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

Options

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

Answer

A

Explanation

`N_(2)O_(4)(g)hArr 2NO_(2)(g)`
`{:("Initial",1"mol",,,),("At eqm.",(1-0.2)" mol",,,0.4"mol"):}`
No. of moles at eqm. `=1-0.2+0.4=1.2`
Initially `:`
`PV=Nrt`
`1xxV=1xx RT`
`implies V=RT` ,brgt At eqm.,
`PV=nRT`
`PV=1.2 RT`
`P=(1.2RT)/(V)=(1.2RT)/(RT)=1.2 "atm. "`