University Physics III · unit UPIII-06
Temperature and Kinetic Theory
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Temperature, Equilibrium & Thermometry: infer an input with uncertainty
Temperature is the number shared by every body in mutual thermal equilibrium, made measurable by choosing a property that changes with hotness and fixing its scale at reproducible states; the kelvin takes that property to be the pressure of a dilute gas at constant volume. Use the supplied topic model Y_out = b*(1+t/273.15), where Y_out is p_B AT t (kPa). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. t is Bath temperature t, measured in °C. Keep a (Sample A pressure at 0 °C) = 100 kPa; b (Sample B pressure at 0 °C) = 60 kPa fixed. For this exercise use only -212.813 °C ≤ t ≤ 62.8125 °C 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 = 50.185338 kPa, with a bounded reading uncertainty ±3.5418 kPa. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0275625 °C, 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.
p_B AT t against t
Predictions from the supplied Temperature, Equilibrium & Thermometry 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): p_B AT t / kPa
Horizontal axis (x): t: Bath temperature t / °C
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Bath temperature t / °C | p_B AT t / kPa |
|---|---|
| -212.8125 | 13.253706754530473 |
| -143.90625 | 28.389621087314662 |
| -75 | 43.52553542009885 |
| -6.09375 | 58.66144975288303 |
| 62.8125 | 73.79736408566723 |
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(t) = 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(-212.813) = 13.2537 kPa and Y_out(62.8125) = 73.7974 kPa. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = -212.813 + (50.1853 − 13.2537)(62.8125 − -212.813)/(73.7974 − 13.2537) = -44.6812 °C. 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 t = -44.6812 °C; substituting back gives Y_out = 50.185338 kPa.
- The interpolation error in the input is |t_line − t| = 2.84217e-14 °C. This is within the requested numerical tolerance.
- Invert both limits Y_out = 46.643538 kPa and 53.727138 kPa. Sort the roots even if the response decreases: -60.8053 °C ≤ t ≤ -28.5572 °C.
- The lower and upper input deviations are 16.124 °C and 16.124 °C. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer t ≈ -44.6812 °C; compatible interval [-60.8053, -28.5572] °C. Endpoint interpolation gives -44.6812 °C.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Thermal Expansion & Thermal Stress: infer an input with uncertainty
The linear expansion coefficient α is the fractional change in length per kelvin, α = (1/L)(dL/dT), so a free member of initial length L₀ warmed by ΔT grows by ΔL ≈ αL₀ΔT. Use the supplied topic model Y_out = gap/(0.012*len), where Y_out is GAP CLOSES AT ΔT (K). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. len is Rod length L₀, measured in m. Keep dt (Temperature rise ΔT) = 60 K; gap (Clearance s) = 0.5 mm fixed. For this exercise use only 1.425 m ≤ len ≤ 4.575 m 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 = 12.450819 K, with a bounded reading uncertainty ±0.5015 K. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown len by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000315 m, 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.
GAP CLOSES AT ΔT against len
Predictions from the supplied Thermal Expansion & Thermal Stress 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): GAP CLOSES AT ΔT / K
Horizontal axis (x): len: Rod length L₀ / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| len: Rod length L₀ / m | GAP CLOSES AT ΔT / K |
|---|---|
| 1.425 | 29.239766081871345 |
| 2.2125 | 18.832391713747647 |
| 3 | 13.888888888888888 |
| 3.7874999999999996 | 11.001100110011002 |
| 4.574999999999999 | 9.107468123861569 |
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(len) = 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(1.425) = 29.2398 K and Y_out(4.575) = 9.10747 K. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives len_line = 1.425 + (12.4508 − 29.2398)(4.575 − 1.425)/(9.10747 − 29.2398) = 4.05188 m. 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 len = 3.3465 m; substituting back gives Y_out = 12.450819 K.
- The interpolation error in the input is |len_line − len| = 0.705383 m. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 11.949319 K and 12.952319 K. Sort the roots even if the response decreases: 3.21693 m ≤ len ≤ 3.48695 m.
- The lower and upper input deviations are 0.129573 m and 0.140449 m. 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 len ≈ 3.3465 m; compatible interval [3.21693, 3.48695] m. Endpoint interpolation gives 4.05188 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The ideal-gas equation of state: infer an input with uncertainty
The ideal-gas equation of state, pV = nRT, is the single relation among pressure, volume, absolute temperature and amount of substance that every sufficiently dilute gas obeys, with R = 8.314 J mol⁻¹ K⁻¹ the same for all of them. Use the supplied topic model Y_out = 60.22/v, where Y_out is NUMBER DENSITY N/V (×10²⁵ m⁻³). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. v is Volume V, measured in L. Keep t (Absolute temperature T) = 300 K fixed. For this exercise use only 7.125 L ≤ v ≤ 22.875 L 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 = 3.598984 ×10²⁵ m⁻³, with a bounded reading uncertainty ±0.14496 ×10²⁵ m⁻³. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown v by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.001575 L, 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.
NUMBER DENSITY N/V against v
Predictions from the supplied The ideal-gas equation of state 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): NUMBER DENSITY N/V / ×10²⁵ m⁻³
Horizontal axis (x): v: Volume V / L
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Volume V / L | NUMBER DENSITY N/V / ×10²⁵ m⁻³ |
|---|---|
| 7.125 | 8.451929824561404 |
| 11.0625 | 5.44361581920904 |
| 15 | 4.014666666666667 |
| 18.9375 | 3.17993399339934 |
| 22.875 | 2.632568306010929 |
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(v) = 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(7.125) = 8.45193 ×10²⁵ m⁻³ and Y_out(22.875) = 2.63257 ×10²⁵ m⁻³. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives v_line = 7.125 + (3.59898 − 8.45193)(22.875 − 7.125)/(2.63257 − 8.45193) = 20.2594 L. 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 v = 16.7325 L; substituting back gives Y_out = 3.598984 ×10²⁵ m⁻³.
- The interpolation error in the input is |v_line − v| = 3.52691 L. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.454024 ×10²⁵ m⁻³ and 3.743944 ×10²⁵ m⁻³. Sort the roots even if the response decreases: 16.0846 L ≤ v ≤ 17.4347 L.
- The lower and upper input deviations are 0.647858 L and 0.702237 L. 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 v ≈ 16.7325 L; compatible interval [16.0846, 17.4347] L. Endpoint interpolation gives 20.2594 L.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Kinetic model of gas pressure: infer an input with uncertainty
The kinetic model states that gas pressure is the momentum delivered to unit wall area per second by molecular collisions, which for point molecules in random motion comes to p = ⅓ n m ⟨v²⟩, with ⟨v²⟩ the mean of the squared speeds. Use the supplied topic model Y_out = exp(-0.5*(vp/sx)^2), where Y_out is COUNT AT PROBE ÷ AT vₓ = 0 (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. vp is Probe class vₓ, measured in m s⁻¹. Keep sx (Rms x-speed σₓ) = 300 m s⁻¹; n (Number density n) = 2.5 ×10²⁵ m⁻³ fixed. For this exercise use only 170 m s⁻¹ ≤ vp ≤ 1430 m s⁻¹ 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 = 0.0074893179, with a bounded reading uncertainty ±0.0011217. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown vp by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.126 m s⁻¹, 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.
COUNT AT PROBE ÷ AT vₓ = 0 against vp
Predictions from the supplied Kinetic model of gas pressure 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): COUNT AT PROBE ÷ AT vₓ = 0
Horizontal axis (x): vp: Probe class vₓ / m s⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| vp: Probe class vₓ / m s⁻¹ | COUNT AT PROBE ÷ AT vₓ = 0 |
|---|---|
| 170 | 0.851670507229441 |
| 485 | 0.27068335971794333 |
| 800 | 0.028565500784550377 |
| 1115 | 0.0010009501745566799 |
| 1430 | 0.000011645909499213329 |
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(vp) = 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(170) = 0.851671 and Y_out(1430) = 0.0000116459. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives vp_line = 170 + (0.00748932 − 0.851671)(1430 − 170)/(0.0000116459 − 0.851671) = 1418.94 m s⁻¹. 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 vp = 938.6 m s⁻¹; substituting back gives Y_out = 0.0074893179.
- The interpolation error in the input is |vp_line − vp| = 480.337 m s⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.0063676179 and 0.0086110179. Sort the roots even if the response decreases: 925.121 m s⁻¹ ≤ vp ≤ 954.031 m s⁻¹.
- The lower and upper input deviations are 13.4793 m s⁻¹ and 15.4311 m s⁻¹. 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 vp ≈ 938.6 m s⁻¹; compatible interval [925.121, 954.031] m s⁻¹. Endpoint interpolation gives 1418.94 m s⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.