Assume that the decomposition of $HN{{O}_{3}}$ can be represented by the following equation:
$4HN{{O}_{3}}(g)\rightleftharpoons 4N{{O}_{2}}(g)+2{{H}_{2}}O(g)+{{O}_{2}}(g)$.
The reaction approaches equilibrium at 400 K temperature and 30 atm pressure. At equilibria, the partial pressure of $HN{{O}_{3}}$ is $2atm$.
The value of ${{K}_{c}}$ $(mol/{{L}^{3)}}$ at 400 K is: (Use: $R=0.08atm-L/mol-K$ )

Hint: The reaction involving gas molecules attain equilibrium at a certain temperature and pressure. The partial pressure as well as the concentration of the components can be used to express the equilibrium constant.

Complete step by step answer:
We are given, a decomposition reaction of nitric acid gas as follows:
$4HN{{O}_{3}}(g)\rightleftharpoons 4N{{O}_{2}}(g)+2{{H}_{2}}O(g)+{{O}_{2}}(g)$ -------- (a)
The above reaction reaches equilibrium at pressure 30 atm (that is the total partial pressure of the system) and at temperature of 400 K.

So, the equilibrium constant, ${{K}_{P}}$ of the reaction in terms of partial pressure is given as the partial pressure of the products to the partial pressure of the reactant. That is as follows:
${{K}_{P}}=\dfrac{{{\left( {{P}_{N{{O}_{2}}}} \right)}^{4}}{{\left( {{P}_{{{H}_{2}}O}} \right)}^{2}}{{\left( {{P}_{{{O}_{2}}}} \right)}^{1}}}{{{\left( {{P}_{HN{{O}_{3}}}} \right)}^{4}}}$ --------- (b)
where the power over the partial pressure term is the coefficient of the component in the balanced reaction.

It is given that, at the equilibrium, the partial pressure of the reactant $HN{{O}_{3}}$ is 2atm pressure. Then, from the reaction (a), we can obtain the relation between the partial pressures of the components, whose number of moles taking part in the reaction is equal to the coefficient.
So, we get, ${{P}_{N{{O}_{2}}}}=4{{P}_{{{O}_{2}}}}$ and ${{P}_{{{H}_{2}}O}}=2{{P}_{{{O}_{2}}}}$ --------- (c)
Thus, the total partial pressure of the system will be ${{P}_{total}}={{P}_{HN{{O}_{3}}}}+{{P}_{N{{O}_{2}}}}+{{P}_{{{O}_{2}}}}+{{P}_{{{H}_{2}}O}}=30\,atm$

Substituting the values of ${{P}_{N{{O}_{2}}}}$and ${{P}_{{{H}_{2}}O}}$from equation (c), we get,
${{P}_{total}}={{P}_{HN{{O}_{3}}}}+4{{P}_{{{O}_{2}}}}+{{P}_{{{O}_{2}}}}+2{{P}_{{{O}_{2}}}}={{P}_{HN{{O}_{3}}}}+7{{P}_{{{O}_{2}}}}$
Given the values ${{P}_{HN{{O}_{3}}}}=2\,atm$and ${{P}_{total}}=30\,atm$, we get the above equation as:
${{P}_{total}}={{P}_{HN{{O}_{3}}}}+7{{P}_{{{O}_{2}}}}=2+7{{P}_{{{O}_{2}}}}=30$
${{P}_{{{O}_{2}}}}=\dfrac{30-2}{7}=4\,atm$
Substituting this value of \[{{P}_{{{O}_{2}}}}=4\,atm\]in equation (c), we get,
${{P}_{N{{O}_{2}}}}=4{{P}_{{{O}_{2}}}}=16\,atm$and ${{P}_{{{H}_{2}}O}}=2{{P}_{{{O}_{2}}}}=4\,atm$.

Now, substituting these values of partial pressure in the equilibrium constant ${{K}_{P}}$, we get,
${{K}_{P}}=\dfrac{{{\left( {{P}_{N{{O}_{2}}}} \right)}^{4}}{{\left( {{P}_{{{H}_{2}}O}} \right)}^{2}}{{\left( {{P}_{{{O}_{2}}}} \right)}^{1}}}{{{\left( {{P}_{HN{{O}_{3}}}} \right)}^{4}}}$
${{K}_{P}}=\dfrac{{{\left( 16 \right)}^{4}}{{\left( 4 \right)}^{2}}{{\left( 4 \right)}^{1}}}{{{\left( 2 \right)}^{4}}}={{(16)}^{5}}={{\left( 2 \right)}^{20}}$
It is also known that the equilibrium constant of the reaction can be expressed in terms of the concentration of the components, given by ${{K}_{c}}$. The relation between the two equilibrium constants ${{K}_{P}}$ and ${{K}_{c}}$is as follows:
${{K}_{c}}=\dfrac{{{K}_{p}}}{{{\left( RT \right)}^{\Delta n}}}$
where R is the gas constant and $\Delta n$ is the change in the number of moles of the gas (that is, moles of product – moles of reactant).
Then, $\Delta n=\,(4+1+1-3)=3\,\text{moles}$

Thus, the equilibrium constant in terms of the concentration will be, ${{K}_{c}}=\dfrac{{{K}_{p}}}{{{\left( RT \right)}^{\Delta n}}}$
${{K}_{c}}=\dfrac{{{K}_{p}}}{{{\left( 0.08\times 400 \right)}^{3}}}=\dfrac{{{\left( 2 \right)}^{20}}}{{{\left( 32 \right)}^{3}}}=\,32$

Therefore, the equilibrium constant is equal to $32\,mol/{{L}^{3}}$ .

Note: It should be noted that, if in the reaction the change in the moles of the gas molecules is zero. Then, both the equilibrium constants are equal to each other, that is, ${{K}_{P}}={{K}_{c}}$.
Also, equilibrium is attained when the rate of forward reaction is equal to the rate of reverse reaction.