Q7 / 160 · TS EAMCET 2018 (05 May Shift 2) · Chemistry

The ratio between the RMS velocity of $\mathrm{N}_2$ at $200 \mathrm{~K}$ and that of $\mathrm{CO}$ at $800 \mathrm{~K}$ is (molecular mass of $\mathrm{N}_2=28 \mathrm{~g} \mathrm{~mol}^{-1}$, molecular mass of $\mathrm{CO}=28 \mathrm{~g} \mathrm{~mol}^{-1}$ )

  • A
    1.00
  • B
    0.75
  • C
    0.25
  • Correct answer
    D
    0.50

Solution

RMS velocity $\left(v_{\mathrm{rms}}\right)=\sqrt{3 R T / M}$ $$ \begin{aligned} & \text { where, } R=\text { molar gas constant } \\ & T=\text { temperature in Kelvin } \\ & \begin{aligned} & M_m=\text { molar mass in kilograms } / \mathrm{mol} \\ & {\left[\because M_{\left(\mathrm{N}_2\right)}=M_{(\mathrm{CO})}\right] } \\ & v_{\mathrm{rms}}\left(\mathrm{N}_2\right)=\sqrt{3 \times 8.31 \times 200} \\ & v_{\mathrm{rms}}(\mathrm{CO})=\sqrt{3 \times 8.31 \times 800} \\ & \therefore \frac{v_{\mathrm{rms}}\left(\mathrm{N}_2\right)}{v_{\mathrm{rms}}(\mathrm{CO})}=\frac{\sqrt{200}}{\sqrt{800}} \\ &=\frac{14.14}{28.28}=0.50 \end{aligned} \end{aligned} $$

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

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