Problem 2.32 — Heat Capacity of Mixture

Problem Statement

Solve the thermodynamics problem: Find $C_v$ and $C_p$ of a mixture of $\nu_1=1\ \text{mol}$ He and $\nu_2=2\ \text{mol}$ N₂. Total internal energy: $U = \nu_1\frac{3}{2}RT + \nu_2\frac{5}{2}RT = (1.5+5.0)RT = 6.5RT$. Total molar heat capacity at constant volume (per total mole $\nu=3$): $$C_v^{mix} = \frac{1}{\nu}\frac{dU}{dT} = \f

Given Information

• $U = \nu_1\frac{3}{2}RT + \nu_2\frac{5}{2}RT = (1.5+5.0)RT = 6.5RT$

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

$$C_v^{mix} = \frac{1}{\nu}\frac{dU}{dT} = \f

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

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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.