Answer :

Let the initial pressure of the chamber A and B be Pa and Pb respectively.


Let the final pressure of the chamber A and B be P’a and P’b respectively.


Let the curved surface area of tube be A


Given:


Length of chamber A=20cm=0.2m


Length of chamber B=10cm=0.1m


Volume =areaheight


Initial volume of chamber A=Va=0.2A


Initial volume of chamber B=Vb=0.1A


Initial Temperature of chamber A=Ta=400K


Initial Temperature of chamber B=Tb=100K


For first (momentary) equilibrium, pressure of both chamber will be same.


Pa=Pb


Let the final temperature at equilibrium be T.


We know that ideal gas equation


PV=nRT


Where V= volume of gas


R=gas constant =8.3JK-1mol-1


T=temperature


n=number of moles of gas


P=pressure of gas.


Then,


For chamber A, number of moles, before and after final equilibrium will be same as no new gas has been added. So, applying ideal gas equation, before and after final equilibrium and equation nR, we get,




Similarly, for chamber B




At second equilibrium pressures on both sides will be same again.


P’a=P’b




Now Pa=Pb so,


V’b =2V’a ……(iii)


Volume of chamber A plus volume of chamber B will be equal to total volume. So,


V’b+V’a=V=0.3A


2V’a+V’a=0.3A (from (iii))


3V’a=0.3A


V’a=0.1A


Now we know that volume=lengtharea. So,


V’a=lA


Where l=length of chamber A after equilibrium


lA=0.1A


l=0.1m=10cm


length of chamber A after equilibrium is 10 cm.


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