AP EAMCET · Chemistry · States of Matter
At 133.33 K . the RMS velocity of an ideal gas is \(\left(\mathrm{M} \equiv 0.083 \mathrm{~kg} \mathrm{~mol}^{-1}: \mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)\)
- A \(200 \mathrm{~ms}^{-1}\)
- B \(150 \mathrm{~ms}^{-1}\)
- C \(2000 \mathrm{~ms}^{-1}\)
- D \(400 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(A) \(200 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
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- \(v_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}\)…
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