A ring shaped tube contains two ideal gases with equal masses and atomic mass numbers \[{{M}_{1}}=32\] and \[{{M}_{2}}=28.\] The gases are separated by one fixed partition P and another movable conducting partition S which can move freely without friction inside the ring. The angle a as shown in the figure in equilibrium is - |
A) \[\frac{7\pi }{8}\]
B) \[\frac{8\pi }{7}\]
C) \[\frac{15\pi }{16}\]
D) \[\frac{16\pi }{15}\]
Correct Answer: D
Solution :
\[{{P}_{1}}={{P}_{2}}\] \[{{T}_{1}}={{T}_{2}}\] \[\Rightarrow \frac{{{V}_{1}}}{{{n}_{1}}}=\frac{{{V}_{2}}}{{{n}_{2}}}\] \[\Rightarrow \frac{2\pi -\alpha }{{{n}_{1}}}=\frac{\alpha }{{{n}_{2}}}\] \[\Rightarrow {{M}_{1}}(2\pi -\alpha )={{M}_{2}}\alpha \] or \[\alpha =\frac{2\pi {{M}_{1}}}{{{M}_{1}}+{{M}_{2}}}=\frac{16\pi }{15}\]You need to login to perform this action.
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