Problem Statement
Solve the thermodynamics problem: One mole of diatomic ideal gas ($\gamma=1.4$) expands adiabatically from $T_1=400\ \text{K}$ to $T_2=250\ \text{K}$. Find the work done. In an adiabatic process $Q=0$, so by the first law $W = -\Delta U$: $$W = -\nu C_v(T_2-T_1) = -1.0\times\frac{5}{2}\times8.314\times(250-400)$$ $$= -1.0\times20.78
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$W = -\nu C_v(T_2-T_1) = -1.0\times\frac{5}{2}\times8.314\times(250-400)$$
$$= -1.0\times20.78
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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