Air-standard (Otto) efficiency expression: For an ideal Otto cycle, the air-standard efficiency in terms of compression ratio r and specific heat ratio γ is

Introduction / Context:
The air-standard efficiency of the Otto cycle relates thermal efficiency to compression ratio r and the ratio of specific heats γ. This canonical formula explains why higher compression ratios (subject to knock limits) yield better efficiency in spark-ignition engines.

Given Data / Assumptions:

• Idealised air-standard analysis (working fluid is air, constant γ).
• Processes: two isentropic (compression/expansion) and two constant-volume heat exchange steps.
• No real losses (friction, heat transfer to walls, dissociation) considered.

Concept / Approach:

Air-standard analysis gives η_Otto = 1 - (T4 - T1)/(T3 - T2). Using isentropic relations for compression and expansion, temperature ratios collapse into r^(γ - 1), leading to the famous closed-form dependence on r and γ.

Step-by-Step Solution:

Verification / Alternative check:

Limiting behaviour: as r increases, 1 / r^(γ - 1) → 0 and η → 1, consistent with expectations (though real engines are knock-limited). For γ → 1, efficiency tends to zero, matching thermodynamic intuition.

Why Other Options Are Wrong:

Forms with 1 - r^(γ - 1) or 1 + 1 / r^(γ - 1) are algebraically incorrect. “None of these” is false because the correct closed form is present. r^(γ - 1) - 1 is not an efficiency and can exceed 1.

Common Pitfalls:

Forgetting the exponent is (γ - 1); mixing Otto and Diesel efficiency formulae; misapplying polytropic exponents.

Final Answer:

η = 1 - 1 / r^(γ - 1)