At 300K , the partial pressures of `N_2O_4`(g) and `NO_2`(g) in an equilibrium mixture of the reaction: `N_2O_4(g)

Explanation

`N_2O_4(g) `K_p=(p_(NO_2)^2)/(p_(N_2O_4))=((1.1)^2)/(0.28)=4.32atm`
If the volume is doubled, then the partial pressures of `NO_2` and `N_2O_4` will become half of their initial values.
Therefore, `p_(NO_2)=(1.1)/(2)=0.55` atm and `p_(N_2O_4)=(0.28)/2`
=0.14 atm
Let at new equilibrium partial pressure of `N_2O_4` is reduce by p.
`N_2O_4(g) At new equilibrium : `(0.14-p)` atm (0.55+2p)atm
`K_p=4.32=((0.55+2p)^2)/(0.14-p):.p=0.045atm`
So, at new equilibrium, `p_(N_2O_4)=(0.14-0.045)atm`
=0.19atm