ii. One mole of `N_(2)O_(4)(g)` at `100 K` is kept in a closed container at `1.0` atm pressure. It is heated to `300 K`, where `30%` by mass of `N_(2)O_(4)(g)` decomposes to `NO_(2)(g)`. The resultant pressure will be

Options

A. `3.9 atm`
B. `1.95 atm`
C. `1.0 atm`
D. `3.0 atm`

Explanation

ii. a. First method:
`V` is constant (closed container)
On tripling the temperature rArr Pressure becomes three times
`((P_(1)V_(1))/T_(1)=(P_(2)V_(2))/T_(2)) rArr P=3 "atm"`
Decomposition of `N_(2)O_(4) 30%` by mass `prop 30%` by moles `("mass" prop "mole")`
`:. alpha=30/100=0.3`
`{:(,N_(2)O_(4),hArr,2NO_(2),("At pressure" =3 "atm")),("Intial",1,,0,),("At eq",1-0.3=0.7,,2xx0.3=0.6,),("Total moles"=0.7+0.6=1.3,,,,):}`
`("Intial mole")/("moles at equilibrium")= (("Intial pressure after"),("change of temperature"))/("Equilibrium pressure")`
`1/1.3=3/PrArr P=3.9 "atm"`
Second method:
`alpha=(T_(1)P_(2)-T_(2)P_(1))/(T_(2)P_(1))`
`0.3=(100xxP_(2)-300xx1)/(300xx1)`
`rArr P_(2)=3.9 "atm"`