If $H e$ $He$ (g) has an average kinetic energy of 5930 J/mol under certain conditions, what is the root mean square speed of $N_{2}$ $N_2$ (g) molecules under the same conditions?

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If $H e$ $He$ (g) has an average kinetic energy of 5930 J/mol under certain conditions, what is the root mean square speed of $N_{2}$ $N_2$ (g) molecules under the same conditions?

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Answer 1 · 2018-08-06T20:15:18

Well, what's in common? Not their degrees of freedom, but their temperature.

$v_{R M S} (N_{2}) = 650.7 m/s$ $v_(RMS)("N"_2) = "650.7 m/s"$

The average kinetic energy of $N_{2}$ $"N"_2$ would be higher, because it has more ways to move. But its RMS speed is lower due to its higher molar mass.

Helium is an atom, which translates in 3 dimensions with zero rotational and vibrational degrees of freedom.

Hence, according to the equipartition theorem,

$⟨ κ ⟩ \equiv \frac{K}{n} = \frac{N}{2} R T$ $<< kappa >> -= K/n = N/2RT$

is the average kinetic energy, where we have that $N = 3$ $N = 3$ for helium atom's linear degrees of freedom.

What temperature is it at?

$T = \frac{2}{3} \frac{1}{R} ⟨ κ ⟩$ $T = 2/3 1/R << kappa >>$

$= \frac{2}{3} \cdot \frac{1}{8.314 J/mol \cdot K} \cdot 5930 J/mol$ $= 2/3 cdot 1/("8.314 J/mol"cdot"K") cdot "5930 J/mol"$

$=$ $=$ $475.5 K$ $"475.5 K"$

Now, a pitfall would be to assume that $N$ $N$ is the same for $N_{2}$ $"N"_2$ ... it's not. $N_{2}$ $"N"_2$ is a MOLECULE, which rotates and vibrates. As it turns out,

Rotational degrees of freedom are non-negligible at room temperature.
Vibrational degrees of freedom are negligible for diatomic molecules at room temperature.

So, what we find is that

$N = N_{trans} + N_{rot} + N_{vib} \approx 3 + 2$ $N = N_("trans") + N_("rot") + N_"vib" ~~ 3 + 2$

because diatomic molecules rotate using two angles in spherical coordinates ( $θ, ϕ$ $theta,phi$ ).

Fortunately, this does not matter because all we want is the root-mean-square speed, which depends only on molar mass and temperature.

$v_{R M S} = \sqrt{\frac{3 R T}{M}}$ $v_(RMS) = sqrt((3RT)/M)$

where $M$ $M$ is the molar mass in $kg/mol$ $"kg/mol"$ . Why is that necessary? Why not $g/mol$ $"g/mol"$ ? Well, what are the units of $R$ $R$ ?

$v_{R M S} (N_{2}) = \sqrt{\frac{3 R T}{M}}$ $color(blue)(v_(RMS)("N"_2)) = sqrt((3RT)/M)$

$= \sqrt{\frac{3 \cdot 8.314 kg \cdot m^{2} {/s}^{2} / mol / K \cdot 475.5 K}{0.028014 kg/mol}}$ $= sqrt((3cdot"8.314 kg"cdot"m"^2"/s"^2//"mol"//"K" cdot "475.5 K")/"0.028014 kg/mol")$

$=$ $=$ $650.7 m/s$ $color(blue)("650.7 m/s")$

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If $H e$ $He$ (g) has an average kinetic energy of 5930 J/mol under certain conditions, what is the root mean square speed of $N_{2}$ $N_2$ (g) molecules under the same conditions?

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If HeHe (g) has an average kinetic energy of 5930 J/mol under certain conditions, what is the root mean square speed of N2N_2 (g) molecules under the same conditions?

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If $H e$ $He$ (g) has an average kinetic energy of 5930 J/mol under certain conditions, what is the root mean square speed of $N_{2}$ $N_2$ (g) molecules under the same conditions?