Category: Part 2: Thermodynamics
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Problem 2.50 — Clausius Inequality
Problem Statement State and explain the Clausius inequality. How does it relate to the second law of thermodynamics? Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving…
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Problem 2.48 — Entropy Change: Mixing of Gases
Problem Statement Solve the thermodynamics problem: Find the entropy change when $\nu_1=1\ \text{mol}$ of N₂ and $\nu_2=1\ \text{mol}$ of O₂ (each at same $T$ and $p$) are mixed isothermally and isobarically. Entropy of mixing (Gibbs mixing entropy) for ideal gases: $$\Delta S_{mix} = -R\sum_i \nu_i\ln x_i$$ where $x_i = \nu_i/(\nu_1+\nu_2)$ is the m Given Information…
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Problem 2.49 — Entropy in Free Expansion
Problem Statement Solve the thermodynamics problem: One mole of ideal gas undergoes free (irreversible) expansion from $V_1$ to $V_2=2V_1$. Find $\Delta S$. In free expansion into a vacuum: $Q=0$, $W=0$, $\Delta U=0$ (ideal gas), so $T$ unchanged. But the process is irreversible — entropy still increases. To find $\Delta S$, use any reversible path b…
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Problem 2.46 — Entropy Change: Isothermal Expansion
Problem Statement Solve the thermodynamics problem: Find the entropy change when $\nu=1\ \text{mol}$ of ideal gas expands isothermally from $V_1=1.0\ \text{L}$ to $V_2=5.0\ \text{L}$. In a reversible isothermal expansion $dQ = T\,dS$, and all heat absorbed becomes work: $$\Delta S = \frac{Q}{T} = \frac{W}{T} = \frac{\nu RT\ln(V_2/V_1)}{T} = \nu R\ln\ Given Information See problem statement…
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Problem 2.47 — Entropy Change: Heating at Constant Pressure
Problem Statement Solve the thermodynamics problem: Find the entropy change of $\nu=1\ \text{mol}$ of diatomic gas heated isobarically from $T_1=300\ \text{K}$ to $T_2=600\ \text{K}$. $$\Delta S = \int_{T_1}^{T_2}\frac{\nu C_p\,dT}{T} = \nu C_p\ln\frac{T_2}{T_1}$$ For diatomic gas: $C_p = \frac{7}{2}R = 29.1\ \text{J/mol·K}$. $$\Delta S = 1.0\times29 Given Information See problem statement for all given quantities. Physical…
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Problem 2.44 — Efficiency of Otto Cycle
Problem Statement Find the efficiency of an ideal Otto cycle (petrol engine) with compression ratio $r = V_1/V_2 = 8$. Assume $\gamma=1.4$. Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations…
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Problem 2.45 — Diesel Cycle Efficiency
Problem Statement Find the efficiency of an ideal Diesel cycle with compression ratio $r=16$ and cutoff ratio $\alpha=2.0$. Assume $\gamma=1.4$. Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion,…
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Problem 2.43 — Refrigerator: Coefficient of Performance
Problem Statement A Carnot refrigerator operates between $T_1=250\ \text{K}$ (cold) and $T_2=300\ \text{K}$ (hot). Find the COP. Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically…
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Problem 2.41 — Thermal Efficiency of Direct Carnot Cycle
Problem Statement Solve the thermodynamics problem: A Carnot engine operates between $T_1=500\ \text{K}$ (hot) and $T_2=300\ \text{K}$ (cold). Find the efficiency. The Carnot efficiency is the maximum possible efficiency for any heat engine operating between two temperatures: $$\eta = 1 – \frac{T_2}{T_1} = 1 – \frac{300}{500} = 1 – 0.6 = 0.4 = 40\%$$ Given…
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Problem 2.42 — Carnot Cycle: Work and Heat
Problem Statement Solve the thermodynamics problem: A Carnot engine absorbs $Q_1=1000\ \text{J}$ per cycle from a source at $T_1=400\ \text{K}$ and rejects heat to a sink at $T_2=300\ \text{K}$. Find $W$, $Q_2$, and $\eta$. $$\eta = 1-\frac{T_2}{T_1} = 1-\frac{300}{400} = 0.25 = 25\%$$ $$W = \eta Q_1 = 0.25\times1000 = 250\ \text{J}$$ $$Q_2 = Q_1…