University Physics II · unit UPII-E3
Institutional Extension · The Laws of Thermodynamics
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.
Internal Energy & the First Law: infer an input with uncertainty
Internal energy is the total microscopic kinetic and potential energy a system holds in its own rest frame, and the first law says its change equals the heat added minus the work the system does: ΔU = Q − W. Use the supplied topic model Y_out = q-w, where Y_out is STORED ΔU = Q − W (J). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. q is Heat into the system Q, measured in J. Keep w (Work done by the system W) = 250 J fixed. For this exercise use only -708.75 J ≤ q ≤ 708.75 J 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 = -94.075 J, with a bounded reading uncertainty ±82.924 J. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown q by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.14175 J, 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.
STORED ΔU = Q − W against q
Predictions from the supplied Internal Energy & the First Law 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): STORED ΔU = Q − W / J
Horizontal axis (x): q: Heat into the system Q / J
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q: Heat into the system Q / J | STORED ΔU = Q − W / J |
|---|---|
| -708.75 | -958.75 |
| -354.37500000000006 | -604.375 |
| -1.1368683772161603e-13 | -250.0000000000001 |
| 354.3749999999998 | 104.37499999999977 |
| 708.7499999999998 | 458.7499999999998 |
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(q) = 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(-708.75) = -958.75 J and Y_out(708.75) = 458.75 J. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives q_line = -708.75 + (-94.075 − -958.75)(708.75 − -708.75)/(458.75 − -958.75) = 155.925 J. 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 q = 155.925 J; substituting back gives Y_out = -94.075 J.
- The interpolation error in the input is |q_line − q| = 5.68434e-14 J. This is within the requested numerical tolerance.
- Invert both limits Y_out = -176.999 J and -11.151 J. Sort the roots even if the response decreases: 73.001 J ≤ q ≤ 238.849 J.
- The lower and upper input deviations are 82.924 J and 82.924 J. 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 q ≈ 155.925 J; compatible interval [73.001, 238.849] J. Endpoint interpolation gives 155.925 J.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Work by gases and p-V diagrams: infer an input with uncertainty
The work done by a gas is the integral of its own pressure over its change in volume, W = ∫ p dV, which on a pressure-volume diagram is the signed area between the path and the volume axis. Use the supplied topic model Y_out = (v2-2)*(s*300+(1-s)*p2), where Y_out is WORK BY THE GAS (J). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. v2 is Final volume V₂, measured in L. Keep p2 (Final pressure p₂) = 100 kPa; s (Corner fraction s) = 0.5 fixed. For this exercise use only 1.74375 L ≤ v2 ≤ 7.25625 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 = 621.275 J, with a bounded reading uncertainty ±64.496 J. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown v2 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00055125 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.
WORK BY THE GAS against v2
Predictions from the supplied Work by gases and p-V diagrams 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): WORK BY THE GAS / J
Horizontal axis (x): v2: Final volume V₂ / L
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v2: Final volume V₂ / L | WORK BY THE GAS / J |
|---|---|
| 1.74375 | -51.250000000000014 |
| 3.1218749999999997 | 224.37499999999994 |
| 4.5 | 500 |
| 5.878124999999999 | 775.6249999999998 |
| 7.25625 | 1051.25 |
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(v2) = 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(1.74375) = -51.25 J and Y_out(7.25625) = 1051.25 J. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives v2_line = 1.74375 + (621.275 − -51.25)(7.25625 − 1.74375)/(1051.25 − -51.25) = 5.10637 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 v2 = 5.10637 L; substituting back gives Y_out = 621.275 J.
- The interpolation error in the input is |v2_line − v2| = 8.88178e-16 L. This is within the requested numerical tolerance.
- Invert both limits Y_out = 556.779 J and 685.771 J. Sort the roots even if the response decreases: 4.78389 L ≤ v2 ≤ 5.42885 L.
- The lower and upper input deviations are 0.32248 L and 0.32248 L. 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 v2 ≈ 5.10637 L; compatible interval [4.78389, 5.42885] L. Endpoint interpolation gives 5.10637 L.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Isothermal, isobaric, isochoric and adiabatic processes: infer an input with uncertainty
A named gas process is a quasi-static path on which one thing is held fixed — the volume, the pressure, the temperature, or the heat flow at zero — and that single constraint fixes both the shape of the p-V curve and the split of energy between heat and work. Use the supplied topic model Y_out = 1/(v2^gamma), where Y_out is ADIABAT p₂/p₁ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. v2 is Final volume V₂ / V₁, measured in × V₁. Keep gamma (Adiabatic index γ = C_p/C_V) = 1.4 fixed. For this exercise use only 1.87187 × V₁ ≤ v2 ≤ 4.62812 × V₁ 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 = 0.16948631, with a bounded reading uncertainty ±0.007863. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown v2 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000275625 × V₁, 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.
ADIABAT p₂/p₁ against v2
Predictions from the supplied Isothermal, isobaric, isochoric and adiabatic processes 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): ADIABAT p₂/p₁
Horizontal axis (x): v2: Final volume V₂ / V₁ / × V₁
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v2: Final volume V₂ / V₁ / × V₁ | ADIABAT p₂/p₁ |
|---|---|
| 1.871875 | 0.415731008000429 |
| 2.5609374999999996 | 0.2680658114927191 |
| 3.25 | 0.1920274410793859 |
| 3.9390624999999995 | 0.14670671060248142 |
| 4.628125 | 0.11706656833371051 |
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(v2) = 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(1.87187) = 0.415731 and Y_out(4.62812) = 0.117067. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives v2_line = 1.87187 + (0.169486 − 0.415731)(4.62812 − 1.87187)/(0.117067 − 0.415731) = 4.14436 × V₁. 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 v2 = 3.55319 × V₁; substituting back gives Y_out = 0.16948631.
- The interpolation error in the input is |v2_line − v2| = 0.591177 × V₁. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.16162331 and 0.17734931. Sort the roots even if the response decreases: 3.43994 × V₁ ≤ v2 ≤ 3.67582 × V₁.
- The lower and upper input deviations are 0.113252 × V₁ and 0.122633 × V₁. 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 v2 ≈ 3.55319 × V₁; compatible interval [3.43994, 3.67582] × V₁. Endpoint interpolation gives 4.14436 × V₁.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Heat Engines & Thermal Efficiency: infer an input with uncertainty
A heat engine is a device that runs a working substance around a closed cycle, absorbing heat Q_H from a hot reservoir, delivering net work W, and rejecting the rest to a cold reservoir; its thermal efficiency is the ratio e = W/Q_H. Use the supplied topic model Y_out = 100*(p2-100)*(v2-1)/(1.5*(p2-100)+2.5*p2*(v2-1)), where Y_out is EFFICIENCY e (%). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. v2 is Maximum volume V₂, measured in L. Keep p2 (High pressure p₂) = 300 kPa fixed. For this exercise use only 2.31875 L ≤ v2 ≤ 4.68125 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 = 23.291006 %, with a bounded reading uncertainty ±0.11431 %. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown v2 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00023625 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.
EFFICIENCY e against v2
Predictions from the supplied Heat Engines & Thermal Efficiency 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): EFFICIENCY e / %
Horizontal axis (x): v2: Maximum volume V₂ / L
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v2: Maximum volume V₂ / L | EFFICIENCY e / % |
|---|---|
| 2.31875 | 20.46060606060606 |
| 2.9093750000000003 | 22.047812359043753 |
| 3.5 | 22.988505747126435 |
| 4.090625 | 23.610862429125635 |
| 4.68125 | 24.053088310362426 |
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(v2) = 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(2.31875) = 20.4606 % and Y_out(4.68125) = 24.0531 %. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives v2_line = 2.31875 + (23.291 − 20.4606)(4.68125 − 2.31875)/(24.0531 − 20.4606) = 4.18009 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 v2 = 3.75988 L; substituting back gives Y_out = 23.291006 %.
- The interpolation error in the input is |v2_line − v2| = 0.420212 L. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 23.176696 % and 23.405316 %. Sort the roots even if the response decreases: 3.65638 L ≤ v2 ≤ 3.87063 L.
- The lower and upper input deviations are 0.103498 L and 0.110753 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 v2 ≈ 3.75988 L; compatible interval [3.65638, 3.87063] L. Endpoint interpolation gives 4.18009 L.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.