Question

what is the density of argon gas (ar) at 3.35 bar and 39.4 °c? express your answer in g/l

Explanation:

Step1: Convert pressure to SI - unit

1 bar = 100000 Pa, so 3.35 bar = 3.35×100000 Pa = 335000 Pa.

Step2: Convert temperature to Kelvin

T(K)=T(°C)+273.15, so T = 39.4 + 273.15=312.55 K.

Step3: Use the ideal - gas law and density formula

The ideal - gas law is PV = nRT, where n = m/M (m is mass and M is molar mass), and density ρ=m/V. Rearranging gives ρ = PM/RT.
The molar mass of argon M = 39.95 g/mol, R = 8.314 J/(mol·K).
Substitute the values: ρ=(335000 Pa×39.95 g/mol)/(8.314 J/(mol·K)×312.55 K).
Since 1 J = 1 Pa·m³, ρ=(335000 Pa×39.95 g/mol)/(8.314 Pa·m³/(mol·K)×312.55 K).
ρ=(335000×39.95 g)/(8.314×312.55 m³).
ρ = 514.9 g/m³.
To convert to g/L, since 1 m³ = 1000 L, ρ = 5.15 g/L.

Answer:

5.15 g/L