Study of Gas Laws
Introduction
Gases have unique properties that can be described by mathematical relationships between pressure, volume, and temperature. ICSE Class 9 covers Boyle's Law, Charles' Law, and their applications.
Boyle's Law (Pressure-Volume Relationship)
Statement: At constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure.
V ∝ 1/P (at constant T)
or
P₁V₁ = P₂V₂
Where:
- P₁, V₁ = initial pressure and volume
- P₂, V₂ = final pressure and volume
Charles' Law (Temperature-Volume Relationship)
Statement: At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute (Kelvin) temperature.
V ∝ T (at constant P)
or
V₁/T₁ = V₂/T₂
Kelvin and Celsius Scale
For gas law calculations, temperature must be in Kelvin.
T(K) = t(°C) + 273
Absolute zero: -273°C or 0 K (the lowest possible temperature)
<ICSEExample title="Charles Law Problem"> A gas occupies 300 cm³ at 27°C. What volume will it occupy at 127°C, pressure remaining constant? <Solution> T₁ = 27 + 273 = 300 K, V₁ = 300 cm³ T₂ = 127 + 273 = 400 K, V₂ = ? V₁/T₁ = V₂/T₂ 300/300 = V₂/400 V₂ = 400 cm³ </Solution> </ICSEExample> <ICSEExample title="Finding Temperature"> A gas occupies 200 cm³ at 27°C. At what temperature (in °C) will it occupy 300 cm³ if pressure is constant? <Solution> T₁ = 27 + 273 = 300 K, V₁ = 200 cm³ V₂ = 300 cm³, T₂ = ? V₁/T₁ = V₂/T₂ 200/300 = 300/T₂ T₂ = (300 × 300)/200 = 450 K t = 450 - 273 = 177°C </Solution> </ICSEExample>Graphical Representation
Boyle's Law
Plot of V vs 1/P: Straight line through origin Plot of V vs P: Hyperbola
Charles' Law
Plot of V vs T: Straight line that extrapolates to V = 0 at T = -273°C (absolute zero)
Combined Gas Equation
When all three variables (P, V, T) change:
P₁V₁/T₁ = P₂V₂/T₂
<ICSEExample title="Combined Gas Law"> A gas occupies 400 cm³ at 27°C and 760 mm Hg pressure. Find its volume at 127°C and 855 mm Hg. <Solution> P₁ = 760, V₁ = 400, T₁ = 27 + 273 = 300 K P₂ = 855, V₂ = ?, T₂ = 127 + 273 = 400 K P₁V₁/T₁ = P₂V₂/T₂ (760 × 400)/300 = (855 × V₂)/400 V₂ = (760 × 400 × 400)/(300 × 855) V₂ = 121600000/256500 V₂ = 474.07 cm³ </Solution> </ICSEExample>Standard Temperature and Pressure (STP)
Standard conditions for gas measurements:
- Standard Temperature (T): 0°C = 273 K
- Standard Pressure (P): 760 mm Hg = 1 atm = 1.013 × 10⁵ Pa
Common Mistakes With Fixes
| Mistake | Correction |
|---|---|
| Using Celsius instead of Kelvin | Always convert °C to K (add 273) before calculations |
| Boyle's Law: P and V directly proportional | P and V are INVERSELY proportional |
| Forgetting that temperature must be absolute | Use Kelvin scale for ALL gas law calculations |
| Mixing pressure units | Maintain consistent units throughout the problem |
ICSE Exam Focus
| Topic | Marks (approx.) | Frequency |
|---|---|---|
| Boyles Law numericals | 4-5 marks | Very common |
| Charles Law numericals | 4-5 marks | Very common |
| Combined gas equation | 5 marks | Common |
| STP and temperature conversion | 2-3 marks | Frequently asked |
Self-Test
Q1: State Boyles Law. A gas at 720 mm Hg pressure occupies 250 mL. What volume will it occupy at 800 mm Hg?
Q2: A gas occupies 150 cm³ at 27°C. What is its volume at 87°C if pressure is constant?
Q3: Convert: (i) 47°C to Kelvin (ii) 373 K to °C
Q4: A gas has volume 300 cm³ at 27°C and 740 mm Hg. What volume will it occupy at STP?
Q5: Explain what happens to the volume of a gas if: (i) pressure is doubled at constant T (ii) temperature is halved at constant P
