If the RMS speed of nitrogen at a certain temperature is $3000 \mathrm{~ms}^{-1}$, the approximate kinetic…
If the RMS speed of nitrogen at a certain temperature is $3000 \mathrm{~ms}^{-1}$, the approximate kinetic energy of one mole of nitrogen at that temperature in $\mathrm{kJ}$ is (assume nitrogen as ideal gas)
9.0
126.0
90.0
12.6
Solution
Given, $U_{\mathrm{rms}}=3000 \mathrm{~m} / \mathrm{s}$
$
\begin{aligned}
& \because U_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}} \\
& 9 \times 10^6 \mathrm{~m}^2 \mathrm{~s}^2=\frac{3 R T}{28 \mathrm{~g} / \mathrm{mol}} \\
& \quad\left[\because \text { Molar mass of } \mathrm{N}_2=28 \mathrm{~g} / \mathrm{mol}\right] \\
& \therefore R T=28 \times 3 \times 10^3 \mathrm{~J} / \mathrm{mol} \\
& \quad\left[\because \frac{\mathrm{m}^2}{\mathrm{~s}^2}=\mathrm{J} / \mathrm{kg}, 1 \mathrm{~kg}=1000 \mathrm{~g}\right]
\end{aligned}
$
As we know that,
K.E for 1 mole of an ideal gas $=\frac{3}{2} R T$
$
\therefore \mathrm{K} . \mathrm{E}=\frac{3}{2} \times \frac{28 \times 10^3 \times 3}{1000} \mathrm{~kJ}=126 \mathrm{~kJ}
$