A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change

Q: Why is the work done in process $b \rightarrow c$ zero?

On a $p-V$ diagram, the work done by a gas is represented by the area under the curve. A vertical line means the volume is constant ($\Delta V = 0$). Since work is defined as $W = \int p\,dV$, if there is no change in volume, the gas does no work.

Q: Why is $\Delta U_{abca} = 0$?

Internal energy ($U$) is a 'state function'. This means its value depends only on the current state of the system (its pressure, volume, and temperature), not on the path taken to get there. In a cyclic process, the initial and final states are exactly the same, so the net change in internal energy is zero.

Q: Can we find the work done in process $c \rightarrow a$ with the given information?

Not directly. We know $\Delta U_{ca} = -180\text{ J}$, but we don't have the heat exchanged ($\Delta Q_{ca}$) or the exact mathematical function of the curve to calculate the area under it. If we had $\Delta Q_{ca}$, we could use the First Law to find $W_{ca}$.

Q: Why did we add the internal energies instead of the heat or work to find the missing value?

We can add heat and work too, but $\Delta U$ is special because its sum over a closed loop is exactly zero. Heat and work are 'path functions', meaning their net sum over a cycle is not zero (it equals the net work done or net heat absorbed). The equation $\Delta U_{cycle} = 0$ gives us a powerful, guaranteed constraint to find missing $\Delta U$ values for individual segments.

Q: Is the net work done in this entire cycle positive or negative?

The cycle $a \rightarrow b \rightarrow c \rightarrow a$ is traced in an anti-clockwise direction on the $p-V$ diagram. Therefore, the net work done by the gas in the entire cycle will be negative. This means net work is done *on* the gas by the surroundings.

Question

A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the internal energy of the gas along the path ca is $-180	ext{ J}$. The gas absorbs $250	ext{ J}$ of heat along the path ab and $60	ext{ J}$ along the path bc. The work done by the gas along the path abc is

Options:

(A) $120	ext{ J}$

(B) $130	ext{ J}$

(C) $100	ext{ J}$

(D) $140	ext{ J}$

Sigma Percentile · Accepted Answer

### Decoding the Cyclic Process: A Journey Through the First Law of Thermodynamics

Thermodynamics is often perceived as a dense jungle of equations, but at its core, it is simply the ultimate accounting system of the universe. Energy cannot be created or destroyed; it can only change forms. In this problem, we are tasked with auditing the energy flow of an ideal gas as it undergoes a cyclic process $a ightarrow b ightarrow c ightarrow a$.

Let's break down this thermodynamic journey step by step, using the First Law of Thermodynamics as our guiding compass.

### The Setup: Visualizing the Cycle

Imagine you are tracking the state of a gas on a $p-V$ (pressure-volume) diagram. The gas starts at state $a$, expands to state $b$, gets pressurized at constant volume to state $c$, and finally returns to its original state $a$.

We are armed with three crucial pieces of data:
1. The heat absorbed during the expansion $a ightarrow b$ is $\Delta Q_{ab} = 250	ext{ J}$.
2. The heat absorbed during the pressurization $b ightarrow c$ is $\Delta Q_{bc} = 60	ext{ J}$.
3. The change in internal energy during the return path $c ightarrow a$ is $\Delta U_{ca} = -180	ext{ J}$.

Our mission is to find the total work done by the gas along the path $a ightarrow b ightarrow c$, which is simply the sum of the work done in each segment: $W_{abc} = W_{ab} + W_{bc}$.

### The Isochoric Shortcut: Path $b ightarrow c$

Let's start with the easiest segment. Look closely at the path from $b$ to $c$ on the $p-V$ diagram. It is a perfectly vertical line. What does this geometric feature tell us physically?

A vertical line means the volume of the gas is not changing. In thermodynamic terms, this is an **isochoric process**. Since work done by a gas is defined as the area under the $p-V$ curve (or mathematically, $W = \int p\,dV$), a process with zero change in volume ($\Delta V = 0$) results in absolutely zero work done.

$$W_{bc} = 0$$

Now, let's apply the First Law of Thermodynamics ($\Delta Q = \Delta U + W$) to this specific path:

$$\Delta Q_{bc} = \Delta U_{bc} + W_{bc}$$
$$60	ext{ J} = \Delta U_{bc} + 0$$
$$\Delta U_{bc} = 60	ext{ J}$$

This tells us that all the heat supplied to the gas during this phase goes entirely into increasing its internal energy (making the gas molecules move faster and increasing the temperature).

### The Cyclic Secret: Closing the Loop

Here is where the magic of state functions comes into play. Internal energy ($U$) is a state function. It doesn't care how you got to a particular state; it only cares about where you are.

Because our gas undergoes a **cyclic process** and returns exactly to state $a$, its net change in internal energy for the entire cycle must be zero.

$$\Delta U_{abca} = 0$$

This means the sum of the internal energy changes along all three individual paths must perfectly balance out:

$$\Delta U_{ab} + \Delta U_{bc} + \Delta U_{ca} = 0$$

We already know $\Delta U_{bc} = 60	ext{ J}$, and the problem generously gave us $\Delta U_{ca} = -180	ext{ J}$. Let's plug these in to uncover the hidden internal energy change for path $a ightarrow b$:

$$\Delta U_{ab} + 60	ext{ J} + (-180	ext{ J}) = 0$$
$$\Delta U_{ab} - 120	ext{ J} = 0$$
$$\Delta U_{ab} = 120	ext{ J}$$

### The Final Stretch: Path $a ightarrow b$

We are now fully equipped to tackle the path $a ightarrow b$. We know the heat absorbed ($\Delta Q_{ab} = 250	ext{ J}$) and we just deduced the change in internal energy ($\Delta U_{ab} = 120	ext{ J}$).

Once again, we call upon the First Law of Thermodynamics:

$$\Delta Q_{ab} = \Delta U_{ab} + W_{ab}$$

Substituting our known values:

$$250	ext{ J} = 120	ext{ J} + W_{ab}$$
$$W_{ab} = 250	ext{ J} - 120	ext{ J}$$
$$W_{ab} = 130	ext{ J}$$

This positive value makes perfect physical sense. Looking at the graph, the volume is increasing from $a$ to $b$, meaning the gas is expanding and doing positive work on its surroundings.

### The Grand Finale: Total Work Done

The question asks for the total work done along the combined path $a ightarrow b ightarrow c$. This is simply the algebraic sum of the work done in the two segments:

$$W_{abc} = W_{ab} + W_{bc}$$
$$W_{abc} = 130	ext{ J} + 0	ext{ J}$$
$$W_{abc} = 130	ext{ J}$$

And there we have it! By carefully auditing the energy using the First Law and leveraging the path-independent nature of internal energy, we've successfully navigated the cycle. The correct option is **(b)**.

Select Answer:

Visualized Solution

$p - V Diagram of Cyclic Process$

$First Law & Cyclic Process$

$Analyzing Path b \to c$

$Internal Energy of Cycle$

$Calculating Δ U_{ab}$

$Analyzing Path a \to b$

$Calculating W_{ab}$

$Total Work Done along ab c$

$The Way Forward$

The Sigma Insight: First Law of Thermodynamics

Decoding the Cyclic Process

The Setup

The Isochoric Shortcut

The Cyclic Secret

The Final Stretch

The Grand Finale

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