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Figure shows a cylindridcal tube with a adibatic walls and fitted with an adiabatic separtor. The separator can be slid into the tube by an external mechanism. An ideal gas`(gamma=1.5)` is injected in the two sides at equal pressures and temperatures . The separator remains in equilibrium at the middel. It is now slid to a position where it divides the tube in the ratio 1:3 Find the ratio of the tempertures in the two parts of the vessel.

Text Solution

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Given `gamma=1.5`
We know for adiabatic process `TV^(gamma-1)`
`=constant`
So, `T_1 V_1^(gamma-1) = T_2 V_2^(gamma-1)` …(equation)
As, it is an adiabatic process and all the other conditions are same, hence the above equation can be applied.
So, `T_1 xx ((3v)/(4)) = T_2 xx ((v)/(4))`
So, `T_1:T_2 = 1:3`
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