A gas can be taken from A to B via two different processes ACB and ADB. When path ACB is used 60 J o

Q: Why can we use the same $\Delta U$ for both paths? Aren't they different processes?

Yes, they are different processes, which is why the work done and heat supplied are different. However, internal energy ($U$) is a 'state function'. This means it only depends on the current state of the gas (its pressure, volume, and temperature), not how it got there. Since both paths start at $A$ and end at $B$, the change in internal energy ($\Delta U = U_B - U_A$) must be identical for both.

Q: How do we know if the heat and work values should be positive or negative?

We follow standard sign conventions. The problem states 'heat flows into the system', which means the system is gaining energy, so $\Delta Q$ is positive. It also states 'work is done by the system' (expansion), which means the system is expending energy to push its surroundings, so $\Delta W$ is positive.

Q: Why is the work done in path $ADB$ less than the work done in path $ACB$?

Work done by a gas is geometrically equal to the area under the curve on a $p-V$ diagram. If you look at the diagram, the area under path $ACB$ (which goes up to a higher pressure before expanding) is a large rectangle. The area under path $ADB$ (which expands at a lower pressure before going up) is a much smaller rectangle. Hence, $W_{ACB} > W_{ADB}$.

Q: What would happen if the gas went from $A$ to $B$ via $ACB$, and then returned from $B$ to $A$ via $BDA$?

This would form a complete thermodynamic cycle. For any complete cycle, the initial and final states are the exact same point ($A$), so the net change in internal energy $\Delta U_{cycle}$ would be exactly zero. The net work done would be the area enclosed inside the loop.

Q: Can the change in internal energy ever be negative?

Absolutely! If a gas is cooled or compressed in such a way that its final temperature is lower than its initial temperature, its internal energy decreases, making $\Delta U$ negative. In this specific problem, the gas is moving to a state of higher pressure and volume, meaning its temperature (and thus internal energy) increased.

Question

A gas can be taken from $A$ to $B$ via two different processes $ACB$ and $ADB$.
When path $ACB$ is used 60 J of heat flows into the system and 30 J of work is done by the system. If path $ADB$ is used work done by the system is 10 J the heat flow into the system in path $ADB$ is

Options:

(A) 80 J

(B) 40 J

(C) 100 J

(D) 20 J

Sigma Percentile · Accepted Answer

### The Tale of Two Paths: Mastering the First Law of Thermodynamics

Imagine you are standing at the base of a mountain (State $A$) and you need to reach the summit (State $B$). You could take the steep, rugged trail, or you could take the longer, scenic route. Regardless of which path you choose, your change in altitude—your potential energy—will be exactly the same once you reach the top.

This beautiful concept is the heart of thermodynamics, and it is exactly what we need to solve this classic JEE problem.

### The Setup: Visualizing the p-V Diagram

In our problem, a gas is taken from an initial state $A$ to a final state $B$ via two distinct paths on a pressure-volume ($p-V$) diagram: path $ACB$ and path $ADB$.

We are given specific data for the first journey. When the gas travels along path $ACB$, $60	ext{ J}$ of heat flows into the system ($\Delta Q_{ACB} = 60	ext{ J}$), and the system does $30	ext{ J}$ of work ($\Delta W_{ACB} = 30	ext{ J}$).

For the second journey along path $ADB$, we only know that the work done by the system is $10	ext{ J}$ ($\Delta W_{ADB} = 10	ext{ J}$). Our mission is to find the heat flow for this second path.

### The First Law: Our Trusty Tool

To bridge the gap between these two paths, we must invoke the **First Law of Thermodynamics**, which is essentially the law of conservation of energy for thermal systems. It states:

$$ \Delta Q = \Delta W + \Delta U $$

Here, $\Delta Q$ is the heat supplied to the system, $\Delta W$ is the work done by the system, and $\Delta U$ is the change in the system's internal energy.

The absolute magic of this equation lies in $\Delta U$. Internal energy is a **state function**. This means that $\Delta U$ depends *only* on the initial state $A$ and the final state $B$. It does not care whether you took path $ACB$, path $ADB$, or did a loop-de-loop before arriving at $B$. The value of $\Delta U$ will be identical for all paths connecting $A$ and $B$.

### Journey through ACB: Unlocking the Secret

Let's use the data from our first path to unlock the secret value of $\Delta U$. We substitute our knowns into the First Law:

$$ \Delta Q_{ACB} = \Delta W_{ACB} + \Delta U $$

$$ 60 = 30 + \Delta U $$

By simply rearranging the terms, we find the change in internal energy:

$$ \Delta U = 60 - 30 = 30	ext{ J} $$

We have now discovered that moving the gas from state $A$ to state $B$ intrinsically requires an internal energy increase of $30	ext{ J}$.

### Journey through ADB: Reaping the Rewards

Now, we shift our focus to the second path, $ADB$. We know the work done is $10	ext{ J}$. But more importantly, because the initial and final states are the same, we know that $\Delta U$ must still be $30	ext{ J}$!

We apply the First Law one more time:

$$ \Delta Q_{ADB} = \Delta W_{ADB} + \Delta U $$

Substitute our known work and our newly discovered internal energy:

$$ \Delta Q_{ADB} = 10 + 30 $$

$$ \Delta Q_{ADB} = 40	ext{ J} $$

And there we have it! The heat flow into the system along path $ADB$ is $40	ext{ J}$.

### The Way Forward

Problems like this are JEE favorites because they test your conceptual clarity rather than your ability to crunch massive numbers. Always remember: Work and Heat are path-dependent, but Internal Energy is the steadfast anchor that depends only on where you start and where you finish.

Select Answer:

Visualized Solution

Visualizing the Thermodynamic Paths

The First Law of Thermodynamics

Analyzing Path $A C B$

Calculating $Δ U$ from Path $A C B$

Shifting Focus to Path $A D B$

Setting up the Equation for Path $A D B$

Final Calculation

The Way Forward: Cyclic Processes

The Sigma Insight: First Law of Thermodynamics

The Tale of Two Paths

The Setup

The First Law

Journey through ACB

Journey through ADB

The Way Forward

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