Question

A thermodynamic cycle $xyzx$ is shown on a $V-T$ diagram .

The$P-V$ diagram that best describes this cycle is : (Diagrams are schematic and not upto scale)

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Step 1: Given Data,

In the $V-T$ diagram the given variables are,

$V$ is Volume,

$T$ is Temperature,

Let $P$ is pressure,

Step 2: Consider the path $xy$,

Since the path, $xy$ is a straight line hence, the volume is directly proportional to the temperature.

$$\begin{gathered} V\propto T \\ V=KT \end{gathered}$$

(Where, $K$ is proportionality constant.)
Also, we know that, Ideal gas equation,

(Where, $P$ is pressure, $V$ is volume, $n$ is the number of gas, $R$ is the universal gas constant and $T$ is temperature.)

$$\begin{gathered} PV=nRT \\ P=\frac{nRT}{V} \end{gathered}$$

Putting $V=KT$ in the above equation,

$$\begin{gathered} P=\frac{nRT}{KT} \\ P=\frac{nR}{K} \end{gathered}$$

Since all the right-hand side terms of the above equations are constant. Therefore, $P$is a constant.

The isobaric process is the process in which the pressure is constant.
Therefore, Path $xy$ corresponds to the isobaric process.

Step 3: Consider the path $yz$,

Along the path, $yz$ the volume is not changing hence, Volume $V$ is constant.

Step 4. Consider the path $zx$,

Along path, $zx$ the temperature is not changing hence, the temperature $T$ is constant (isothermal process).

The process in which temperature is constant is called the isothermal process.

$PV=$ constant

$P=VK\prime$ [ $K\text{'}$is proportionality constant.]

Only the diagram of option ′ C ′ satisfies these conditions.

Hence, the correct option is (C).