1. Use the ideal gas law initially

For the initial state, $P_1 V_1 = nRT_1$ with $T_1 = 250\,\text{K}.$

So, $P_1 V_1 = nR(250).$

2. Find the change in number of moles due to dissociation

Let the initial number of moles of diatomic gas be $n$.

A diatomic molecule dissociates as $A_2 \to 2A.$

Given that 25% of the molecules dissociate, so moles dissociated = $0.25n.$

• Undissociated diatomic molecules left: $n - 0.25n = 0.75n$

• Monatomic molecules formed: $2(0.25n)=0.50n$

Hence total final moles, $n_f = 0.75n + 0.50n = 1.25n = \frac{5n}{4}.$

3. Apply ideal gas law in the final state

Final temperature: $T_2 = 2000\,\text{K}$ Final volume: $V_2 = 2V_1$ Final moles: $n_f = \frac{5n}{4}$

Thus, $P_2(2V_1)=\frac{5n}{4}R(2000).$

So, $P_2 = \frac{\frac{5n}{4}R(2000)}{2V_1}.$

4. Form the ratio $\dfrac{P_2}{P_1}$

From the initial state, $P_1 = \frac{nR(250)}{V_1}.$

Therefore, \frac{P_2}{P_1}=rac{\frac{5n}{4}R(2000)}{2V_1}\cdot \frac{V_1}{nR(250)}.

Now simplify: $\frac{P_2}{P_1} = \frac{5}{4}\cdot \frac{2000}{2\cdot 250}$

Since $2\cdot 250 = 500,$ we get $\frac{2000}{500}=4.$

Hence, $\frac{P_2}{P_1} = \frac{5}{4}\cdot 4 = 5.$

5. Final answer

$\boxed{5}$

The derived answer matches the stored correct answer.