JEE MainPhysicsToughKinetic Theory of Gases

A ring shaped tube contains two ideal gases with equal masses
and atomic mass numbers M_1=32 and M_2=28. The gases are separated by
one fixed partition X and another movable conducting partition Y which
can move freely without friction inside the ring. The angle is $\alpha$ ,
in equilibrium as shown in thefigure (in degrees). Find the value
of $\frac{\alpha}{48}$ .

Show answer and solution

At equilibrium, pressure and temperature are same.

\therefore P_{1} = P_{2}

and

T_{1} = T_{2}

Also, let cross-sectional area of tube be A and radius of ring be r.

\therefore\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}

\Rightarrow \frac{360 - \alpha}{n_{1}} = \frac{\alpha}{n_{2}}

\left( \therefore V_{1} = (360 - \alpha)rA \right.\

and

V_{2} = \alpha rA

\Rightarrow M_{1}(360 - \alpha) = M_{2}\alpha\lbrack\therefore n_{1} = \frac{m}{M_{1}}andn_{2} = \frac{m}{M_{2}}\rbrack

\alpha = \frac{360M_{1}}{M_{1} + M_{2}} = 192^{\circ}

Source: Grass-G / P-08 Physics paper 23 October.docx

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