Problem Statement
Solve the thermodynamics problem: One mole of ideal gas undergoes free (irreversible) expansion from $V_1$ to $V_2=2V_1$. Find $\Delta S$. In free expansion into a vacuum: $Q=0$, $W=0$, $\Delta U=0$ (ideal gas), so $T$ unchanged. But the process is irreversible — entropy still increases. To find $\Delta S$, use any reversible path b
Given Information
- $V_2=2V$
Physical Concepts & Formulas
Entropy is a state function measuring the dispersal of energy. The Second Law requires that the total entropy of an isolated system never decreases. For a reversible process, $dS = dQ_\text{rev}/T$; for irreversible processes, the actual entropy change exceeds $dQ/T$.
- $\Delta S = \int dQ_{\text{rev}}/T$ — entropy change for reversible process
- $\Delta S = nC_V \ln(T_2/T_1) + nR\ln(V_2/V_1)$ — ideal gas entropy change
- $\Delta S \geq 0$ for isolated system (Second Law)
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
$$\eta = 1 – \frac{300}{600} = 1 – 0.5 = 0.50 = 50\%$$
$$\boxed{\eta_{\text{Carnot}} = 1 – \dfrac{T_C}{T_H}}$$
Answer
$$\boxed{\eta_{\text{Carnot}} = 1 – \dfrac{T_C}{T_H}}$$
Physical Interpretation
Entropy increases in any irreversible process. The numerical result confirms the Second Law: the universe’s entropy has increased, and the process cannot spontaneously reverse.
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