Question

A ring shaped tube contains two ideal gases with equal masses and molar masses $M_1=32$ and $M_2=28$. The gases are separates by one fixed partition and another movable stopper $S$ which can move freely without friction inside the ring. The angle $\alpha$ as shown in the figure is _________ degrees.

• A. $291$
• B. $219$
• C. $129$
• D. $192$

Answer 1

The correct option is D $192$

Pressure on both the sides will be equal

$\Rightarrow P_1=P_2$

$\Rightarrow \frac{n_{1}RT}{V_1}=\frac{n_{2}RT}{V_2}$

$\Rightarrow \frac{n_1}{V_1}=\frac{n_2}{V_2}$

We know, n=$\frac{m}{M}$

$\Rightarrow \frac{m}{M_1{V}_1}=\frac{m}{M_2{V}_2}$

$\Rightarrow \frac{V_2}{V_1}=\frac{M_2}{M_1}=\frac{32}{28}=\frac{8}{7}$

But, $\frac{V_2}{V_1}=\frac{\alpha }{360°-\alpha }$

Therefore,$\frac{8}{7}=\frac{\alpha }{360°-\alpha }$

$\Rightarrow 8(360°-\alpha ))=7(\alpha )$

$\Rightarrow 15\alpha =360°\times 8$

therefore, $\alpha =192$

Hence, the correct option is D.