At 133.33 K . the RMS velocity of an ideal gas is $\left(\mathrm{M} \equiv 0.083 \mathrm{~kg}…

At 133.33 K . the RMS velocity of an ideal gas is $\left(\mathrm{M} \equiv 0.083 \mathrm{~kg} \mathrm{~mol}^{-1}: \mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)$
  1. $200 \mathrm{~ms}^{-1}$
  2. $150 \mathrm{~ms}^{-1}$
  3. $2000 \mathrm{~ms}^{-1}$
  4. $400 \mathrm{~ms}^{-1}$

Solution

Given $\mathrm{T}=133.33 \mathrm{K}$ $\mathrm{M}=0.083 \mathrm{~kg} \mathrm{~mol}^{-1} \Rightarrow \mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol} \mathrm{~K}^{-1}$ To Find the RMS velocity of an ideal gas the formula is- $v_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ putting the value in the formula. $v_{\mathrm{rms}}=\frac{\sqrt{3 \times 8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 133.33 \mathrm{k}}}{0.083 \mathrm{~kg} / \mathrm{mol}}$ $v_{\mathrm{rms}}=\sqrt{\frac{3323.99}{0.083}} \Rightarrow v_{\mathrm{rms}}=\sqrt{40048.14 \mathrm{~J} \mathrm{~kg}^{-1}}$ $v_{\mathrm{rms}}=200.12 \mathrm{~m} / \mathrm{s}$

Asked in: AP EAMCET 2024 (20 May Shift 1)

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