A ring shaped tube contains two ideal gases with equal masses and molar masses M1 = 32 and M2 = 28 . The gases are separated by one fixed partition and another movable stopper S which can move freely without friction inside the ring. The angle α in degrees

The correct answer is 192.

Step-by-Step Explanation:

Let the ring-shaped tube have a uniform cross-sectional area A. The tube is divided into two sections by a fixed partition and a freely movable, frictionless stopper S. Let the volume of the first gas (with molar mass $M_1=32$) be $V_1$ and the volume of the second gas (with molar mass $M_2=28$) be $V_2$.

Since the stopper can move freely without friction, in equilibrium, the pressures of both gases on either side of the stopper must be equal:
$P_1=P_2=P$
Both gases are also in thermal equilibrium with the surroundings, so their temperatures are equal:
$T_1=T_2=T$

Using the ideal gas law ($PV=nRT$) for both gases, we can write:
$\frac{V_1}{V_2}=\frac{n_1}{n_2}$
where $n_1$ and $n_2$ are the number of moles of the two gases.

The number of moles of a gas is given by the ratio of its mass to its molar mass:
$n_1=\frac{m}{M_1}$
and
$n_2=\frac{m}{M_2}$
Since the two gases have equal masses ($m_1=m_2=m$), the ratio of their moles is:
$\frac{n_1}{n_2}=\frac{M_2}{M_1}=\frac{28}{32}=\frac{7}{8}$

Because the tube is circular with a uniform cross-section, the volume occupied by each gas is directly proportional to the central angle subtended by that gas sector. Let ${\alpha }_1$ and ${\alpha }_2$ be the angles in degrees for the two sectors, where:
${\alpha }_1+{\alpha }_2=360°$
Therefore, the ratio of the angles is equal to the ratio of the volumes:
$\frac{{\alpha }_1}{{\alpha }_2}=\frac{V_1}{V_2}=\frac{n_1}{n_2}=\frac{7}{8}$

Solving for the angle occupied by the second gas (${\alpha }_2=\alpha$):
$\alpha =360°\times \frac{8}{7+8}=360°\times \frac{8}{15}=24°\times 8=192°$

Thus, the angle $\alpha$ in degrees is 192.