Answer
🧪 Final Pressure of Methane Gas Using Combined Gas Law
Objective:
To calculate the final pressure (P₂) of methane gas when:
- Initial temperature: 17.0 °C
- Final temperature: −12.0 °C
- Volume decreases by 30%
- Initial pressure: 4.8 atm
Step 1: Confirm Ideal Gas Behavior
The boiling point of methane (CH₄) is −161 °C. Since the given temperatures are above this value, we can treat methane as an ideal gas.
Step 2: Convert Temperatures to Kelvin
T₁ = 273.15 + 17 = 290.15 K
T₂ = 273.15 + (−12) = 261.15 K
T₂ = 273.15 + (−12) = 261.15 K
Step 3: Determine Volumes
Let initial volume V₁ = V. Volume decreases by 30%, so:
V₂ = V − 0.3V = 0.7V
Step 4: Apply Combined Gas Law
(P₁ × V₁) / T₁ = (P₂ × V₂) / T₂
Substitute known values:
(4.8 × V) / 290.15 = (P₂ × 0.7V) / 261.15
Cancel V from both sides and solve for P₂:
4.8 / 290.15 = (P₂ × 0.7) / 261.15
P₂ = (4.8 × 261.15) / (290.15 × 0.7) ≈ 6.17 atm
P₂ = (4.8 × 261.15) / (290.15 × 0.7) ≈ 6.17 atm
✅ Final Answer:
P₂ ≈ 6.2 atm (rounded to 2 significant figures)
Key Concepts:
- The combined gas law relates pressure, volume, and temperature.
- Decreasing volume increases pressure (inverse relationship).
- Decreasing temperature reduces pressure (direct relationship).
- In this scenario, the effect of volume decrease outweighs the temperature drop, leading to a higher final pressure.