Q6 / 160 · AP EAMCET 2018 (24 Apr Shift 1) · Chemistry

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)

  • A
    9.0
  • Correct answer
    B
    126.0
  • C
    90.0
  • D
    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 $=\frac{3}{2} R T$ $$ \therefore \mathrm{K} . \mathrm{E}=\frac{3}{2} \times \frac{28 \times 10^3 \times 3}{1000} \mathrm{~kJ}=126 \mathrm{~kJ} $$

Hence, the correct answer is option: 'B'.

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