Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

One mole of N2O4(g) at 100K is kept in a closed container at 1.0atm pressure. It is heated to 400K ,where 30% by mass of N2O4(g) decomposes of NO2(g) . The resultant pressure will be:
(A) 4.2
(B) 5.2
(C) 3.2
(D) 6.2

Answer
VerifiedVerified
471k+ views
like imagedislike image
Hint: Any fluid is assumed to obey the gas laws at an equilibrium state during which the fluid and its properties are related as PV=nRT . Also, in this equilibrium state, when one of the properties of the fluid are considered to be constant, this derives a relation between the other properties.
When the volume of a fluid system is equilibrium is assumed to be constant, then the pressure is directly proportional to the temperature of the fluid.
 PT(Vconst)
Since the volume during the process is constant, the process is called isochoric.

Formulas used: We will be using the formula to equate the pressure and temperature at the initial stage and at equilibrium, P1T1=P2T2 where P1 is the pressure experience by the fluid at the initial stage, while T1 is the temperature of the liquid at the initial stage, P2 is the pressure experience by the fluid at equilibrium, T2 is the temperature of the fluid at equilibrium. We equate the ratio of pressure to the temperature at the initial and the stages after equilibrium because at constant volume, PT=k where k is a constant.

Complete Step by Step answer
We know that volume is constant since the dissociation happens in the same container. When the column is constant, we know that the pressure and temperature vary directly. Thus, we can find the change in pressure when the container is heated to 400K .
 P1T1=P2T2 where P1=1.0atm,T1=100K,T2=400K,P2=x
Now substituting the values, we have, we get,
 1.0100K=x400K
Solving for x we get,
 x=4.0atm
Now we have the pressure after the container heated, P2=4.0atm
We also know that the compound N2O4(g) is dissociated to NO2(g) which can be given by the chemical equation,

Initially the container contains 1 mole of N2O4(g) and 0 moles of NO2(g) .After heating the container to temperature T2 at pressure P2 we reach a state of equilibrium , when α moles of N2O4(g) dissociates from the reactant to form 2α moles of NO2(g) .
We also know that during heating N2O4 by 30% by mass, so the dissociative constant will be α=30100=0.3 .
Thus, at equilibrium there will be 1α=10.3=0.7 moles of N2O4 and 2α=2×0.3=0.6 moles of NO2 . Thus, at equilibrium the total number of moles in the container will be 0.7+0.6=1.3 .
Again, by rules of has law, we can equate the ratio of pressure to the number moles before and after equilibrium.
 P2n2=P2n2 where P2=x,n2=1.3,n2=1,P2=4atm
Substituting the values, we know and solving for P2 we get,
 P2=P2×n2n2
 P2=4×1.31
 P2=5.2atm
Thus, the resultant pressure will be 5.2atm .
Option B is correct.

Note
We can also find the resultant pressure by using the formula,
 α=T1P2T2P1T2P1
Now substituting the values that we have already found out, we get
 0.3=100P2400400
Solving for P2 ,
 P2=5.2atm .