Ideal-gas density — Assuming CO₂ behaves as a perfect gas, compute its density (kg/m³) at 263°C and 2 atm.

Introduction:
Estimating gas density from state variables using the ideal-gas relation is a frequent step in sizing equipment and calculating mass flow rates. Here we evaluate CO₂ at elevated temperature and moderate pressure.

Given Data / Assumptions:

• T = 263°C = 263 + 273 = 536 K.
• P = 2 atm = 2 × 101325 Pa = 202650 Pa.
• Molar mass of CO₂ = 44 g/mol = 0.044 kg/mol.
• Ideal-gas behavior (Z = 1).

Concept / Approach:
Use ρ = (P * M) / (R * T), where R = 8.314 kJ/(kmol·K) expressed as 8.314 J/(mol·K) for SI consistency → 8.314 (Pa·m³)/(mol·K).

Step-by-Step Solution:
Compute numerator: P * M = 202650 Pa * 0.044 kg/mol = 8916.6 Pa·kg/mol.Compute denominator: R * T = 8.314 * 536 = 4459.9 Pa·m³/(mol).Compute density: ρ = 8916.6 / 4459.9 ≈ 2.00 kg/m³.

Verification / Alternative check:
A quick check with ρ ≈ (P/RT) * M: at 1 atm and 300 K, CO₂ is about 1.84 kg/m³; doubling pressure and raising T to 536 K yields approximately (2/ (536/300)) × 1.84 ≈ (2 * 300 / 536) × 1.84 ≈ 2.06 kg/m³ → consistent with 2.

Why Other Options Are Wrong:

• 1 or 3 or 4 or 5 kg/m³ do not satisfy the ideal-gas calculation with given T and P.

Common Pitfalls:
Forgetting to convert Celsius to Kelvin; mixing units (atm vs. Pa); using molar mass in g/mol without converting to kg/mol.

Final Answer:
2