We consider tobacco virus diffusing along a 0.47 m long tube filled with water. The cross-sectional area of the tube is unknown. The diffusion coefficient is found in Table 10.5, p. 319 in the textbook. A total amount of virus of 2.6 x 10-16 kg is transported in a steady state along the tube in 0.20 minutes. If the difference in the density levels of virus at the two ends of the tube is 0.07 g/cm3, what is the cross-sectional area of the tube?
I don't have your textbook. You should look for the diffusion coefficient yourself, since your problem is URGENT. Call it D. It should have dimensions of something like cm^2/s.
The transport rate of the virus will be
D*(density2 - density1)*Area/L = 2.17*10^-13 g/sec
Solve for the area A.
To find the cross-sectional area of the tube, we can use Fick's Law of Diffusion, which states that the rate of diffusion is proportional to the cross-sectional area and the concentration gradient.
First, let's convert the total amount of virus transported into kilograms:
2.6 x 10^(-16) kg
Next, let's convert the units of the density difference:
0.07 g/cm^3 = 70 kg/m^3
Now, let's calculate the rate of diffusion:
Rate of diffusion = Total amount of virus / Time = (2.6 x 10^(-16) kg) / (0.20 minutes)
To calculate the diffusion coefficient (D), you should refer to Table 10.5 on page 319 of the textbook mentioned.
Once you have the diffusion coefficient, you can use the formula for the rate of diffusion:
Rate of diffusion = (D * A * ΔC) / L
Where:
D = Diffusion coefficient
A = Cross-sectional area of the tube
ΔC = Concentration difference (C2 - C1)
L = Length of the tube
We have the following values:
Rate of diffusion = (D * A * ΔC) / L
Total amount of virus = 2.6 x 10^(-16) kg
Time = 0.20 minutes
ΔC = 70 kg/m^3
L = 0.47 m
Rearranging the formula, we get:
A = (Rate of diffusion * L) / (D * ΔC)
Now, plug in the given values to calculate the cross-sectional area of the tube.