Path-independence of state properties: A system goes from state 1 to state 3 via path 1–2–3 with Q = +100 kJ and W = +60 kJ. Along an alternative path 1–4–3, the work is W = +20 kJ. What heat Q is absorbed along 1–4–3 (kJ)?
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A-140 kJ
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B-80 kJ
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C-40 kJ
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D+60 kJ
Answer
Correct Answer: +60 kJ
Explanation
Introduction / Context:This problem tests recognition that internal energy is a state property and therefore path independent. While heat and work depend on the path, their difference (for a closed system neglecting KE/PE changes) equals the change in internal energy between the same end states.
Given Data / Assumptions:
- Closed system; sign convention: heat to system positive, work by system positive.
- Path 1–2–3: Q_123 = +100 kJ, W_123 = +60 kJ.
- Path 1–4–3: W_143 = +20 kJ; Q_143 = ?
Concept / Approach:First law for a closed system: ΔU = Q − W. Since states 1 and 3 are the same regardless of path, ΔU_123 = ΔU_143. Compute ΔU from the known path, then solve for the unknown heat on the alternative path using the same ΔU value.
Step-by-Step Solution:
Compute ΔU from 1–2–3: ΔU = Q − W = 100 − 60 = +40 kJ.Apply first law to 1–4–3: ΔU = Q − W.Substitute ΔU = +40 kJ and W = +20 kJ: 40 = Q − 20.Solve for Q: Q = 40 + 20 = +60 kJ.Verification / Alternative check:Check consistency: Larger work on the first path (60 kJ) required larger heat input (100 kJ) to reach the same ΔU; with smaller work (20 kJ) on the second path, proportionally smaller heat (60 kJ) is needed to maintain ΔU = +40 kJ.
Why Other Options Are Wrong:
- −140, −80, −40 kJ: Incorrect sign/magnitude; they would imply a different ΔU, violating state-function path independence.
Common Pitfalls:Using Q + W instead of Q − W; forgetting the sign convention (work by system positive).
Final Answer:+60 kJ