times 100 }{273} = \dfrac{5 \times 1140 \times \text{V}_2}{1638} \\[1em] \text{V}_2 = \dfrac{760 \times 100 \times 1638}{273 \times 5 \times 1140} \\[1em] \text{V}_2 = 80 \text{ cm}^3

∴ Final volume of the gas = 80 cm3

Chapter Overview: Study of Gas Laws

Gas laws describe the behaviour of gases in terms of pressure (P), volume (V), and temperature (T). Boyle's Law states that at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure (PV = constant). Charles' Law states that at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature (V/T = constant). These laws combine to give the Combined Gas Law: P1V1/T1 = P2V2/T2. Temperature must always be in Kelvin (K = °C + 273). Standard Temperature and Pressure (STP) is defined as 0°C (273 K) and 760 mm Hg (1 atm). The chapter also introduces the concept of absolute zero (−273°C or 0 K), where theoretically a gas would have zero volume. Students must solve numerical problems using these laws, convert between temperature scales, and understand the graphical representations of these relationships.

Key Formulas

Law / Formula Expression & Conditions
Boyle's LawP1V1 = P2V2 (at constant T and mass)
Charles' LawV1/T1 = V2/T2 (at constant P and mass); T in Kelvin
Combined Gas LawP1V1/T1 = P2V2/T2
Kelvin ConversionT(K) = T(°C) + 273
STP0°C (273 K) and 760 mm Hg (1 atm or 101.3 kPa)
Absolute Zero−273°C (0 K); theoretically the lowest possible temperature
Pressure Units1 atm = 760 mm Hg = 76 cm Hg = 101325 Pa

Must-Know Concepts

  • Boyle's Law graph: P vs V is a hyperbola; PV vs P is a straight horizontal line
  • Charles' Law graph: V vs T(K) is a straight line through the origin; V vs T(°C) intercepts at −273°C
  • Always convert temperature to Kelvin before using Charles' Law or Combined Gas Law
  • Gas laws apply to ideal gases; real gases deviate at high pressure and low temperature
  • When solving problems, identify which variable is constant to choose the correct law
  • At STP, 1 mole of any gas = 22.4 L (this connects to the mole concept in Class X)

Important Diagrams to Practice

  • P vs V graph (hyperbola) and PV vs P graph (horizontal line) for Boyle's Law
  • V vs T(K) graph (straight line through origin) for Charles' Law
  • V vs T(°C) graph showing extrapolation to absolute zero (−273°C)

Common Mistakes

  • Using temperature in °C instead of Kelvin in Charles' Law calculations
  • Confusing inversely proportional (Boyle's) with directly proportional (Charles')
  • Not converting pressure units consistently (mixing mm Hg with atm)
  • Applying gas laws to liquids or solids (they apply only to gases)
  • Forgetting to check that mass is constant when applying gas laws

Scoring Tips

  • Step 1 in every problem: convert temperature to Kelvin
  • Write down given values clearly with proper units before applying the formula
  • State which law you are using and the condition that makes it applicable
  • For graph questions, label axes correctly and mention the nature of the curve
  • Practice combined gas law problems - these are the most frequently asked

Frequently Asked Questions

Why must temperature be in Kelvin for gas law calculations?

The Celsius scale has an arbitrary zero point (freezing point of water). Charles' Law requires an absolute scale where 0 represents zero molecular motion (and theoretically zero volume). The Kelvin scale starts at absolute zero (−273°C), making it a true proportional scale for gas behaviour.

Can a gas actually reach absolute zero?

No. According to the Third Law of Thermodynamics, absolute zero can never be reached, only approached. At temperatures near absolute zero, gases liquefy and then solidify. Gas laws do not apply to liquids or solids.

What is the difference between an ideal gas and a real gas?

An ideal gas perfectly obeys all gas laws - its molecules have zero volume and no intermolecular forces. Real gases deviate from ideal behaviour at high pressures (molecules are close together) and low temperatures (intermolecular forces become significant). At normal conditions, most gases behave approximately ideally.