Answer :

Given:

V_{1}= 2.0litre, V_{2}=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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