When a system is taken from state i to state f along the path iaf, it is found that Q = 50 cal and W

Q: Why is $\Delta U$ the same for both paths?

Internal energy is a property of the state of the system (like temperature or pressure). Since both paths start at state $i$ and end at state $f$, the change in internal energy must be identical, regardless of how the system got there.

Q: Can work done ($W$) be the same for both paths?

Generally, no. Work done is a path function and represents the area under the $p-V$ curve. Since path $iaf$ and path $ibf$ trace different curves, the area under them (and thus the work done) will be different.

Q: What if the process was cyclic, going $i \to a \to f \to b \to i$?

In a complete cycle, the system returns to its initial state, so the net change in internal energy $\Delta U_{cycle}$ is zero. The net work done would be the area enclosed by the loop, and by the First Law, it would equal the net heat exchanged.

Q: Do we need to convert calories to Joules to solve this?

Not necessarily! As long as all your energy terms (Heat, Work, Internal Energy) are in the same units (calories in this case), the First Law equation $\Delta Q = \Delta U + W$ holds perfectly true.

Q: Why is the work done along path $ibf$ less than path $iaf$?

If you look at a standard $p-V$ diagram for this setup, path $iaf$ typically lies above path $ibf$. Since work done is the area under the $p-V$ curve, the higher path ($iaf$) will naturally have a larger area and thus more work done compared to the lower path ($ibf$).

Question

When a system is taken from state $i$ to state $f$ along the path $iaf$, it is found that $Q = 50	ext{ cal}$ and $W = 20	ext{ cal}$. Along the path $ibf$, $Q = 36	ext{ cal}$. $W$ along the path $ibf$ is

Options:

(A) $6	ext{ cal}$

(B) $16	ext{ cal}$

(C) $66	ext{ cal}$

(D) $14	ext{ cal}$

Sigma Percentile · Accepted Answer

### The Tale of Two Paths

Imagine you are standing at the base of a mountain (state $i$) and your goal is to reach the summit (state $f$). You have two choices: a steep, rugged trail that goes over a high ridge (path $iaf$), or a gentle, winding path that meanders through the valley before ascending (path $ibf$).

No matter which path you choose, your total change in altitude—your potential energy—will be exactly the same once you reach the top. However, the amount of effort you exert (work done) and the amount of food you need to consume (heat supplied) will vary drastically depending on the route. This beautiful analogy perfectly captures the essence of the First Law of Thermodynamics.

### The Master Equation: The First Law

In thermodynamics, the First Law is our ultimate accounting tool. It states that energy cannot be created or destroyed, only transferred. Mathematically, it is expressed as:

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

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

The most critical concept to grasp here is the nature of internal energy ($U$). Internal energy is a **state function**. It is completely blind to history. It does not care whether the system took path $iaf$ or path $ibf$; it only cares about the coordinates of state $i$ and state $f$. Therefore, the change in internal energy, $\Delta U$, is identical for both paths.

### Decoding Path $iaf$

Let's analyze the first route, path $iaf$. The problem provides us with two crucial pieces of data for this journey:

$$Q_{iaf} = 50 	ext{ cal}$$
$$W_{iaf} = 20 	ext{ cal}$$

By substituting these values into our First Law equation, we can uncover the hidden state variable, $\Delta U$:

$$50 = \Delta U + 20$$

With a simple algebraic rearrangement, we find:

$$\Delta U = 50 - 20 = 30 	ext{ cal}$$

This $30 	ext{ cal}$ is the golden key. Because internal energy is path-independent, this value is now locked in for *any* process that starts at state $i$ and ends at state $f$.

### Conquering Path $ibf$

Now, let's shift our focus to the alternative route, path $ibf$. For this path, we are given the heat supplied:

$$Q_{ibf} = 36 	ext{ cal}$$

We need to find the work done, $W_{ibf}$. Thanks to our previous calculation, we already know that $\Delta U$ must be $30 	ext{ cal}$. We plug these values back into the First Law:

$$36 = 30 + W_{ibf}$$

Solving for the work done:

$$W_{ibf} = 36 - 30 = 6 	ext{ cal}$$

And there we have it! The work done along the lower path is exactly $6 	ext{ cal}$.

### The Geometric Perspective

If we look at this from a geometric standpoint on a $p-V$ (pressure-volume) diagram, the work done by a gas is represented by the area under the curve. Path $iaf$ is the upper curve, meaning it encompasses a larger area down to the volume axis, which perfectly aligns with the higher work value of $20 	ext{ cal}$. Path $ibf$ is the lower curve, enclosing a smaller area, which mathematically validates our result of $6 	ext{ cal}$. Physics and geometry are always in perfect harmony!

Animated Solution for Physics - Thermodynamics: When a system is taken from state $i$ to state $f$ along the path $ia f$ , it is found that $Q = 50 cal$ and $W = 20 cal$ . Along the path $ib f$ , $Q = 36 cal$ . $W$ along the path $ib f$ is

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Visualized Solution

\text{Thermodynamic Paths}

\text{First Law of Thermodynamics}

\text{Analyzing Path } iaf

\text{Calculating } \Delta U

\text{Analyzing Path } ibf

\text{Calculating } W_{ibf}

\text{Cyclic Process Insight}

The Sigma Insight: First Law of Thermodynamics

The Tale of Two Paths

The Master Equation

Decoding Path $ia f$

Conquering Path $ib f$

The Geometric Perspective

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