Analyzing the Setup
Imagine you are looking at two massive tanks of water, connected by a narrow pipe. The tank on the left has a slightly higher concentration of potassium permanganate (n1) than the tank on the right (n2). Because nature loves balance, the molecules will naturally want to flow from the crowded left tank to the spacious right tank. This process is called diffusion.
The problem gives us a beautiful simplification: we can treat these solute molecules as if they were a dilute ideal gas. This means they bounce around independently, and we can apply the laws of thermodynamics to them.
The Master Equation
Since we are treating the molecules as an ideal gas, we can use the ideal gas law to find the partial pressure they exert. The pressure
P is given by the number of molecules per unit volume multiplied by Boltzmann's constant and temperature:
P=nkBT
Because the left vessel has a higher concentration, it has a higher partial pressure. The pressure difference between the two vessels is the engine driving this whole process:
ΔP=P1−P2=(n1−n2)kBT=ΔnkBT
The Force of Diffusion
Pressure is simply force spread over an area. To find the total driving force pushing the molecules through the connecting tube, we multiply this pressure difference by the cross-sectional area
S of the tube:
Fdriving=ΔP⋅S=ΔnkBTS
This perfectly matches our first option. The pressure difference acts like an invisible hand, steadily pushing the molecules across the bridge.
The Viscous Resistance
But the molecules don't just zoom across instantly. As they move through the water, they experience a viscous drag—a fluid friction that slows them down. The problem tells us that each individual molecule feels a resistive force of βv, where v is their drift velocity.
To find the total resistance, we need to know how many molecules are currently inside the tube. The volume of the tube is its area times its length (S⋅ℓ). Since the concentration difference Δn is extremely small compared to n1, we can safely assume the concentration inside the tube is roughly n1.
Therefore, the total number of molecules in the tube is:
Ntube=n1Sℓ
Multiplying this by the drag on a single molecule gives us the total viscous force:
Fviscous=(n1Sℓ)⋅βv
The Force Balance
When the molecules reach a steady drift velocity, they are no longer accelerating. This means the invisible hand pushing them forward is perfectly balanced by the fluid friction pulling them back. Let's equate the two forces:
ΔnkBTS=n1Sℓβv
Notice how the area
S beautifully cancels out from both sides! Rearranging this gives us the force balance equation:
n1βvℓ=ΔnkBT
This confirms our second option. It's a profound statement: the macroscopic pressure difference is directly balanced by the microscopic drag forces.
The Final Calculation
Now, we want to find the actual rate of transfer—how many molecules cross the tube every second. In physics, this flux is calculated by multiplying the concentration, the velocity, and the area:
dtdN=n1⋅v⋅S
We already found the velocity
v from our force balance equation:
v=n1βℓΔnkBT
Let's substitute this velocity back into our rate equation:
dtdN=n1(n1βℓΔnkBT)S
The
n1 terms cancel out, leaving us with a clean, elegant expression for the transfer rate:
dtdN=(ℓΔn)(βkBT)S
This confirms our third option. Finally, what happens as time goes on? As molecules migrate from left to right, the left tank becomes less concentrated and the right tank becomes more concentrated. The difference Δn shrinks. Looking at our rate equation, if Δn decreases, the rate of transfer must also decrease. The flow slows down as the system approaches equilibrium, making the final option incorrect.
