Diffusion Rate and Force Balance of Potassium Permanganate Solution

To determine which of the statements are correct, we analyze the system step-by-step using the principles of ideal gas kinetic theory and fluid dynamics applied to dilute solutions.

1. Driving Force (Option A)

The system of $\text{KMnO}_4$ molecules behaves as a dilute ideal gas. The partial pressure of the molecules in a region with concentration $n$ at temperature $T$ is given by: $P = n k_B T$

The partial pressure difference between the two vessels across the connecting tube of cross-sectional area $S$ is: $\Delta P = P_1 - P_2 = (n_1 - n_2) k_B T = \Delta n k_B T$

The driving force $F_{\text{driving}}$ acting on the fluid within the tube due to this pressure difference is: $F_{\text{driving}} = \Delta P \cdot S = \Delta n k_B T S$

Thus, Option A is correct.

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2. Force Balance (Option B)

Since $\Delta n = n_1 - n_2 \ll n_1$, the average concentration of $\text{KMnO}_4$ molecules inside the tube can be approximated as $n \approx n_1$.

The volume of the connecting tube is $V_{\text{tube}} = S l$. Therefore, the total number of molecules inside the tube is: $N_{\text{tube}} \approx n_1 S l$

Each molecule experiences a viscous retarding force of magnitude $\beta v$, where $v$ is the drift speed. The total retarding force on all molecules in the tube is: $F_{\text{viscous}} = N_{\text{tube}} (\beta v) = n_1 S l \beta v$

At steady drift motion, the driving force balances the viscous retarding force: $F_{\text{driving}} = F_{\text{viscous}}$ $\Delta n k_B T S = n_1 S l \beta v$

Dividing both sides by $S$: $n_1 \beta v l = \Delta n k_B T$

Thus, Option B is correct.

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3. Rate of Molecule Transfer (Option C)

The drift velocity $v$ of the molecules can be expressed using the force balance relation from Option B: $v = \frac{\Delta n k_B T}{n_1 \beta l}$

The total rate of molecules transferring through the tube per second (flux) is given by: $\frac{dN}{dt} = n_1 S v$

Substituting the value of $v$: $\frac{dN}{dt} = n_1 S \left( \frac{\Delta n k_B T}{n_1 \beta l} \right) = \left( \frac{\Delta n}{l} \right) \left( \frac{k_B T}{\beta} \right) S$

Thus, Option C is correct.

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4. Time Dependence of the Rate (Option D)

As molecules diffuse from the left vessel to the right vessel, the concentration $n_1$ decreases while $n_2$ increases. Consequently, the concentration difference $\Delta n = n_1 - n_2$ decreases over time.

Since the rate of molecule transfer is directly proportional to $\Delta n$: $\frac{dN}{dt} \propto \Delta n$

The transfer rate must decrease exponentially over time as equilibrium is approached. Therefore, it does not remain constant.

Thus, Option D is incorrect.

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Conclusion

The correct options are A, B, and C.