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Karnataka Board PUCPUC Science Class 11

Figure Shows a Cylindrical Tube with Adiabatic Walls and Fitted with an Adiabatic Separator.

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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.

Answer in Brief
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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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Interpretation of Temperature in Kinetic Theory - Introduction of Kinetic Theory of an Ideal Gas
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Chapter 27: Specific Heat Capacities of Gases - Exercises [Page 79]

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HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 27 Specific Heat Capacities of Gases
Exercises | Q 26 | Page 79

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