Problem Statement
Solve the thermodynamics problem: Show that for an ideal gas with $f$ degrees of freedom: $C_v = fR/2$, $C_p = (f+2)R/2$, and $\gamma = (f+2)/f$. By the equipartition theorem, each quadratic degree of freedom has average energy $\frac{1}{2}k_BT$ per molecule. The molar internal energy: $$U = \nu\frac{f}{2}RT \implies C_v = \frac{1}{
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Capacitors store electric charge on conducting plates separated by an insulator (dielectric). The capacitance $C = Q/V$ depends on geometry and dielectric constant. Energy stored is $U = Q^2/(2C) = CV^2/2 = QV/2$. Series and parallel combinations follow rules opposite to resistors.
- $C = Q/V$ — definition of capacitance
- $U = \frac{1}{2}CV^2 = \frac{Q^2}{2C}$ — energy stored
- $C_{\text{parallel}} = C_1 + C_2$ — parallel combination
- $1/C_{\text{series}} = 1/C_1 + 1/C_2$ — series combination
- $C = \varepsilon_0\varepsilon_r A/d$ — parallel plate capacitor
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$U = \nu\frac{f}{2}RT \implies C_v = \frac{1}{
Given Information
- Temperatures, pressures, volumes, and process type as given
- Universal gas constant $R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}$
- $C_p$ and $C_v$ or $\gamma = C_p/C_v$ as applicable
Physical Concepts & Formulas
The First Law of Thermodynamics $\Delta U = Q – W$ is an energy balance: internal energy increases when heat flows in and decreases when the gas does work. For ideal gases, internal energy depends only on temperature: $\Delta U = nC_v \Delta T$. Different processes have different constraints: isothermal ($T = \text{const}$, $W = nRT\ln(V_f/V_i)$), adiabatic ($Q=0$, $PV^\gamma = \text{const}$), isobaric ($P = \text{const}$, $W = P\Delta V$), isochoric ($V = \text{const}$, $W = 0$). Carnot efficiency sets the upper bound for any heat engine: $\eta = 1 – T_C/T_H$.
- $\Delta U = Q – W$ — First Law
- $PV = nRT$ — Ideal Gas Law
- $W_{\text{isothermal}} = nRT\ln(V_f/V_i)$
- $PV^\gamma = \text{const}$ — adiabatic process
- $\eta_{\text{Carnot}} = 1 – T_C/T_H$ — maximum efficiency
Step-by-Step Solution
Step 1 — Identify the process (isothermal, adiabatic, isobaric, isochoric).
Step 2 — Write the appropriate work expression and compute $W$.
Step 3 — Find $\Delta U = nC_v\Delta T$.
Step 4 — Apply First Law: $Q = \Delta U + W$.
Worked Calculation
Substituting all values with units:
Carnot engine: $T_H = 600\,\text{K}$, $T_C = 300\,\text{K}$:
$$
$$
Answer
$$
Answer
$$\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.
Leave a Reply