University Physics III · topic UPIII-06.07
Collision rate, mean free path, and diffusion
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Collision Rate, Mean Free Path & Diffusion: predict a finite change
The mean free path λ is the average distance a molecule travels between successive collisions, λ = 1/(√2 n σ), where n is the number density and σ = πd² the collision cross-section for an effective diameter d. Use the supplied topic model Y_out = 3.1076*temp/(d^2*p), where Y_out is MEAN FREE PATH λ (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. p is Pressure p, measured in kPa. Keep temp (Temperature T) = 300 K; d (Effective diameter d) = 0.37 nm fixed. For this exercise use only 76.5625 kPa ≤ p ≤ 273.438 kPa on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at p = 119.875 kPa and 147.438 kPa. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δp across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
MEAN FREE PATH λ against p
Predictions from the supplied Collision Rate, Mean Free Path & Diffusion model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): MEAN FREE PATH λ / nm
Horizontal axis (x): p: Pressure p / kPa
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Pressure p / kPa | MEAN FREE PATH λ / nm |
|---|---|
| 76.5625 | 88.94608011210329 |
| 125.78125 | 54.14109224214983 |
| 175 | 38.913910049045185 |
| 224.21875 | 30.371832233401125 |
| 273.4375 | 24.90490243138892 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of p separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At p = 119.875 kPa, substitute into Y_out = 3.1076*temp/(d^2*p): Y₀ = 56.8086 nm.
- At p = 147.438 kPa, the same fixed-parameter model gives Y₁ = 46.1886 nm.
- Signed change: ΔY_out = Y₁ − Y₀ = -10.62 nm; the input change is Δp = 27.5625 kPa.
- Average sensitivity is ΔY_out/Δp = -0.385307 nm per kPa. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 56.8086 nm, Y₁ = 46.1886 nm; ΔY_out = -10.62 nm; average sensitivity = -0.385307 nm per kPa.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Collision Rate, Mean Free Path & Diffusion: infer an input with uncertainty
The mean free path λ is the average distance a molecule travels between successive collisions, λ = 1/(√2 n σ), where n is the number density and σ = πd² the collision cross-section for an effective diameter d. Use the supplied topic model Y_out = 3.1076*temp/(d^2*p), where Y_out is MEAN FREE PATH λ (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. p is Pressure p, measured in kPa. Keep temp (Temperature T) = 300 K; d (Effective diameter d) = 0.37 nm fixed. For this exercise use only 76.5625 kPa ≤ p ≤ 273.438 kPa on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 34.628619 nm, with a bounded reading uncertainty ±1.4586 nm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0196875 kPa, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
MEAN FREE PATH λ against p
Predictions from the supplied Collision Rate, Mean Free Path & Diffusion model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): MEAN FREE PATH λ / nm
Horizontal axis (x): p: Pressure p / kPa
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Pressure p / kPa | MEAN FREE PATH λ / nm |
|---|---|
| 76.5625 | 88.94608011210329 |
| 125.78125 | 54.14109224214983 |
| 175 | 38.913910049045185 |
| 224.21875 | 30.371832233401125 |
| 273.4375 | 24.90490243138892 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(p) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(76.5625) = 88.9461 nm and Y_out(273.438) = 24.9049 nm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives p_line = 76.5625 + (34.6286 − 88.9461)(273.438 − 76.5625)/(24.9049 − 88.9461) = 243.545 kPa. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives p = 196.656 kPa; substituting back gives Y_out = 34.628619 nm.
- The interpolation error in the input is |p_line − p| = 46.8887 kPa. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 33.170019 nm and 36.087219 nm. Sort the roots even if the response decreases: 188.708 kPa ≤ p ≤ 205.304 kPa.
- The lower and upper input deviations are 7.9486 kPa and 8.64765 kPa. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer p ≈ 196.656 kPa; compatible interval [188.708, 205.304] kPa. Endpoint interpolation gives 243.545 kPa.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.