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JEE Advanced · Physics · 14. KTG

A mixture of 2 moles of helium gas (atomic mass \(=4 \mathrm{amu}\) ) and 1 mole of argon gas (atomic mass \(=40 \mathrm{amu}\) ) is kept at \(300 \mathrm{~K}\) in a container. The ratio of the rms speeds \(\left(\frac{v_{\mathrm{rms}}(\text { helium })}{v_{\mathrm{rms}}(\operatorname{argon})}\right)\) is

  1. A \(0.32\)
  2. B \(0.45\)
  3. C \(2.24\)
  4. D \(3.16\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(3.16\)

Step-by-step Solution

Detailed explanation

Using \(V_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}} \Rightarrow V_{r m s} \propto \frac{1}{\sqrt{M}}\) \(\frac{v_{\text {rms (helium) }}}{v_{\text {rms (argon) }}}=\sqrt{\frac{M_{\text {argon }}}{M_{\text {helium }}}}=\sqrt{\frac{40}{4}}=\sqrt{10} \approx 3.16\)
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