Problem Statement
Solve the thermodynamics problem: Find the work done when $\nu=1\ \text{mol}$ of a van der Waals gas expands isothermally from $V_1$ to $V_2$ at temperature $T$. From the van der Waals equation: $p = \frac{\nu RT}{V-\nu b} – \frac{a\nu^2}{V^2}$. Work (with $\nu=1$): $$W = \int_{V_1}^{V_2} p\,dV = RT\ln\frac{V_2-b}{V_1-b} + a\left(\f
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
The van der Waals equation of state corrects the ideal gas law for finite molecular volume and intermolecular attractions. The parameter $a$ accounts for molecular attraction (reducing effective pressure) and $b$ for excluded volume (reducing available volume).
- $(P + a/V_m^2)(V_m – b) = RT$ — van der Waals equation (per mole)
- Critical point: $T_c = 8a/(27Rb)$, $P_c = a/(27b^2)$, $V_{m,c} = 3b$
- Compressibility factor: $Z = PV_m/RT$
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$W = \int_{V_1}^{V_2} p\,dV = RT\ln\frac{V_2-b}{V_1-b} + a\left(\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: $T_H = 600\,\text{K}$, $T_C = 300\,\text{K}$:
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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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