1.00 mole of oxygen at 60 degrees celcius and 5 bar is flowing through a 20cm diameter pipe (inner diameter). What is the velocity of the gas in the pipe? Can use relation u = V/A = m / p x A

where u = velocity in m/x, V = volumetric florate, A = cross sectional area, m = mass flow rate, p = density

u = 0.176 m/x

Explanation:

∴ n O2 = 1 mol

∴ mass O2 = 1 mol * 32 g/mol = 32 g

∴ T = 60°C = 333 K

∴ P = 5 bar

⇒ V = RTn/P = (83.14 bar. cm³/mol. K) * (333 K) * (1 mol)) / (5 bar)

⇒ V = 5537.124 cm³

∴ d = 20cm

⇒ A = (1/4π) * d² = 314.16 cm²

velocity of the gas (m/x):

u = m / ρ*A let time (t) = x sec

∴ ρ = 32 g / 5537.124 cm³ = 5.78 E-3 g/cm³

∴ mass flow rate (m) = 32g / x

⇒ u = (32 g/x) / ((5.78 E-3 g/cm³) * (314.16cm²))

⇒ u = 17.625 cm/x * (m/100cm) = 0.176 m/x