University Physics III · unit UPIII-07
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.
Thermodynamic systems and the first law: infer an input with uncertainty
The first law states that the internal energy of a closed system changes only by the heat that enters it minus the work it does on its surroundings, ΔU = Q − W, whatever path connects the two equilibrium end states. Use the supplied topic model Y_out = (q-w)/(1.5*n*8.314), where Y_out is ΔT, monatomic ideal gas (K). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. n is Argon in the system n, measured in mol. Keep q (Heat into the system Q) = 400 J; w (Work done by the system W) = 150 J fixed. For this exercise use only 0.978125 mol ≤ n ≤ 4.52187 mol 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 = 6.3846194 K, with a bounded reading uncertainty ±0.29271 K. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown n by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000354375 mol, 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.
ΔT, monatomic ideal gas against n
Predictions from the supplied Thermodynamic systems and 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): ΔT, monatomic ideal gas / K
Horizontal axis (x): n: Argon in the system n / mol
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| n: Argon in the system n / mol | ΔT, monatomic ideal gas / K |
|---|---|
| 0.9781249999999999 | 20.49483235611411 |
| 1.8640624999999997 | 10.754203734222491 |
| 2.75 | 7.289639235754222 |
| 3.6359375 | 5.513435777794341 |
| 4.521875 | 4.433229113658408 |
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(n) = 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(0.978125) = 20.4948 K and Y_out(4.52187) = 4.43323 K. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives n_line = 0.978125 + (6.38462 − 20.4948)(4.52187 − 0.978125)/(4.43323 − 20.4948) = 4.09133 mol. 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 n = 3.13981 mol; substituting back gives Y_out = 6.3846194 K.
- The interpolation error in the input is |n_line − n| = 0.951518 mol. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 6.0919094 K and 6.6773294 K. Sort the roots even if the response decreases: 3.00217 mol ≤ n ≤ 3.29068 mol.
- The lower and upper input deviations are 0.137638 mol and 0.150865 mol. 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 n ≈ 3.13981 mol; compatible interval [3.00217, 3.29068] mol. Endpoint interpolation gives 4.09133 mol.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Gas work, p-V paths, and path dependence: infer an input with uncertainty
The work done by a gas in a quasi-static process is W = ∫p dV along its path in the p-V plane: the signed area under the curve, positive for expansion, negative for compression, and different for different paths between the same two states. Use the supplied topic model Y_out = 100*p1*(1-1/v2), where Y_out is W ROUTE B (V₁ THEN p₂) (J). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. v2 is Final volume V₂ (V₁ = 1.0 L), measured in L. Keep p1 (Initial pressure p₁) = 2 × 10⁵ Pa; f (External pressure p_ext / p₂) = 0.6 fixed. For this exercise use only 1.76563 L ≤ v2 ≤ 3.73438 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 = 132.5819 J, with a bounded reading uncertainty ±2.0792 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 0.000196875 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.
W ROUTE B (V₁ THEN p₂) against v2
Predictions from the supplied Gas work, p-V paths, and path dependence 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): W ROUTE B (V₁ THEN p₂) / J
Horizontal axis (x): v2: Final volume V₂ (V₁ = 1.0 L) / 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 (V₁ = 1.0 L) / L | W ROUTE B (V₁ THEN p₂) / J |
|---|---|
| 1.765625 | 86.72566371681415 |
| 2.2578125 | 111.41868512110726 |
| 2.75 | 127.27272727272727 |
| 3.2421875 | 138.3132530120482 |
| 3.734375 | 146.44351464435147 |
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(1.76563) = 86.7257 J and Y_out(3.73438) = 146.444 J. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives v2_line = 1.76563 + (132.582 − 86.7257)(3.73438 − 1.76563)/(146.444 − 86.7257) = 3.27739 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 = 2.96656 L; substituting back gives Y_out = 132.5819 J.
- The interpolation error in the input is |v2_line − v2| = 0.310829 L. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 130.5027 J and 134.6611 J. Sort the roots even if the response decreases: 2.87781 L ≤ v2 ≤ 3.06096 L.
- The lower and upper input deviations are 0.0887528 L and 0.0944013 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 ≈ 2.96656 L; compatible interval [2.87781, 3.06096] L. Endpoint interpolation gives 3.27739 L.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Molar heat capacities and equipartition: infer an input with uncertainty
The molar heat capacity is the heat needed per mole per kelvin of temperature rise, C_V with the volume held fixed and C_p with the pressure held fixed; for an ideal gas C_p − C_V = R exactly. Use the supplied topic model Y_out = 8.314*(1.5+(2*tr/10^lt)^2*exp(-2*tr/10^lt)/max((1-exp(-2*tr/10^lt))^2,1e-9)+(tv/10^lt)^2*exp(-tv/10^lt)/max((1-exp(-tv/10^lt))^2,1e-9)), where Y_out is C_V (J K⁻¹ mol⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; max(a,b) selects the larger value. tv is Vibrational temperature θ_vib, measured in K. Keep lt (Temperature log₁₀(T / K)) = 2.5; tr (Rotational temperature θ_rot) = 3 K fixed. For this exercise use only 958.75 K ≤ tv ≤ 5841.25 K 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 = 20.789799 J K⁻¹ mol⁻¹, with a bounded reading uncertainty ±0.00075322 J K⁻¹ mol⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown tv by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.48825 K, 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.
C_V against tv
Predictions from the supplied Molar heat capacities and equipartition 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): C_V / J K⁻¹ mol⁻¹
Horizontal axis (x): tv: Vibrational temperature θ_vib / K
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| tv: Vibrational temperature θ_vib / K | C_V / J K⁻¹ mol⁻¹ |
|---|---|
| 958.75 | 24.853355093413025 |
| 2179.375 | 21.186810787571044 |
| 3400 | 20.80532669765694 |
| 4620.625 | 20.78555121386278 |
| 5841.25 | 20.784777542264607 |
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(tv) = 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(958.75) = 24.8534 J K⁻¹ mol⁻¹ and Y_out(5841.25) = 20.7848 J K⁻¹ mol⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives tv_line = 958.75 + (20.7898 − 24.8534)(5841.25 − 958.75)/(20.7848 − 24.8534) = 5835.22 K. 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 tv = 3937.06 K; substituting back gives Y_out = 20.789799 J K⁻¹ mol⁻¹.
- The interpolation error in the input is |tv_line − tv| = 1898.16 K. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 20.789046 J K⁻¹ mol⁻¹ and 20.790552 J K⁻¹ mol⁻¹. Sort the roots even if the response decreases: 3884.6 K ≤ tv ≤ 3997.85 K.
- The lower and upper input deviations are 52.4598 K and 60.7848 K. 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 tv ≈ 3937.06 K; compatible interval [3884.6, 3997.85] K. Endpoint interpolation gives 5835.22 K.
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 holds a single quantity fixed along the whole path — temperature, pressure, volume, or the heat transferred, which is zero — so that the first law fixes the remaining energy terms from the path alone. Use the supplied topic model Y_out = v2^(-gamma), where Y_out is p₂ ON THE ADIABAT (×10⁵ Pa). 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 gamma (Heat-capacity ratio γ) = 1.4 fixed. For this exercise use only 0.7125 L ≤ v2 ≤ 2.2875 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 = 0.48642382 ×10⁵ Pa, with a bounded reading uncertainty ±0.025867 ×10⁵ Pa. 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 0.0001575 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.
p₂ ON THE ADIABAT 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): p₂ ON THE ADIABAT / ×10⁵ Pa
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 | p₂ ON THE ADIABAT / ×10⁵ Pa |
|---|---|
| 0.7125 | 1.607315688903108 |
| 1.10625 | 0.8681712559718779 |
| 1.5 | 0.5668553336114626 |
| 1.8937499999999998 | 0.4090235331274138 |
| 2.2874999999999996 | 0.31397455686962167 |
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(0.7125) = 1.60732 ×10⁵ Pa and Y_out(2.2875) = 0.313975 ×10⁵ Pa. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives v2_line = 0.7125 + (0.486424 − 1.60732)(2.2875 − 0.7125)/(0.313975 − 1.60732) = 2.0775 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 = 1.67325 L; substituting back gives Y_out = 0.48642382 ×10⁵ Pa.
- The interpolation error in the input is |v2_line − v2| = 0.404245 L. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.46055682 ×10⁵ Pa and 0.51229082 ×10⁵ Pa. Sort the roots even if the response decreases: 1.61246 L ≤ v2 ≤ 1.73985 L.
- The lower and upper input deviations are 0.0607929 L and 0.0666007 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 ≈ 1.67325 L; compatible interval [1.61246, 1.73985] L. Endpoint interpolation gives 2.0775 L.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.