Question

One mole of $ {N_2}{O_4}\left( g \right) $ at $ 100K $ is kept in a closed container at $ 1.0{\text{atm}} $ pressure. It is heated to $ 400K $ ,where $ 30\% $ by mass of $ {N_2}{O_4}\left( g \right) $ decomposes of $ N{O_2}\left( g \right) $ . The resultant pressure will be:
(A) $ 4.2 $
(B) $ 5.2 $
(C) $ 3.2 $
(D) $ 6.2 $

Answer 1

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.
 $ P \propto T\left( {Vconst} \right) $
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, $ \dfrac{{{P_1}}}{{{T_1}}} = \dfrac{{{P_2}}}{{{T_2}}} $ where $ {P_1} $ is the pressure experience by the fluid at the initial stage, while $ {T_1} $ is the temperature of the liquid at the initial stage, $ {P_2} $ is the pressure experience by the fluid at equilibrium, $ {T_2} $ 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, $ \dfrac{P}{T} = 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 $ .
 $ \dfrac{{{P_1}}}{{{T_1}}} = \dfrac{{{P_2}}}{{{T_2}}} $ where $ {P_1} = 1.0{\text{atm}},{T_1} = 100K,{T_2} = 400K,{P_2} = x $
Now substituting the values, we have, we get,
 $ \dfrac{{1.0}}{{100K}} = \dfrac{x}{{400K}} $
Solving for $ x $ we get,
 $ x = 4.0{\text{atm}} $
Now we have the pressure after the container heated, $ {P_2} = 4.0{\text{atm}} $
We also know that the compound $ {N_2}{O_4}\left( g \right) $ is dissociated to $ N{O_2}\left( g \right) $ which can be given by the chemical equation,

Initially the container contains 1 mole of $ {N_2}{O_4}\left( g \right) $ and 0 moles of $ N{O_2}\left( g \right) $ .After heating the container to temperature $ {T_2} $ at pressure $ {P_2} $ we reach a state of equilibrium , when $ \alpha $ moles of $ {N_2}{O_4}\left( g \right) $ dissociates from the reactant to form $ 2\alpha $ moles of $ N{O_2}\left( g \right) $ .
We also know that during heating $ {N_2}{O_4} $ by $ 30\% $ by mass, so the dissociative constant will be $ \alpha = \dfrac{{30}}{{100}} = 0.3 $ .
Thus, at equilibrium there will be $ 1 - \alpha = 1 - 0.3 = 0.7 $ moles of $ {N_2}{O_4} $ and $ 2\alpha = 2 \times 0.3 = 0.6 $ moles of $ N{O_2} $ . 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.
 $ \dfrac{{{P_2}^\prime }}{{{n_2}^\prime }} = \dfrac{{{P_2}}}{{{n_2}}} $ where $ {P_2}^\prime = x,{n_2}^\prime = 1.3,{n_2} = 1,{P_2} = 4{\text{atm}} $
Substituting the values, we know and solving for $ {P_2}^\prime $ we get,
 $ {P_2}^\prime = \dfrac{{{P_2} \times {n_2}^\prime }}{{{n_2}}} $
 $ {P_2}^\prime = \dfrac{{4 \times 1.3}}{1} $
 $ \Rightarrow {P_2}^\prime = 5.2{\text{atm}} $
Thus, the resultant pressure will be $ 5.2{\text{atm}} $ .
Option B is correct.

Note
We can also find the resultant pressure by using the formula,
 $ \alpha = \dfrac{{{T_1}{P_2} - {T_2}{P_1}}}{{{T_2}{P_1}}} $
Now substituting the values that we have already found out, we get
 $ 0.3 = \dfrac{{100{P_2} - 400}}{{400}} $
Solving for $ {P_2} $ ,
 $ \Rightarrow {P_2} = 5.2{\text{atm}} $ .