With that problem
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Final pressure of the gas remaining in the first container is 3.5 atm

Explanation:

Since it is given that temperature is constant , we can apply -

PV=constant , where

P = Pressure of the gas

V = Volume of the container in which the gas is contained

Initially,

For container 1 -

= 4 atm

= 6 L

Finally,

For container 2 -

= 3 atm

= 1 L

For container 1 -

= ?

= 6 L

∴ = +

∴ 4×6 =  (×6) +(3×1)

∴ = 3.5 atm

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