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**Hint:**There is a direct relationship between degree of dissociation for dissociation of type \[A \to B + C\]and total pressure at equilibrium; we are going to use the relation here. We will get two equations one for \[50\% \] dissociation and other for \[80\% \] dissociation. Degree of dissociation is in fraction so we need to convert the percentage into fraction. Dividing both the equations we will get the value of pressure.

**Formula used:**\[{{\text{K}}_{\text{p}}} = \dfrac{{{\alpha ^2} \times {{\text{P}}_{\text{T}}}}}{{(1 - {\alpha ^2})}}\] for reaction of type \[A \to B + C\]

Where, ${\alpha}_1$ is degree of dissociation, \[{{\text{K}}_P}\] is equilibrium constant in terms of pressure, \[{{\text{P}}_{\text{T}}}\] is total pressure at equilibrium.

**Complete step by step solution:**

α is the degree of dissociation, it is the ratio of dissociated moles to the total no of moles of the reactant. \[{{\text{K}}_P}\] is equilibrium constant which is a function of temperature, if temperature remains constant then it will not change. Hence for both the dissociation \[{{\text{K}}_P}\] will remain the same as temperature is constant. To convert α into fraction we need to divide it with 100.

So ${\alpha}_1$ will be \[\dfrac{{50}}{{100}} = 0.5\] and pressure is given to us that is \[{\text{4 atm}}\].

Putting the above value in formula:

\[{K_p} = \dfrac{{{{(0.5)}^2} \times 4}}{{(1 - {{(0.5)}^2})}} = \dfrac{1}{{0.75}}\]

Similarly ${\alpha}_1$ will be \[\dfrac{{80}}{{100}} = 0.8\]

\[{K_p} = \dfrac{{{{(0.8)}^2} \times {{\text{P}}_{\text{t}}}}}{{(1 - {{(0.8)}^2})}} = 1.777{{\text{P}}_{\text{t}}}\]

Dividing or equating both the equation we will get:

\[\dfrac{1}{{0.75}} = 1.7777{{\text{P}}_{\text{t}}}\]

Rearranging we will get the value of pressure:

\[{{\text{P}}_{\text{t}}}{\text{ = 0}}{\text{.75 atm}}\]

**Hence, option C is correct.**

**Note:**The dissociation of \[{\text{PC}}{{\text{l}}_5}\] follows the following reaction: \[{\text{PC}}{{\text{l}}_{\text{5}}}{\text{ }} \rightleftharpoons {\text{ PC}}{{\text{l}}_3}{\text{ + C}}{{\text{l}}_2}{\text{ }}\]. As we can see that the number of moles of both the product is same, this implies that pressure of these reactants will be same at equilibrium and pressure of \[{\text{PC}}{{\text{l}}_5}\] will be different.

\[{\text{a}}\] is initial moles of reactant and \[x\] is amount of reactant dissociated. In this case \[\alpha \] will be \[\dfrac{x}{{\text{a}}}\].

\[{\text{time PC}}{{\text{l}}_{\text{5}}}{\text{ }} \rightleftharpoons {\text{ PC}}{{\text{l}}_3}{\text{ + C}}{{\text{l}}_2}{\text{ }}\]

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