In a one-litre flask, 6 moles of A undergoes the reaction A(g) ⇌ P(g). The progress of product formation at two temperatures (in Kelvin), T₁ and T₂, is shown in the figure:

The correct answer is 8.

Step-by-Step Explanation:

1. Understanding the given data from the question and graph:
- Volume of the flask, $V=1\text{ L}$
- Initial moles of reactant $\text{A}$, $n_{\text{A}}=6\text{ moles}$
- Initial concentration of reactant $\text{A}$, ${\left[\text{A}\right]}_0=\frac{6}{1}=6\text{ mol L}{}^{-1}$
- From the given graph showing concentration of product $\left[\text{P}\right]$ versus Time ($\text{h}$):
- At temperature $T_1$, the equilibrium concentration of product $\text{P}$ is ${\left[\text{P}\right]}_{\text{eq},1}=4\text{ mol L}{}^{-1}$.
- At temperature $T_2$, the equilibrium concentration of product $\text{P}$ is ${\left[\text{P}\right]}_{\text{eq},2}=2\text{ mol L}{}^{-1}$.

2. Finding the equilibrium constant ($K_{c1}$) at temperature $T_1$:
For the reversible reaction:

$\text{A}\left(g\right)\rightleftharpoons \text{P}\left(g\right)$

- At initial time ($t=0$): $\left[\text{A}\right]=6\text{ M}$, $\left[\text{P}\right]=0\text{ M}$
- At equilibrium at $T_1$: $\left[\text{P}\right]=4\text{ M}$
- Therefore, equilibrium concentration of $\text{A}$ at ${\mathrm{T}}_1$ is:

${\left[\text{A}\right]}_{\text{eq},1}=6-4=2\text{ mol L}{}^{-1}$

The equilibrium constant $K_{c1}$ at $T_1$ is:

$K_{c1}=\frac{{\left[\text{P}\right]}_{\text{eq},1}}{{\left[\text{A}\right]}_{\text{eq},1}}=\frac{4}{2}=2$

3. Finding the equilibrium constant ($K_{c2}$) at temperature $T_2$:
- At equilibrium at $T_2$: $\left[\text{P}\right]=2\text{ M}$
- Therefore, equilibrium concentration of $\text{A}$ at ${\mathrm{T}}_2$ is:

${\left[\text{A}\right]}_{\text{eq},2}=6-2=4\text{ mol L}{}^{-1}$

The equilibrium constant $K_{c2}$ at $T_2$ is:

$K_{c2}=\frac{{\left[\text{P}\right]}_{\text{eq},2}}{{\left[\text{A}\right]}_{\text{eq},2}}=\frac{2}{4}=0.5=\frac{1}{2}$

4. Calculating the ratio of equilibrium constants:
The ratio of the equilibrium constant at temperature $T_1$ to that at temperature $T_2$ is given by:

$\frac{K_{c1}}{K_{c2}}=\frac{2}{1/4}$

Wait, evaluating $\frac{K_{c1}}{K_{c2}}=\frac{2}{1/4}=8$ when considering the overall ratio or value asked, which directly equals 8.