Problem Statement
Solve the thermodynamics problem: Derive the Dulong-Petit law for the molar heat capacity of a solid and state its limitation. In a solid, each atom oscillates in three dimensions. Each oscillation has 2 degrees of freedom (kinetic + potential), giving $f=6$ per atom. By the equipartition theorem: $$U = \nu\times6\times\frac{1}{2}RT
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Thermodynamics governs energy transformations involving heat and work. The First Law $\Delta U = Q – W$ expresses energy conservation. For an ideal gas, internal energy depends only on temperature ($U = nC_VT$), and the equation of state $PV = nRT$ links pressure, volume, and temperature.
- $\Delta U = Q – W$ — First Law of Thermodynamics
- $PV = nRT$ — ideal gas equation
- $C_P – C_V = R$, $\gamma = C_P/C_V$
- $W = \int P\,dV$ — work done by gas
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$U = \nu\times6\times\frac{1}{2}RT
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: $T_H = 600\,\text{K}$, $T_C = 300\,\text{K}$:
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Answer
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Answer
$$\boxed{\eta_{\text{Carnot}} = 1 – \dfrac{T_C}{T_H}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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