Anonymous
Anonymous asked in Science & MathematicsChemistry · 1 year ago

What is the density (g/L) of N2 gas at 889.2 Torr and 173 °C?

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  • 1 year ago
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    n / V = P / RT = (889.2 Torr) / ((62.36367 L Torr/K mol) x (173 + 273) K) = 0.031969 mol/L

    (28.01344 g N2/mol) x (0.031969 mol/L) = 0.89556 g/L = 0.896 g/L

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  • 1 year ago

    Gas law: PV = nRT

    No. of moles of 1 L of the gas, n = PV/(RT) = (889.2/760) × 1 / [0.08206 × (273 + 173)] mol = 0.03197 mol

    Molar mass of N₂ = 14.0×2 g/mol = 28.0 g/mol

    Density of N₂ = (0.03197 mol/L) × (28.0 g/mol) = 0.895 g/L

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  • 1 year ago

    m/V = (MMP)/(RT)

    Density = (molar mass of N2) (pressure in atm) / ( (ideal gas constant 0.0821) (temperature in Kelvin))

    So for temperature in kelvin 173 +273 = K

    For pressure in atm, 889.2 Torr / 760 = pressure in atm

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