Answer
🧪 Methane Gas Volume Change Using the Combined Gas Law
Concept:
Methane (CH₄) behaves as an ideal gas at temperatures above −161 °C. The gas law used is:
PV = nRT
When the number of moles (n) is constant, we apply the combined gas law:
(P₂ × V₂) / (P₁ × V₁) = T₂ / T₁
Rearranged to solve for volume ratio:
V₂ / V₁ = (T₂ / T₁) × (P₁ / P₂)
Step 1: Convert Temperatures to Kelvin
T₁ = −36.0°C = 273.15 − 36.0 = 237.15 K
T₂ = 0.0°C = 273.15 K
T₂ = 0.0°C = 273.15 K
Step 2: Pressure Change
Initial pressure: P₁ = P₀
Final pressure: P₂ = 0.95 × P₀ (5% decrease)
Step 3: Substitute into Volume Ratio Formula
V₂ / V₁ = (273.15 / 237.15) × (P₀ / 0.95P₀)
= (273.15 / 237.15) × (1 / 0.95)
≈ 1.151 × 1.0526 ≈ 1.212
= (273.15 / 237.15) × (1 / 0.95)
≈ 1.151 × 1.0526 ≈ 1.212
Step 4: Calculate Percentage Increase in Volume
Percentage Increase = [(V₂ − V₁) / V₁] × 100
= (1.212 − 1) × 100 = 21.2%
= (1.212 − 1) × 100 = 21.2%
✅ Final Answer:
The volume of methane gas increases by approximately 21.2%.
Summary:
- Temperature increases → volume increases
- Pressure decreases → volume increases
- Both factors lead to expansion of the gas
- Ideal gas law holds as methane is above its boiling point of −161°C