A ring shaped tube contain two ideal gases with equal masses and relative molar masses `M_1=32 and M_2=28.`
The gases are separated by one fixed partiotin and another movable stopper S which can move freely without friction inside the ring. The angle `alpha` as shown in the figure is ...... degrees.

Options

A. `150^(@)`
B. `172^(@)`
C. `192^(@)`
D. `82^(@)`

Answer

C

Explanation

In the position of equilibrium, pressures on both sides of the partitions F and S will be equal.
i.e. `P_(1)=P_(2)" "thereforePV=nRT,P=(nRT)/V`
`therefore(n_(1)RT)/V_(1)=(n_(2)RT)/V_(2)` but `n = m/M` and m is same for both
`thereforem/(M_(1)V_(1))=m/(M_(2)V_(2))`
`thereforeV_(2)/V_(1)=M_(1)/M_(2)=32/28=8/7`
`thereforeV_(2)=8/7V_(1)" "thereforeV_(1)+8/7V_(1)=V`
`therefore` Total volume `V=15/7V_(1)`. V corresponds to `360^(@)`.
`therefore15/7V_(1)` will correspond to `360^(@)`.
`thereforeV_(1)` will correspond to `(360xx7)/15=168^(@)`
and `V_(2)` will correspond to 360 - 168 = `192^(@)`
`thereforealpha=192^(@)`