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Question
Figure shows a cylindrical tube with adiabatic walls and fitted with an adiabatic separator. The separator can be slid into the tube by an external mechanism. An ideal gas (γ = 1.5) is injected in the two sides at equal pressures and temperatures. The separator remains in equilibrium at the middle. It is now slid to a position where it divides the tube in the ratio 1 : 3. Find the ratio of the temperatures in the two parts of the vessel.
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Solution
Given:
γ = 1.5
For an adiabatic process, TVγ−1 = constant.
So, T1 V1γ−1 = T2 V2γ−1
As it is an adiabatic process and all the other conditions are same, the above equation can be applied.
In the new position, the slid is dividing the tube in the ratio 3:1.
So, if the total volume is V, then one side will occupy a volume of `3/4 "V"`
and the other side will occupy `"V"/4`.
`So, "T"_1 xx ((3"v")/4)^(gamma -1) = "T"_2 xx ("v"/4)^ (gamma -1)`
`=> "T"_1 xx ((3"v")/4)^( 1.5 -1) = "T"_2 xx ("v"/4)^(1.5 -1)`
`"T"_1/"T"_2 = sqrt3/1`
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