A thermodynamic system is taken from an initial state i with internal energy U_i = 100 J to the fina

Q: Why is the work done along path $ia$ zero?

Because path $ia$ is a vertical line on the p-V diagram, which means the volume is constant (isochoric process). When there is no change in volume, the work done by the gas is zero.

Q: Can we say $Q_{ibf} = Q_{iaf}$ since they start and end at the same states?

No! Heat ($Q$) and work ($W$) are path functions, meaning they depend on the specific route taken from the initial to the final state. Only the change in internal energy ($\Delta U$) is the same for both paths.

Q: How do we know $W_{iaf} = W_{af}$?

The total work done along a path is the sum of the work done along its segments. Since $W_{ia} = 0$, the total work $W_{iaf} = W_{ia} + W_{af} = 0 + W_{af} = W_{af}$.

Q: Why did we need to find $U_f$?

We needed $U_f$ to calculate the change in internal energy for the segment $bf$ ($\Delta U_{bf} = U_f - U_b$). Without $\Delta U_{bf}$, we couldn't use the First Law to find the heat supplied $Q_{bf}$.

Q: Is it possible for internal energy to be negative?

Yes, internal energy is a relative quantity. While the absolute internal energy of an ideal gas is positive (proportional to absolute temperature), the values given in problems are often relative to a chosen reference state, so they can be negative. However, in this problem, all given values are positive.

Question

A thermodynamic system is taken from an initial state $i$ with internal energy $U_i = 100	ext{ J}$ to the final state $f$ along two different paths $iaf$ and $ibf$, as schematically shown in the figure. The work done by the system along the paths $af$, $ib$ and $bf$ are $W_{af} = 200	ext{ J}$, $W_{ib} = 50	ext{ J}$ and $W_{bf} = 100	ext{ J}$ respectively. The heat supplied to the system along the path $iaf$, $ib$ and $bf$ are $Q_{iaf}$, $Q_{ib}$ and $Q_{bf}$ respectively. If the internal energy of the system in the state $b$ is $U_b = 200	ext{ J}$ and $Q_{iaf} = 500	ext{ J}$, the ratio $Q_{bf} / Q_{ib}$ is

Sigma Percentile · Accepted Answer

Welcome to a classic puzzle from the world of thermodynamics! In this problem, we are tasked with analyzing a system that transitions from an initial state $i$ to a final state $f$ through two distinct paths on a $p-V$ diagram.

Our goal is to find the ratio of the heat supplied along two specific segments of the lower path. Let's break this down step by step.

### Analyzing the Setup

Look closely at the $p-V$ diagram provided. We have a thermodynamic system starting at state $i$ and ending at state $f$. It can take two routes: the upper path $iaf$ or the lower path $ibf$.

Notice the geometry of these paths. Vertical lines represent isochoric processes (constant volume), where the work done is zero. Horizontal lines represent isobaric processes (constant pressure), where work is done as the volume changes.

### The Master Equation

To navigate these paths, our master tool is the **First Law of Thermodynamics**. This fundamental principle states that the heat supplied to a system ($Q$) equals the change in its internal energy ($\Delta U$) plus the work done by it ($W$).

$$Q = \Delta U + W$$

A crucial concept to remember here is that while heat and work are **path functions** (they depend on the specific route taken), internal energy is a **state function**. This means the change in internal energy ($\Delta U$) depends only on the initial and final states, regardless of the path taken!

### Navigating the Upper Path

Let's focus on the upper path $iaf$. The segment $ia$ is vertical, meaning the volume is constant, so the work done is zero ($W_{ia} = 0$). The total work done for the path $iaf$ is just the work done along $af$, which is given as $200	ext{ J}$.

$$W_{iaf} = W_{ia} + W_{af} = 0 + 200	ext{ J} = 200	ext{ J}$$

We are given that the total heat supplied along $iaf$ is $500	ext{ J}$. Using the First Law, we can find the change in internal energy for this path:

$$\Delta U_{iaf} = Q_{iaf} - W_{iaf} = 500	ext{ J} - 200	ext{ J} = 300	ext{ J}$$

Since the initial internal energy is $U_i = 100	ext{ J}$, the final internal energy $U_f$ must be:

$$U_f = U_i + \Delta U_{iaf} = 100	ext{ J} + 300	ext{ J} = 400	ext{ J}$$

### Exploring the Lower Path

Now let's move to the lower path, looking first at segment $ib$. We are given $U_b = 200	ext{ J}$. So, the change in internal energy from $i$ to $b$ is:

$$\Delta U_{ib} = U_b - U_i = 200	ext{ J} - 100	ext{ J} = 100	ext{ J}$$

The work done along $ib$ is given as $50	ext{ J}$. Applying the First Law to segment $ib$, the heat $Q_{ib}$ will be the sum of the internal energy change and the work done:

$$Q_{ib} = \Delta U_{ib} + W_{ib} = 100	ext{ J} + 50	ext{ J} = 150	ext{ J}$$

Next, let's look at segment $bf$. The change in internal energy here will be $U_f$ minus $U_b$:

$$\Delta U_{bf} = U_f - U_b = 400	ext{ J} - 200	ext{ J} = 200	ext{ J}$$

The work done along $bf$ is given as $100	ext{ J}$. Again, using the First Law for segment $bf$, the heat $Q_{bf}$ will be:

$$Q_{bf} = \Delta U_{bf} + W_{bf} = 200	ext{ J} + 100	ext{ J} = 300	ext{ J}$$

### Final Calculation

Finally, we need to find the ratio of $Q_{bf}$ to $Q_{ib}$. We found $Q_{bf}$ is $300	ext{ J}$ and $Q_{ib}$ is $150	ext{ J}$.

$$\frac{Q_{bf}}{Q_{ib}} = \frac{300	ext{ J}}{150	ext{ J}} = 2$$

Dividing $300$ by $150$ gives us exactly $2$. And that is our final answer! The elegance of thermodynamics lies in how interconnected these properties are. By simply tracking the energy, we can unravel the entire journey of the system.

Enter Numerical Value:

Visualized Solution

$p - V Diagram Analysis$

$First Law of Thermodynamics$

$Work Done along ia f$

$Internal Energy at f$

$Analysis of Path ib$

$Heat Supplied along ib$

$Analysis of Path b f$

$Heat Supplied along b f$

$Final Ratio$

$State vs Path Functions$

The Sigma Insight: First Law of Thermodynamics

Analyzing the Setup

The Master Equation

Navigating the Upper Path

Exploring the Lower Path

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p−V Diagram Analysis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Law of Thermodynamics

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Work Done along iaf

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Internal Energy at f

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analysis of Path ib

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Supplied along ib

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analysis of Path bf

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Supplied along bf

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } State vs Path Functions