Answer
🧪 Methane Gas Volume Change Under Varying Conditions
Objective:
To determine whether the volume of methane gas increases or decreases, and by what percentage, when both temperature and pressure change.
Given:
- Initial Temperature: T₁ = −46.0°C = 227.15 K
- Final Temperature: T₂ = −80.0°C = 193.15 K
- Pressure is decreased by 5%: P₂ = 0.95 × P₁
Step-by-Step: Use the Ideal Gas Law Ratio
The ideal gas law is:
PV = nRT
Rearranging to solve for volume:
V = (nRT) / P
Volume is proportional to T / P, so:
% Change in Volume = ((T₂ / P₂ − T₁ / P₁) / (T₁ / P₁)) × 100%
Substitute Known Values:
T₂ / P₂ = 193.15 / (0.95 × P₁) ≈ 203.32 / P₁
T₁ / P₁ = 227.15 / P₁
T₁ / P₁ = 227.15 / P₁
Now compute percentage change:
% Change = ((203.32 − 227.15) / 227.15) × 100%
= (−23.83 / 227.15) × 100% ≈ −10.5%
= (−23.83 / 227.15) × 100% ≈ −10.5%
✅ Final Answer:
The volume decreases by approximately 10.5%.
Key Concepts:
- From the ideal gas law: Volume is directly proportional to temperature and inversely proportional to pressure.
- A decrease in temperature reduces volume.
- Even though pressure also decreased, the greater effect came from the temperature drop, resulting in a net volume decrease.