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NCERT Exemplar · Q17

Q.Two molecules of a gas have speeds of 9×106 m s−19 \times 10^{6}\ \text{m s}^{-1} and 1×106 m s−11 \times 10^{6}\ \text{m s}^{-1}, respectively. What is the root mean square speed of these molecules.

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The root mean square speed is the square root of the mean of the squares of the speeds. For these two molecules, it is 41×106 m s−1\sqrt{41} \times 10^{6}\ \text{m s}^{-1}.

The root mean square speed is a statistical measure that tells us the "typical" speed of molecules in a gas, weighted by their kinetic energy. Unlike the simple average speed, the RMS speed gives more weight to faster molecules because kinetic energy depends on the square of speed. This is why the RMS speed is always greater than or equal to the average speed.

For a set of NN molecules with speeds v1,v2,…,vNv_1, v_2, \dots, v_N, the RMS speed is defined as:

vrms=v12+v22+⋯+vN2Nv_{\text{rms}} = \sqrt{\frac{v_1^2 + v_2^2 + \dots + v_N^2}{N}}

  1. Square each speed. The first molecule has speed v1=9×106 m s−1v_1 = 9 \times 10^{6}\ \text{m s}^{-1}, so its square is:

v12=(9×106)2=81×1012 m2s−2v_1^2 = (9 \times 10^{6})^2 = 81 \times 10^{12}\ \text{m}^2\text{s}^{-2}

The second molecule has speed v2=1×106 m s−1v_2 = 1 \times 10^{6}\ \text{m s}^{-1}, so its square is:

v22=(1×106)2=1×1012 m2s−2v_2^2 = (1 \times 10^{6})^2 = 1 \times 10^{12}\ \text{m}^2\text{s}^{-2}

  1. Add the squares.

v12+v22=81×1012+1×1012=82×1012 m2s−2v_1^2 + v_2^2 = 81 \times 10^{12} + 1 \times 10^{12} = 82 \times 10^{12}\ \text{m}^2\text{s}^{-2}

  1. Divide by the number of molecules. There are N=2N = 2 molecules, so the mean of the squares is:

v12+v222=82×10122=41×1012 m2s−2\frac{v_1^2 + v_2^2}{2} = \frac{82 \times 10^{12}}{2} = 41 \times 10^{12}\ \text{m}^2\text{s}^{-2}

  1. Take the square root. …

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