Sucrose is allowed to diffuse along a 8.0-cm length of tubing filled with water. The tube is 6.1 cm2 in cross-sectional area. The diffusion coefficient is equal to 5.0 10-10 m2/s, and 8.0 10-14 kg is transported along the tube in 18 s. What is the difference in the concentration levels of sucrose at the two ends of the tube?
Answer in kg/m3
its impossible
Oh, I bet sucrose is having quite a ride inside that tubing filled with water! Alright, let's calculate the concentration difference.
First, we need to find the mass flux of sucrose. We can use Fick's first law of diffusion:
J = -D * A * (ΔC / Δx)
Where:
J = mass flux (kg/s)
D = diffusion coefficient (m2/s)
A = cross-sectional area (m2)
ΔC = concentration difference (kg/m3)
Δx = length of the tube (m)
Now, we know that a mass of 8.0 * 10-14 kg is transported along the tube in 18 s. So, we can calculate the mass flux:
J = (8.0 * 10-14 kg) / (18 s) = 4.4 * 10-15 kg/s
Next, we plug in the given values into Fick's first law of diffusion and solve for ΔC:
ΔC = (J * Δx) / (D * A)
ΔC = (4.4 * 10-15 kg/s * 0.08 m) / (5.0 * 10-10 m2/s * 6.1 * 10-4 m2)
Calculating this gives us:
ΔC = 0.011 kg/m3
So, the difference in the concentration levels of sucrose at the two ends of the tube is approximately 0.011 kg/m3. Let's hope the sucrose enjoys its journey and doesn't get too dizzy!
To find the difference in concentration levels of sucrose at the two ends of the tube, we can use Fick's Law of Diffusion:
J = -D * ∆C / ∆x
where:
J is the mass flow rate of sucrose per unit area (kg/(m²·s)),
D is the diffusion coefficient (m²/s),
∆C is the difference in concentration levels of sucrose (kg/m³),
and ∆x is the distance along the tube (m).
Given:
D = 5.0 × 10⁻¹⁰ m²/s,
∆x = 8.0 cm = 0.08 m,
M = 8.0 × 10⁻¹⁴ kg,
t = 18 s,
A = 6.1 cm² = 6.1 × 10⁻⁴ m².
First, let's calculate the mass flow rate of sucrose using the given values:
J = M / (A * t)
J = 8.0 × 10⁻¹⁴ kg / (6.1 × 10⁻⁴ m² * 18 s)
J ≈ 7.04 × 10⁻⁷ kg/(m²·s)
Now, let's rearrange Fick's Law to solve for ∆C:
∆C = - J * ∆x / D
∆C = - (7.04 × 10⁻⁷ kg/(m²·s)) * (0.08 m) / (5.0 × 10⁻¹⁰ m²/s)
∆C ≈ - 1.1264 × 10³ kg/m³
The difference in concentration levels of sucrose at the two ends of the tube is approximately 1.1264 × 10³ kg/m³.
To find the difference in concentration levels of sucrose at the two ends of the tube, we can use Fick's Law of Diffusion, which relates the rate of diffusion to the concentration gradient, the cross-sectional area, and the diffusion coefficient.
The formula for Fick's Law of Diffusion is:
Rate of diffusion = (Diffusion coefficient * Cross-sectional area * Concentration gradient) / Length
Rearranging the formula to solve for the concentration gradient, we get:
Concentration gradient = (Rate of diffusion * Length) / (Diffusion coefficient * Cross-sectional area)
Given that the rate of diffusion is 8.0 * 10^(-14) kg, the length is 8.0 cm (or 0.08 m), the diffusion coefficient is 5.0 * 10^(-10) m^2/s, and the cross-sectional area is 6.1 cm^2 (or 6.1 * 10^(-4) m^2), we can substitute these values into the formula:
Concentration gradient = (8.0 * 10^(-14) kg * 0.08 m) / (5.0 * 10^(-10) m^2/s * 6.1 * 10^(-4) m^2)
Simplifying the equation:
Concentration gradient = (6.4 * 10^(-15) kg*m) / (3.05 * 10^(-13) kg*m^2/s)
Concentration gradient ≈ 2.1 * 10^(-2) s^(-1)
To convert the concentration gradient from s^(-1) to kg/m^3, we need to multiply by the molar mass of sucrose (approximately 342 g/mol) and multiply by Avogadro's number (6.022 * 10^23 mol^(-1)):
Concentration gradient in kg/m^3 = (2.1 * 10^(-2) s^(-1) * 342 g/mol * 6.022 * 10^23 mol^(-1)) / (1000 g/kg)
Simplifying the equation:
Concentration gradient ≈ 428 kg/m^3
Therefore, the difference in concentration levels of sucrose at the two ends of the tube is approximately 428 kg/m^3.