Q8 / 160 · TS EAMCET 2018 (04 May Shift 2) · Chemistry

The ratio between the most probable speed of $\mathrm{N}_2$ at $400 \mathrm{~K}$ and $\mathrm{CO}$ at $800 \mathrm{~K}$ is (molar mass of $\mathrm{N}_2=28 \mathrm{~g} \mathrm{~mol}^{-1}$, molar mass of $\mathrm{CO}=28 \mathrm{~g} \mathrm{~mol}^{-1}$ )

  • A
    0.75
  • B
    0.25
  • Correct answer
    C
    0.707
  • D
    1.414

Solution

Most probable speed $=\sqrt{\frac{2 R T}{M}}$ Molar mass of $\mathrm{N}_2=28 \mathrm{~g} \mathrm{~mol}^{-1}$ Molar mass of $\mathrm{CO}=28 \mathrm{~g} \mathrm{~mol}^{-1}$ $\mathrm{N}_2$ at $400 \mathrm{~K}$ $\mathrm{CO}$ at $800 \mathrm{~K}$ Ratio between the most probable speed of $$ \begin{aligned} & \begin{aligned} \mathrm{N}_2 \text { and } \mathrm{CO} & =\frac{\sqrt{\frac{2 R T_1}{M_2}}}{\frac{\sqrt{2 R T_2}}{M_2}}=\frac{\sqrt{\frac{2 R \times 400}{28}}}{\sqrt{\frac{2 R \times 800}{28}}} \\ & =\sqrt{\frac{400}{800}}=\sqrt{\frac{1}{2}}=\sqrt{0.5} \end{aligned} \\ & \text { Remaining ratio } r_{\left(\mathrm{N}_2\right)}: r_{(\mathrm{CO})}=1: 0.707 \end{aligned} $$

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

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