Question

A ring-shaped tube contains two ideal gases with equal masses and relative molar masses $M_{1}=32$ and $M_{2}=28$ . The gases are separated by one fixed partition and another movable stopper $S$ which can move freely without friction inside the ring. What is the value of the angle $\alpha $ (in degree) at equilibrium?

Answer 1

Pressure on both sides will be equal.
i.e., $p_{1}=p_{2}$
$\frac{\text{n}_{1} \text{RT}}{\text{V}_{1}} = \frac{\text{n}_{2} \text{RT}}{\text{V}_{2}}$ $\left(\text{n} = \frac{\text{m}}{\text{M}}\right)$
or $\frac{\text{m}}{\text{M}_{1} \text{V}_{1}} = \frac{\text{m}}{\text{M}_{2} \text{V}_{2}} \text{ or } \frac{\text{V}_{2}}{\text{V}_{1}} = \frac{\text{M}_{1}}{\text{M}_{2}} = \frac{3 2}{2 8} = \frac{8}{7}$
$\therefore $ $\alpha =\left(\frac{360 ^\circ }{8 + 7}\right)\times 8=192^\circ $