Question

A vessel of volume \[5{\text{ }}litre\] contains \[1.4{\text{ }}g\] of nitrogen at a
temperature \[1800K\] . The pressure of the gas is \[30\% \] of its molecules are dissociated into
atoms at this temperature is:
(A) \[4.05atm\]
(B) $2.025atm$
(C) $3.84atm$
(D) $1.92atm$

Answer 1

Hint:To solve this we need the Ideal gas equation. The ideal gas law relates the four independent
physical properties of a gas at any time. These are pressure, volume, temperature, and the number of
moles of gas. Here, the Nitrogen gas dissociates into atoms so we need to calculate the number of
moles undissociated and the number of moles dissociated and then add them to find the total
number of moles. Later, we will calculate the pressure at \[1800K\] the temperature. We know that
the number of moles undissociated at equilibrium:
$No.\,of\,moles\, = \dfrac{{given\,mass}}{{molar\,mass}}$
We know the Ideal Gas Equation is $PV = nRT$

Complete step by step answer:
The equation for the dissociation of nitrogen gas is given as:
${N_2} \Leftrightarrow 2N$
Initial $100\%$
At ${\text{ }}eq.70\% ,30\%$
● As you can see initially there are zero atoms of nitrogen but after dissociation, there are 2
nitrogen atoms. So, now we will calculate

$No.\,\,of\,moles\,\,undissociated\,at\,equilibrium = \dfrac{{1.4 \times 70}}{{28 \times 100}}
= 0.035$
$No.\,of\,moles\,after\,dissociation = \dfrac{{1.4 \times 30}}{{28 \times 100}} = 0.015$
● Since there are $2$ Nitrogen atoms $No.\,of\,moles\,after\,dissociation = 2 \times 0.015 =
0.03$\[Total{\text{ }}No.{\text{ }}of{\text{ }}moles{\text{ }} = {\text{ }}0.035 + 0.03{\text{ }} =
{\text{ }}0.065\]
● Now, By using the Ideal Gas Equation we can calculate the pressure which is given as,
$P = \dfrac{{nRT}}{V}$
● Now, Put $n = 0.065,V = 5,R = 0.082\,and\,T = 1800$
$P = \dfrac{{0.065 + 0.082 + 1800}}{5} = 1.92atm$
● Hence, The pressure of the gas after \[30\% \] its molecules are dissociated is $1.92atm$
Therefore, Option (C) is correct.

Note:
While calculating the number of moles after dissociation always check what type of gas is
dissociated. For example, Nitrogen is a diatomic gas so it will dissociate into two atoms if it was
phosphorus and sulfur then the number of dissociated atoms would not be two because these are
polyatomic gases. The number of atoms dissociated would have been four and eight respectively.
Also, while putting the values in the ideal gas equation, the correct value of the Gas constant $(R)$
should be chosen as there are different values $(R)$ for the different scenarios.