Q. 205.0( 1 Vote )

# The container sho

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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