An ideal gas is taken through the cycle A - B - C - A, as shown in the figure. If the net heat su

Q: Why is the change in internal energy zero for the cycle?

Internal energy is a state function. Since the gas returns to its exact initial state (temperature, pressure, and volume) at the end of the cycle, the net change in internal energy is zero.

Q: Why is the work done in $B \rightarrow C$ zero?

Work done by a gas is given by $W = \int p dV$. In process $B \rightarrow C$, the volume remains constant at $2 \text{ m}^3$. Since $dV = 0$, no work is done.

Q: How do we know $A \rightarrow B$ is isobaric?

Look at the graph carefully. The x-axis represents pressure. For the entire path from A to B, the pressure remains constant at $10 \text{ N/m}^2$, making it an isobaric process.

Q: Why is the work done in $C \rightarrow A$ negative?

In process $C \rightarrow A$, the volume of the gas decreases from $2 \text{ m}^3$ to $1 \text{ m}^3$. A decrease in volume means the gas is being compressed, so work is done *on* the gas, making the work done *by* the gas negative.

Q: What if the axes were swapped (p on y-axis, V on x-axis)?

If the axes were swapped, the area under the curve would be calculated with respect to the x-axis. The physical values of work would remain the same, but the visual representation of the area would change. Always calculate work as $\int p dV$, which is the area relative to the Volume axis.

Question

An ideal gas is taken through the cycle $A \rightarrow B \rightarrow C \rightarrow A$, as shown in the figure. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process $C \rightarrow A$ is

Options:

(A) –5 J

(B) –10 J

(C) –15 J

(D) – 20 J

Sigma Percentile · Accepted Answer

Title: Decoding the Cyclic Process: A Journey Through the First Law

### Analyzing the Setup

When we first glance at this problem, it looks like a standard thermodynamic cycle. However, there is a subtle trap waiting for us! If you look closely at the graph, the y-axis represents **Volume ($V$)** and the x-axis represents **Pressure ($p$)**. This is a $V-p$ graph, not the usual $p-V$ graph.

We are given a cyclic process $A ightarrow B ightarrow C ightarrow A$ and the net heat supplied to the gas, $Q_{	ext{net}} = 5 	ext{ J}$. Our goal is to find the work done in the final leg of the cycle, $C ightarrow A$.

### The Master Equation

To solve this, we need our most trusted tool: the **First Law of Thermodynamics**, which states:
$$ \Delta Q = \Delta W + \Delta U $$

For any complete cyclic process, the gas returns exactly to its initial state. Because internal energy ($U$) is a state function, its net change over the cycle is zero ($\Delta U = 0$).

This simplifies our master equation beautifully:
$$ Q_{	ext{net}} = W_{	ext{net}} $$

Since we know $Q_{	ext{net}} = 5 	ext{ J}$, the total work done by the gas must also be $5 	ext{ J}$. The total work is simply the sum of the work done in each individual step:
$$ W_{	ext{net}} = W_{AB} + W_{BC} + W_{CA} = 5 	ext{ J} $$

### Step-by-Step Execution

Let's break down the cycle and calculate the work done in each step.

**1. Process $A ightarrow B$:**
Looking at the graph, the path from A to B is a vertical line. Since the x-axis is pressure, this means the pressure is constant at $p = 10 	ext{ N/m}^2$. This is an **isobaric** process. The volume increases from $V_A = 1 	ext{ m}^3$ to $V_B = 2 	ext{ m}^3$.

The work done is:
$$ W_{AB} = p \Delta V = 10 	imes (2 - 1) = 10 	ext{ J} $$

**2. Process $B ightarrow C$:**
The path from B to C is a horizontal line. Since the y-axis is volume, the volume remains constant at $V = 2 	ext{ m}^3$. This is an **isochoric** process. Because there is no change in volume ($\Delta V = 0$), no work is done:
$$ W_{BC} = 0 	ext{ J} $$

### Final Calculation

Now we have all the pieces of the puzzle. We substitute our known values back into the net work equation:
$$ W_{AB} + W_{BC} + W_{CA} = 5 $$
$$ 10 + 0 + W_{CA} = 5 $$

Solving for $W_{CA}$, we get:
$$ W_{CA} = 5 - 10 = -5 	ext{ J} $$

The negative sign is perfectly logical! In process $C ightarrow A$, the volume decreases from $2 	ext{ m}^3$ to $1 	ext{ m}^3$. The gas is being compressed, meaning work is done *on* the gas, resulting in a negative value for the work done *by* the gas.

This problem beautifully illustrates the importance of reading graph axes carefully and trusting the fundamental laws of thermodynamics.

Animated Solution for Physics - Thermodynamics: An ideal gas is taken through the cycle $A \to B \to C \to A$ , as shown in the figure. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process $C \to A$ is

Select Answer:

Visualized Solution

$Analyzing the Cycle$

$First Law for Cyclic Process$

$Net Work Done Equation$

$Work Done in A \to B$

$Work Done in B \to C$

$Calculating W_{C A}$

$Conclusion & Takeaways$

The Sigma Insight: First Law of Thermodynamics

Analyzing the Setup

The Master Equation

Step-by-Step Execution

Final Calculation

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