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:

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

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.

Answer 1

The Correct Option is A

Step 1: Set up two simple tests instead of deriving the full S(T,V) formula.
Test 1: a quasistatic adiabatic step has $dS = 0$ by definition, so on a T-S plot it must be a plain VERTICAL segment. Test 2: an isochoric step is not an isothermal step, so it must NOT be a flat horizontal or a straight tilted line, it has to be a curve along which both T and S change together.

Step 2: Find the actual bend of that curve using $C_V$.
At constant volume, $C_V = T \left(\dfrac{\partial S}{\partial T}\right)_V$, so the slope of the isochore on a T-S plot is $\left(\dfrac{dT}{dS}\right)_V = \dfrac{T}{C_V}$. This slope is always positive (T rises as S rises) and it keeps growing as T itself grows, since $C_V$ is roughly constant for an ideal gas. A slope that keeps increasing along the curve means the curve bends upward, it is convex, shaped like the bottom half of a bowl opening up, not a straight tilted line.

Step 3: Grade each of the four panels against the two tests.
Panel (A): two vertical segments (1-2 and 3-4) plus two upward-bending curved segments (2-3 above, 4-1 below), walked in the order compression, heat addition, expansion, heat rejection. Passes both tests.
Panel (B): the same curve and vertical shapes appear, but states 2 and 4 are swapped, so tracing 1 to 2 to 3 to 4 to 1 puts the heat-addition and heat-rejection legs in the wrong place relative to which volume is smaller. Fails the ordering.
Panel (C): the sides connecting the vertical legs are drawn as flat horizontal lines, meaning constant T, that is Test 2 failing outright, since an isochore cannot be isothermal.
Panel (D): same problem as (C), flat horizontal sides where a convex curve should be.

Final Answer:
Panel (A) is the only diagram where both the vertical-adiabat test and the convex-isochore test hold with the states in the correct order. \[ \boxed{\text{Option (A)}} \]