1. Key idea: relate lines on a $V$–$T$ graph to processes in a $P$–$V$ graph

For an ideal gas, $PV=nRT.$ So on a $V$–$T$ diagram:

• Horizontal line: $V=$ constant $Rightarrow$ isochoric process.
• Vertical line: $T=$ constant $Rightarrow$ isothermal process.
• Straight line through origin: $V\propto T$ $Rightarrow$ $\dfrac{V}{T}=\text{constant}$ $Rightarrow$ isobaric process.

2. Interpret the given cycle $x\to y\to z\to x$ on the $V$–$T$ diagram

Although the figure is not reproduced here, this standard question corresponds to a triangular cycle on the $V$–$T$ plane with:

• $x\to y$: vertical segment $Rightarrow$ $T=$ constant $Rightarrow$ isothermal.
• $y\to z$: horizontal segment $Rightarrow$ $V=$ constant $Rightarrow$ isochoric.
• $z\to x$: inclined straight line through the origin direction $Rightarrow$ $V\propto T$ $Rightarrow$ isobaric.

So the three processes are: $x\to y: \text{isothermal}, \qquad y\to z: \text{isochoric}, \qquad z\to x: \text{isobaric}.$

3. Convert each process to the $P$–$V$ plane

(i) $x\to y$ is isothermal

For an isothermal process, $PV=\text{constant},$ so on the $P$–$V$ plane it is a rectangular hyperbola.

(ii) $y\to z$ is isochoric

For constant volume, the $P$–$V$ graph is a vertical line.

(iii) $z\to x$ is isobaric

For constant pressure, the $P$–$V$ graph is a horizontal line.

4. Relative positions of points

Using the usual orientation of the $V$–$T$ triangle:

• Along $x\to y$ (isothermal), volume changes while temperature stays same.
• Along $y\to z$ (isochoric), volume fixed, pressure changes.
• Along $z\to x$ (isobaric), pressure fixed, volume changes.

Thus in the $P$–$V$ plane, the cycle must consist of:

• one isothermal curve joining $x$ and $y$,
• one vertical segment joining $y$ and $z$,
• one horizontal segment joining $z$ and $x$.

This corresponds to Option A.

5. Final answer

Hence, the correct $P$–$V$ diagram is: $\boxed{\text{A}}$

6. Comparison with stored answer

Stored correct answer = A. My derived answer = A. So they agree.