A tank of volume 0.6m3 contains oxygen at an absolute pressure of 2.0×106Nm2 and a temperature of 25.0°C. Assume that oxygen behaves like an ideal gas. (a) How many moles of oxygen are there in the tank? (b) How many kilograms? (c) Find the pressure if thetemperature is increased to 510.0°C. (d) At a temperature of 25.0 C , how many moles can be withdrawn from the tank before the pressure falls to 10% of the original pressure?
Answer
Ideal Gas Law Problem – Oxygen Tank Thermodynamics
🔷 Given Data
- Volume of tank: V = 0.6 m³
- Initial pressure: P₁ = 2.0 × 10⁶ N/m²
- Initial temperature: T₁ = 25°C = 298.15 K
- Gas constant: R = 8.314 J/mol·K
- Molar mass of O₂: 32 g/mol = 0.032 kg/mol
🔹 (a) Number of Moles of Oxygen
Using the Ideal Gas Law:
n = (PV) / (RT)
= (2.0 × 10⁶ × 0.6) / (8.314 × 298.15)
≈ 484.1 moles
🔹 (b) Mass of Oxygen
Mass = n × Molar Mass
= 484.1 × 0.032
= 15.49 kg
🔹 (c) Pressure at 510°C
Convert temperature to Kelvin:
T₂ = 510 + 273.15 = 783.15 K
Using:
P₂ = P₁ × (T₂ / T₁)
= 2.0 × 10⁶ × (783.15 / 298.15)
≈ 5.254 × 10⁶ N/m²
🔹 (d) Moles Withdrawn at 10% Pressure
New pressure:
P₂ = 0.1 × 2.0 × 10⁶ = 2.0 × 10⁵ N/m²
Remaining moles:
n₂ = (P₂ × V) / (R × T)
= (2.0 × 10⁵ × 0.6) / (8.314 × 298.15)
≈ 48.4 moles
Moles withdrawn:
n₁ – n₂ = 484.1 – 48.4 = 435.7 moles
✅ Final Summary
- (a) Moles = 484.1 mol
- (b) Mass = 15.49 kg
- (c) New pressure = 5.254 × 10⁶ N/m²
- (d) Withdrawn moles = 435.7 mol