A cylindrical container of volume 4 . 0 × 10 ^ - 3 m ^ 3 contains...

For gas A

${P}_{A}V = {n}_{A}RT$

${P}_{A} = \dfrac{( 1 ) RT}{V}$

for gas $B$

${P}_{B}V = {n}_{B}RT$

${P}_{B} = \dfrac{( 2 ) RT}{V}$

Now according to the Dalton's law of mixture,

$P = {P}_{A} + {P}_{B}$

$= \dfrac{3RT}{V} = \dfrac{3 \times 8 . 314 ( 400 )}{4 \times {10}^{- 3}} = 24 . 9 \times {10}^{5}Pa$