Questions: Reversible Adiabatic (Isentropic) Processes

Question 1

A gas expands rapidly into a vacuum (free expansion) inside a perfectly insulated container. What happens to the entropy of the gas?

• A. It decreases, because the process is adiabatic and Q = 0
• B. It remains constant, because dS = dQ/T = 0 when Q = 0
• C. It increases, because the process is irreversible even though Q = 0
• D. It remains constant only if the gas is ideal

Answer: (C)

The formula dS = dQ_rev/T only applies to reversible heat transfer — it is not a general statement about all adiabatic processes. Irreversibility generates entropy internally through friction, mixing, or unrestrained expansion, regardless of heat flow. Free expansion is irreversible (the gas cannot spontaneously re-compress) so entropy increases despite Q = 0. Isentropic specifically requires both adiabatic AND reversible conditions to eliminate both sources of entropy change.

Question 2

An ideal gas (γ = 5/3) undergoes an isentropic expansion from volume V₁ to V₂ = 8V₁. If the initial temperature is T₁, what is the final temperature T₂?

• A. T₂ = T₁/8 — temperature drops by the same factor as the volume increase
• B. T₂ = T₁/4 — from TV^(γ−1) = constant with (1/8)^(2/3) = 1/4
• C. T₂ = T₁/2 — temperature drops by the square root of the volume ratio
• D. T₂ = T₁ — temperature is unchanged because Q = 0

Answer: (B)

Using TV^(γ−1) = constant: T₂ = T₁(V₁/V₂)^(γ−1) = T₁(1/8)^(2/3) = T₁/4. Option 3 is the key misconception — Q = 0 means no heat flows, but work is still done by the expanding gas, which decreases internal energy and hence temperature. Option 0 confuses the volume ratio with the temperature ratio directly, ignoring the γ−1 exponent.

Question 3

In a reversible adiabatic process, the work done by the gas equals the decrease in its internal energy.

• T. True
• F. False

Answer: True

From the first law: ΔU = Q − W. For any adiabatic process, Q = 0, so ΔU = −W, meaning W = −ΔU. The work done by the gas equals the decrease in internal energy. This holds for all adiabatic processes — reversible or not — because it follows from the first law alone, not from any reversibility condition.

Question 4

Any adiabatic process is also isentropic, since Q = 0 implies dS = 0.

• T. True
• F. False

Answer: False

This is the central misconception about isentropic processes. Entropy can change through two independent mechanisms: heat transfer (dS = dQ_rev/T) and internal irreversibility (friction, free expansion, turbulence). Setting Q = 0 eliminates the first source but not the second. An irreversible adiabatic process still generates entropy internally. Isentropic requires both conditions simultaneously — zero heat flow and zero irreversible entropy generation — leaving total entropy unchanged. On a T-S diagram, only a reversible adiabatic process is a vertical line.

Question 5

Why is an isentropic process specifically a reversible adiabatic process, and why does irreversibility increase entropy even when no heat is transferred?

Model answer: Entropy can increase from two independent sources: heat flowing into a system, and irreversibilities generated internally (friction, mixing, unrestrained expansion). Q = 0 eliminates the heat-transfer source but not the irreversibility source. A reversible process, by definition, generates no internal entropy — every step can be undone without net change. Combining both conditions (adiabatic + reversible) eliminates both entropy sources, giving ΔS = 0. Free expansion is adiabatic but irreversible: entropy increases from the irreversible mixing/expansion even with no heat flow.

This distinction matters in engineering: real turbines and compressors are approximately adiabatic but not perfectly reversible. Their isentropic efficiency compares actual work to the ideal isentropic work, quantifying the entropy generated by irreversibilities (fluid friction, eddies, heat losses to surroundings). A 90% isentropic efficiency turbine loses 10% of potential work to irreversible entropy generation.