Question:
Consider the Otto cycle for an ideal gas engine consisting of two quasistatic adiabatic and two quasistatic isochoric processes. The correct temperature-entropy (T-S) phase diagram for the cycle is:
Consider the Otto cycle for an ideal gas engine consisting of two quasistatic adiabatic and two quasistatic isochoric processes. The correct temperature-entropy (T-S) phase diagram for the cycle is:
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An adiabatic (isentropic) step is a vertical line on a T-S plot.
An isochoric step is a curved, convex line, not straight, since \(S=C_V\ln T + R\ln V+\text{const}\).
The isochore at the smaller (compressed) volume sits above the one at the larger volume.
An isochoric step is a curved, convex line, not straight, since \(S=C_V\ln T + R\ln V+\text{const}\).
The isochore at the smaller (compressed) volume sits above the one at the larger volume.
Updated On: Jul 28, 2026
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The Correct Option is A
Solution and Explanation
Step 1: Recall what each part of the Otto cycle does.
The Otto cycle has four quasistatic steps: 1 to 2 is an adiabatic (isentropic) compression, 2 to 3 is an isochoric (constant volume) heat addition, 3 to 4 is an adiabatic (isentropic) expansion, and 4 to 1 is an isochoric heat rejection back to the starting state. We need to know what each of these two process types looks like on a temperature versus entropy (T-S) plot.
Step 2: Key Formula or Approach:
For an ideal gas the entropy of one mole can be written as \(S = C_V \ln T + R \ln V + \text{const}\). Along a quasistatic adiabatic process the process is isentropic by definition, so \(dS = 0\): on a T-S plot this is a straight VERTICAL line, since T can change while S stays fixed. Along an isochoric process V is fixed but T changes, so from \(S = C_V \ln T + \text{const}(V)\), which rearranges to \(T = A(V)\, e^{S/C_V}\), this is an increasing, upward-curving (convex) exponential-shaped line, never a straight line.
Step 3: Work out where the two isochores sit relative to each other.
The 2-3 heat addition happens at the smaller, compressed volume \(V_2\), and the 4-1 heat rejection happens at the larger volume \(V_1 \gt V_2\). Since \(S = C_V\ln T + R\ln V + \text{const}\), for the same entropy value a larger volume needs a smaller temperature to keep S the same (the \(R\ln V\) term is larger, so \(C_V\ln T\) must be smaller). So the isochore belonging to the smaller volume \(V_2\) sits above (higher T for the same S) the isochore belonging to the larger volume \(V_1\).
Step 4: Match this to the four panels.
We need a shape with two vertical segments (the adiabats) and two convex upward-curving segments (the isochores), where the upper curve is the 2-3 heat addition and the lower curve is the 4-1 heat rejection, walked in the order 1 (bottom-left) up to 2, across the upper curve to 3, down to 4, and back across the lower curve to 1. This is exactly panel (A).
Step 5: Why the others fail.
Panel (B) draws the identical curve shape but swaps the positions of states 2 and 4, so walking 1 to 2 to 3 to 4 to 1 in that panel traces compression and heat addition in the wrong place relative to the curved isochores, it does not match the Otto sequence. Panels (C) and (D) draw the isochores as straight horizontal sides of a rectangle. A straight horizontal line on a T-S plot means T stays constant while S changes, that is an isothermal process, not an isochoric one, so both misrepresent the physics of the isochoric legs.
Final Answer:
Only panel (A) shows two vertical adiabats joined by two correctly-ordered convex isochores. \[ \boxed{\text{Option (A)}} \]
The Otto cycle has four quasistatic steps: 1 to 2 is an adiabatic (isentropic) compression, 2 to 3 is an isochoric (constant volume) heat addition, 3 to 4 is an adiabatic (isentropic) expansion, and 4 to 1 is an isochoric heat rejection back to the starting state. We need to know what each of these two process types looks like on a temperature versus entropy (T-S) plot.
Step 2: Key Formula or Approach:
For an ideal gas the entropy of one mole can be written as \(S = C_V \ln T + R \ln V + \text{const}\). Along a quasistatic adiabatic process the process is isentropic by definition, so \(dS = 0\): on a T-S plot this is a straight VERTICAL line, since T can change while S stays fixed. Along an isochoric process V is fixed but T changes, so from \(S = C_V \ln T + \text{const}(V)\), which rearranges to \(T = A(V)\, e^{S/C_V}\), this is an increasing, upward-curving (convex) exponential-shaped line, never a straight line.
Step 3: Work out where the two isochores sit relative to each other.
The 2-3 heat addition happens at the smaller, compressed volume \(V_2\), and the 4-1 heat rejection happens at the larger volume \(V_1 \gt V_2\). Since \(S = C_V\ln T + R\ln V + \text{const}\), for the same entropy value a larger volume needs a smaller temperature to keep S the same (the \(R\ln V\) term is larger, so \(C_V\ln T\) must be smaller). So the isochore belonging to the smaller volume \(V_2\) sits above (higher T for the same S) the isochore belonging to the larger volume \(V_1\).
Step 4: Match this to the four panels.
We need a shape with two vertical segments (the adiabats) and two convex upward-curving segments (the isochores), where the upper curve is the 2-3 heat addition and the lower curve is the 4-1 heat rejection, walked in the order 1 (bottom-left) up to 2, across the upper curve to 3, down to 4, and back across the lower curve to 1. This is exactly panel (A).
Step 5: Why the others fail.
Panel (B) draws the identical curve shape but swaps the positions of states 2 and 4, so walking 1 to 2 to 3 to 4 to 1 in that panel traces compression and heat addition in the wrong place relative to the curved isochores, it does not match the Otto sequence. Panels (C) and (D) draw the isochores as straight horizontal sides of a rectangle. A straight horizontal line on a T-S plot means T stays constant while S changes, that is an isothermal process, not an isochoric one, so both misrepresent the physics of the isochoric legs.
Final Answer:
Only panel (A) shows two vertical adiabats joined by two correctly-ordered convex isochores. \[ \boxed{\text{Option (A)}} \]
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