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Intro to Thermodynamics
PHYS 2110 · Thermodynamics — Final Exam (practice)
8 answers, handwritten and photographed, graded against the rubric.
57/100
25 of 44 points · confidence: high
Took 1h 40m of the 2h allowed
An indicative grade from an AI examiner, not an official grade.
By topic
Weak areas
Ranked by points lost. Grading is study guidance, not an official grade.
Question 1 (a)
Correct · 4/4State the first law of thermodynamics for a closed system, defining each term.
Your answer
ΔU = Q − W. ΔU is the change in internal energy, Q is the heat added to the system and W is the work done by the system.
Correct statement with every term defined and the sign convention stated.
What you got right
- Energy conservation
- Sign convention for heat and work
Model answer
For a closed system, ΔU = Q − W, where ΔU is the change in internal energy, Q the heat transferred to the system and W the work done by the system.
Question 1 (b)
Correct · 6/6An ideal gas expands isothermally at 300 K from 2.0 L to 6.0 L (n = 0.50 mol). Calculate the work done by the gas.
Your answer
W = nRT ln(V2/V1) = 0.5 × 8.314 × 300 × ln(3) = 1370 J
Right method, correct substitution and a sensible final value with units.
What you got right
- Isothermal work W = nRT ln(V₂/V₁)
Model answer
W = nRT ln(V₂/V₁) = (0.50)(8.314)(300) ln(3) ≈ 1.37 kJ.
Question 2 (a)
Partly · 3/6State the second law of thermodynamics and explain why no heat engine operating in a cycle can be 100% efficient.
Your answer
Entropy always increases. So you can't turn all the heat into work because some heat has to go to the cold reservoir. That's why efficiency is less than 1.
The conclusion is right, but the law is stated loosely and the reason is asserted rather than argued.
What you got right
- Some heat must be rejected to a cold reservoir
What went wrong / missing concepts
- A formal statement of the second law (Kelvin–Planck)
- Why rejected heat follows from the entropy balance
How to improve
State the Kelvin–Planck form, then apply it to a cyclic engine: over a cycle ΔS of the engine is zero, so heat must leave to the cold reservoir.
Model answer
Kelvin–Planck: no cyclic device can take heat from a single reservoir and convert it entirely into work. Over one cycle the engine's entropy returns to its start, so the entropy it takes in with Q_h at T_h must leave with heat Q_c at T_c > 0. Hence Q_c > 0 and η = 1 − Q_c/Q_h < 1.
Question 2 (b)
Incorrect · 0/6Derive the maximum (Carnot) efficiency of an engine operating between reservoirs at T_h and T_c.
Your answer
No answer submitted.
No answer was submitted.
What went wrong / missing concepts
- Why rejected heat follows from the entropy balance
- Carnot efficiency η = 1 − T_c / T_h
How to improve
Start from ΔS_universe ≥ 0 for one cycle: Q_c/T_c ≥ Q_h/T_h, which gives η ≤ 1 − T_c/T_h, with equality for a reversible engine.
Model answer
Over a cycle, ΔS_universe = −Q_h/T_h + Q_c/T_c ≥ 0, so Q_c/Q_h ≥ T_c/T_h and η = 1 − Q_c/Q_h ≤ 1 − T_c/T_h. The bound is reached only by a reversible (Carnot) engine.
Question 3
Correct · 6/6A refrigerator removes 450 J of heat from its interior for every 150 J of work supplied. Find its coefficient of performance and the heat released to the room.
Your answer
COP = 450/150 = 3. Heat to the room = 450 + 150 = 600 J.
Both parts correct, with energy conservation applied to the heat released.
What you got right
- Refrigerator COP = Q_c / W
- Q_h = Q_c + W
Model answer
COP = Q_c/W = 450/150 = 3.0. By energy conservation Q_h = Q_c + W = 600 J.
Question 4 (a)
Partly · 4/6Calculate the entropy change when 0.20 kg of ice melts at 0 °C (L_f = 334 kJ/kg).
Your answer
ΔS = Q/T = (0.2 × 334000) / 0 = infinite?? Using 273: ΔS = 66800/273 = 245 J/K
The final value is right, but the working first divided by 0 °C — temperature has to be in kelvin from the start.
What you got right
- ΔS = Q / T for a reversible process at constant T
What went wrong / missing concepts
- Using absolute temperature (kelvin) in ΔS = Q/T
How to improve
Convert to kelvin before substituting — ΔS = Q/T only works with absolute temperature.
Model answer
Q = mL_f = (0.20)(334 000) = 66.8 kJ at T = 273 K, so ΔS = Q/T ≈ 245 J/K.
Question 4 (b)
Partly · 1/6Explain, in terms of microstates, why mixing two different ideal gases at the same temperature and pressure increases entropy.
Your answer
Because the gases spread out into each other so there's more disorder.
"More disorder" is the everyday phrase, but it doesn't earn points here — the question asks for microstates.
What went wrong / missing concepts
- Entropy as the number of accessible microstates (S = k ln Ω)
How to improve
Tie the spreading to something countable: after mixing, each molecule can be in the whole volume, so the number of accessible microstates Ω grows and S = k ln Ω increases.
Model answer
After mixing, each gas can occupy the whole volume, so the number of accessible microstates Ω increases. With S = k ln Ω, entropy rises (ΔS = nR ln 2 per gas for equal volumes).
Question 5
Partly · 1/4An ideal gas is compressed adiabatically to half its volume (γ = 1.4). By what factor does its temperature change?
Your answer
TV^(γ−1) is constant, so T2 = T1 × 2^0.4 = 1.32 T1
Right relation and factor, but the reasoning skips why the adiabatic relation applies and how it follows from pV^γ.
What went wrong / missing concepts
- Deriving TV^(γ−1) = const from pV^γ = const and pV = nRT
How to improve
Show the step: combine pV^γ = const with pV = nRT to get TV^(γ−1) = const, then substitute.
Model answer
From pV^γ = const and pV = nRT, TV^(γ−1) = const, so T₂/T₁ = (V₁/V₂)^(γ−1) = 2^0.4 ≈ 1.32.