Q. 265.0( 1 Vote )

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 aides 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 :

Given:


The walls of the cylindrical tube and the separator are made with


adiabatic material. The separator can be slid in the tube by


external mechanism.


An ideal gas of is injected in the two aides of at equal pressure.


It is now slid to a position where it divides tube in the ratio 1:3.


The initial volume of the two aides are equal let’s say V/2,


Where, the total volume of the tube is V.


Now say the, left part of tube has V/4 volume and the right side has 3V/4 volume so that the ratio between them is 1:3.


In adiabatic process, (K = non zero constant)


Where P is the pressure of the gas and V is the volume and =


For ideal gas,


Where P is the pressure, V is the volume, T is the temperature of the gas and R is the gas constant and n is the number of moles of the gas.


Putting this in the adiabatic process condition we get,


(K’ is a non-zero constant)


Therefore,







Again for the other part of the tube,






As initially the gases were at the same pressure and volume, the temperatures would be the same as well.


Therefore,


Therefore, ,


Therefore the ratio of the final temperatures will be


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