A mixture of 2 moles of helium gas (atomic mass =4 amu ) and 1 mole of argon gas (atomic mass =40 amu ) is kept at 300 ~K in a container. The ratio of the rms speeds ( v_ rms ( helium ) v_ rms (argon) ) is

Correct answer: D. 3.16 Using V_ rms = (3 R T)/(M) V_ r m s 1 M v_ rms (helium) v_ rms (argon) = M_ argon M_ helium = (40)/(4) = 10 3.16

JEE Advanced 2012Paper 1

JEE Advanced 2012 · Paper 1 · Physics · Kinetic Theory of Gases

A mixture of 2 moles of helium gas (atomic mass =4 amu ) and 1 mole...

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)

D. $3.16$

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