Problem Statement
Solve the thermodynamics problem: Solve the thermodynamics problem: Show that for any substance: $C_p – C_v = -T\frac{(\partial p/\partial T)_V^2}{(\partial p/\partial V)_T}$. Start from $C_p – C_v = T(\partial p/\partial T)_V(\partial V/\partial T)_p$. Using the triple product rule: $(\partial V/\partial T)_p = -(\partial p/\partia
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Capacitors store electric charge on conducting plates separated by an insulator (dielectric). The capacitance $C = Q/V$ depends on geometry and dielectric constant. Energy stored is $U = Q^2/(2C) = CV^2/2 = QV/2$. Series and parallel combinations follow rules opposite to resistors.
- $C = Q/V$ — definition of capacitance
- $U = \frac{1}{2}CV^2 = \frac{Q^2}{2C}$ — energy stored
- $C_{\text{parallel}} = C_1 + C_2$ — parallel combination
- $1/C_{\text{series}} = 1/C_1 + 1/C_2$ — series combination
- $C = \varepsilon_0\varepsilon_r A/d$ — parallel plate capacitor
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
Full substitution shown in the steps above.
Answer
$$\boxed{\eta_{\text{Carnot}} = 1 – \dfrac{T_C}{T_H}}$$
Physical Interpretation
Capacitors store energy in the electric field between their plates. Doubling the voltage quadruples the stored energy — an important design constraint for high-voltage applications. Charge sharing between capacitors is a lossless process only in the ideal case; real circuits dissipate energy in connecting resistance.
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