n mole of a perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following pro

Q: Why is the work done in process CA exactly zero?

Process $C \rightarrow A$ is an isochoric process, meaning the volume remains strictly constant at $V_1$. Mechanical work in thermodynamics is defined as $W = \int p \, dV$. Since there is no change in volume ($dV = 0$), the gas neither expands to do work nor gets compressed to have work done on it. Thus, $W = 0$.

Q: Why did we get a negative sign for the work done in process BC?

During process $B \rightarrow C$, the gas is being compressed from a larger volume ($2V_1$) to a smaller volume ($V_1$). Whenever a gas is compressed, the displacement of the piston is opposite to the direction of the force exerted by the gas pressure. Therefore, the work done *by* the gas is negative.

Q: How did we know the temperature at state B is T?

The problem explicitly states that the process $A \rightarrow B$ is an *isothermal* expansion at temperature $T$. By definition, an isothermal process maintains a constant temperature throughout. Therefore, if the gas starts at temperature $T$ at state $A$, it must still be at temperature $T$ when it reaches state $B$.

Q: Can we find the net work by just calculating the area of the cycle directly using geometry?

Yes, the net work is exactly the area enclosed by the cycle. However, because the curve $A \rightarrow B$ is a hyperbola ($p \propto 1/V$), it's not a simple geometric shape like a triangle or a circle. You have to use integration to find the area under that curve, which is exactly what the formula $W = nRT \ln(V_f/V_i)$ does for you.

Q: What is the total change in internal energy for this complete cycle?

For any complete cyclic process, the gas returns to its exact initial state (same pressure, volume, and temperature). Since internal energy $U$ is a state function depending only on temperature for an ideal gas, the total change in internal energy $\Delta U_{\text{cycle}}$ is exactly zero.

Question

$n$ mole of a perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following processes.
$A \rightarrow B$: Isothermal expansion at temperature $T$, so that the volume is doubled from $V_1$ to $V_2 = 2V_1$ and pressure changes from $p_1$ to $p_2$.
$B \rightarrow C$: Isobaric compression at pressure $p_2$ to initial volume $V_1$.
$C \rightarrow A$: Isochoric change leading to change of pressure from $p_2$ to $p_1$.
Total work done in the complete cycle ABCA is

Options:

(A) $0$

(B) $nRT \ln 2$

(C) $nRT \left( \ln 2 + \frac{1}{2} \right)$

(D) $nRT \left( \ln 2 - \frac{1}{2} \right)$

Sigma Percentile · Accepted Answer

### The Beauty of Cyclic Processes

Imagine you are tracking the life of a gas trapped inside a cylinder with a movable piston. As you heat it, cool it, expand it, and compress it, the gas goes through a journey. When it returns exactly to its starting state, we call it a **cyclic process**.

The beauty of a cyclic process on a $p-V$ diagram is that the net work done by the gas is simply the area enclosed by the loop. If the cycle is clockwise, the gas does positive net work. If it's counter-clockwise, work is done *on* the gas. In this problem, we have a classic three-step cycle: an isothermal expansion, an isobaric compression, and an isochoric return. Let's break down the physics and the math step-by-step.

### Decoding the Isothermal Expansion ($A ightarrow B$)

Our journey begins at state $A$. The gas expands isothermally to state $B$. "Isothermal" means the temperature remains perfectly constant at $T$. For an ideal gas, if the temperature is constant, the internal energy doesn't change. All the heat you pump into the gas goes entirely into doing work to push the piston out.

The volume doubles from $V_1$ to $2V_1$. The work done during an isothermal process is calculated by integrating $p \, dV$, which yields the famous logarithmic formula:

$$W_{AB} = nRT \ln\left(\frac{V_f}{V_i}ight)$$

Substituting our specific volumes:

$$W_{AB} = nRT \ln\left(\frac{2V_1}{V_1}ight) = nRT \ln 2$$

Geometrically, this positive value represents the entire area under the curve from $A$ to $B$ down to the volume axis.

### The Isobaric Compression ($B ightarrow C$)

Next, the gas is compressed from $B$ to $C$ while keeping the pressure constant at $p_2$. This is an **isobaric** process. Because the volume is decreasing (from $2V_1$ back to $V_1$), the work done by the gas will be negative. The surroundings are doing work on the gas.

The formula for isobaric work is straightforward:

$$W_{BC} = p \Delta V = p_2(V_1 - 2V_1) = -p_2 V_1$$

But wait, our options are all in terms of $nRT$. We need a bridge to connect $p_2 V_1$ to $nRT$. Let's look at state $B$. At state $B$, the gas is at pressure $p_2$, volume $2V_1$, and temperature $T$ (since $A ightarrow B$ was isothermal). Applying the ideal gas law at state $B$:

$$p_2(2V_1) = nRT$$

Dividing both sides by 2, we uncover the hidden relationship:

$$p_2 V_1 = \frac{nRT}{2}$$

Substituting this back into our work equation:

$$W_{BC} = -\frac{nRT}{2}$$

### The Isochoric Return ($C ightarrow A$)

Finally, the gas must return to its initial state $A$. It does this by increasing its pressure from $p_2$ back to $p_1$ while keeping the volume locked at $V_1$. This is an **isochoric** process.

Because the piston doesn't move, the change in volume $\Delta V$ is zero. No movement means no mechanical work:

$$W_{CA} = p \Delta V = 0$$

### Synthesizing the Net Work

We have successfully calculated the work for all three legs of the journey. The net work done in the complete cycle is simply the algebraic sum of these individual contributions:

$$W_{	ext{net}} = W_{AB} + W_{BC} + W_{CA}$$

$$W_{	ext{net}} = nRT \ln 2 - \frac{nRT}{2} + 0$$

Factoring out the common $nRT$ term, we arrive at our elegant final answer:

$$W_{	ext{net}} = nRT \left( \ln 2 - \frac{1}{2} ight)$$

This result perfectly matches the area enclosed by the $ABCA$ loop on the $p-V$ diagram. The positive area under the isotherm is partially canceled out by the negative rectangular area of the isobaric compression, leaving us with the net positive work done by the engine.

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Visualized Solution

$W_{net} = Σ W_{process}$

$A \to B : Isothermal$

$W_{A B} = n R T ln (\frac{V _{f}}{V _{i}})$

$B \to C : Isobaric$

$W_{B C} = p Δ V$

$p_{2} V_{1} = \frac{n R T}{2}$

$C \to A : Isochoric$

$W_{C A} = 0$

$W_{net} = n R T (ln 2 - \frac{1}{2})$

$\oint d W = Area$

The Sigma Insight: Thermodynamic Processes

The Beauty of Cyclic Processes

Decoding the Isothermal Expansion ( $A \to B$ )

The Isobaric Compression ( $B \to C$ )

The Isochoric Return ( $C \to A$ )

Synthesizing the Net Work

Similar Questions

An ideal gas expands isothermally from a volume $V_{1}$ to $V_{2}$ and then compressed to original volume $V_{1}$ adiabatically. Initial pressure is $p_{1}$ and final pressure is $p_{3}$ . The total work done is $W$ . Then,

One mole of an ideal gas in initial state $A$ undergoes a cyclic process $A B C A$ , as shown in the figure. Its pressure at $A$ is $p_{0}$ . Choose the correct option(s) from the following.

Starting with the same initial conditions, an ideal gas expands from volume $V_{1}$ to $V_{2}$ in three different ways, the work done by the gas is $W_{1}$ if the process is purely isothermal, $W_{2}$ if purely isobaric and $W_{3}$ if purely adiabatic, then

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Wnet​=ΣWprocess​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A→B:Isothermal

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } WAB​=nRTln(Vi​Vf​​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B→C:Isobaric

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } WBC​=pΔV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p2​V1​=2nRT​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C→A:Isochoric

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } WCA​=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Wnet​=nRT(ln2−21​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∮dW=Area