A thermally isolated cylindrical closed vessel of height $$8$$ m is kept vertically. It is divided into two equal parts by a diathermic (perfect$$ thermal $$conductor) frictionless partition of mass $$8.3$$ kg. Thus the partition is held initially at a distance of $$4$$ m from the top, as shown in the schematic figure below. Each of the two parts of the vessel contains $$0.1$$ mole of an ideal gas at temperature $$300$$ K. The partition is now released and moves without any gas leaking from one part of the vessel to the other. When equilibrium is reached, the distance of the partition from the top (in$$ $$m) will be $$_______$$ (take$$ the acceleration due to gravity $$=10$$ ms−$$2$$ and the universal gas constant $$=$$ $$8.3$$ J mol−$$1K$$−$$1).

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Infinity Learn · Accepted Answer

Assuming temperature remains constant at $$300$$ KInitially, $$V02=A4In$$ lower chamber,  From $$P1V1=P2V2P1V02T=P1$$′$$V02$$−AxTIn upper chamber,  From $$P1V1=P2V2P2V02T=P2$$′$$V02+AxTFinally$$ during equilibrium, $$P1$$′−$$P2$$′$$A=mg$$⇒$$P1V02V02$$−Ax−$$P2V02V02+AxA=mg$$⇒$$nRT14$$−x−$$14+x=mg$$⇒$$(0.1)(8.3)3004+x$$−$$4+x16$$−$$x2=mg$$⇒$$32x16$$−$$x2=1$$⇒$$6x=16$$−$$x2$$⇒$$x2+6x$$−$$16=0$$⇒$$x=2$$ distance $$=4+2=6m$$

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