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Find the rms speed in air at 0.0 degrees celsius and 1.00 atm of the N2 molecules, the O2 molecules and the CO2 molecules

Respuesta :

Answer:

N₂= 697.24 m/s,

O₂= 652.38 m/s

CO₂= 393.35 m/s

Explanation:

The rms speed, [tex]v_{rms}[/tex] of  a gas is the following:

[tex] v_{rms} = \sqrt \frac{3RT}{M} [/tex]    

where R: is the ideal gas constant, T: is the temperature and M: is the molar mass.

Knowing that the molar mass of the given gases are:

N₂ = 14.0067 g/mol = 14.0067x10⁻³ kg/mol

O₂ = 15.999 g/mol = 15.999x10⁻³ kg/mol

CO₂ = 44.009 g/mol = 44.009x10⁻³ kg/mol

Also that T is 0 °C = 273 K, and R = 8.314 J/ K.mol = 8.314 kg.m². s⁻². K⁻¹. mol⁻¹. The rms speed of the gases are:

For N₂:

[tex] v_{rms} = \sqrt \frac{3 \cdot 8.314 \frac{kg.m^{2}}{s^{2}.K.mol} \cdot 273 K}{14.0067 \cdot 10^{-3} \frac{kg}{mol}} = 697.24 m/s [/tex]

For O₂:

[tex] v_{rms} = \sqrt \frac{3 \cdot 8.314 \frac{kg.m^{2}}{s^{2}.K.mol} \cdot 273 K}{15.999 \cdot 10^{-3} \frac{kg}{mol}} = 652.38 m/s [/tex]    

For CO₂:

[tex] v_{rms} = \sqrt \frac{3 \cdot 8.314 \frac{kg.m^{2}}{s^{2}.K.mol} \cdot 273 K}{44.009 \cdot 10^{-3} \frac{kg}{mol}} = 393.35 m/s [/tex]              

Therefore, the rms speed of N₂ is 697.24 m/s, of O₂ is 652.38 m/s and of CO₂ is 393.35 m/s

I hope it helps you!

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