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 separated by one fixed partition P and another movable stopper S which can move freely without friction inside the ring. The angle $\alpha$ is

• A. ${182}^o$
• B. ${170}^o$
• C. ${192}^o$
• D. ${180}^o$

Answer 1

The correct option is C ${192}^o$

Pressure of one gas$=$Pressure of other gas.

$\Rightarrow \frac{n_{1}RT}{v_1}=\frac{n_{2}RT}{v_2}$

$\Rightarrow \frac{n_1}{v_1}=\frac{n_2}{v_2}$

$\Rightarrow \frac{w}{32}\times \frac{1}{\frac{\left(2\pi -\alpha \right)}{2\pi }\times 2\pi R\times A}$

$\Rightarrow \frac{w}{28}\times \frac{1}{\frac{\alpha }{2\pi }\times 2\pi R\times A}$

$\Rightarrow 32\left(2\pi -\alpha \right)=28\alpha$

$\Rightarrow 8\left(2\pi -\alpha \right)=7\alpha$

$\Rightarrow 2\pi \times 8=15\alpha$

$\Rightarrow \alpha =\frac{2\pi \times 8}{15}=\approx {192}^{\circ }$

hence, we have

$\Rightarrow \alpha ={192}^{\circ }$

therefore, option $C$ is correct.