Q. 205.0( 1 Vote )

The container sho

Answer :

Given:


V1= 2.0litre, V2=3.0litre




When the partition is removed the gases in both the partition mix without any loss of energy and acquire a common pressure P after mixing.


Total volume of gas after equilibrium =


From ideal gas equation, product of pressure p and volume V is



Where


n=no. of moles of gas


….(i)


….(ii)


Also, from kinetic theory of gases,


..(0)


…..(iii)


….(iv)


Where,


n=no. of moles of gas


E=internal kinetic energy of gas


Initially is present


Total initial energy



Using eqn.(iii) and (iv), we get



After removing the partition,


Total moles=


Total volume =


Using eqn-(0) we have final product of pressure p and volume V equal to




Common pressure p is given by



Putting the values of we get



Therefore, total final pressure after the partition is removed is 1.6atm.


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