One mole of `N_(2)O_(4)(g)` at 300 K is kept in a closed container under one atmosphere. It is heated to 600K when 20% by mass of `N_(2)O_(4)`(g) decomposes to `NO_(2)`(g). The resultant pressure is:

Options

A. 1.2 atm
B. 2.4 atm
C. 2.0 atm
D. 1.0 atm

Answer

B

Explanation

`{:(,N_(2)O_(4),rarr,2NO_(2)),("Initial","1 mol",,0),("At eqn.",underset("= 0.8 mol")((1-0.2)"mol"),,"0.4 mole"):}`
Total moles after dissociation, `n_(2)=0.8 + 0.4 =1.2`
Initial temperature = 300 K
`P_(1)V=n_(1)RT_(1) or 1 xx V = 1 xx R xx 300`....(i)
Temperature after dissociation = 600 K
No. of moles after dissociation, `n_(2)-1.2`
`P_(2) xx V = 1.2 xx R xx 600` ....(ii)
Dividing eqn (ii) by eqn (i)
`P_(2)=(1.2xx600)/(300)=2.4 "atm"`.