1.

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 `400 K`, where `30%` by mass of `N_(2)O_(4)(g)` decomposes to `NO_(2)(g)`. The resultant pressure will beA. `4.2`B. `5.2`C. `3.2`D. `6.2`

Answer» b. Since volume is constant. Therefore on increasing the temperature four times, Pressure becomes four times
`:. P=4 "atm"`
Decomposition of `N_(2)O_(4) 30%` by man `alpha 30%` by mole
`:. a=30/100=0.3`
`{:(,N_(2)O_(4),hArr,2NO_(2)),("Initial",1,,0),("at. equilibrium",1-0.3,,2xx0.3),(,=0.7,,=0.6),("Total moles"=0.7+0.6=1.3,,,):}`
`("Initial mole")/("moles at equilibrium")=(("Initial pressure after"),("chabge of temperature"))/("Equilibrium pressure")`
`1/1.3=4/PrArr P=5.2 "atm"`
Alternate method:
`alpha=(T_(1)P_(2)-T_(2)P_(1))/(T_(2)P_(1))`
`0.3=(100xxP_(2)-400xx1)/(400xx1)`
`rArr P_(2)=5.2 "atm"`


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