A sample of an ideal gas goes through the process shown in the figure below. From A to B, the process is adiabatic; from B to C, it is isobaric with 345 k of energy entering the system by heat; from C to D, the process is isothermal; and from D to A, it is isobaric with 371 k of energy leaving the system by heat. Determine the difference in internal energy Ent, - Ent. A P (atm)

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### Understanding the Thermodynamic Process of an Ideal Gas

#### Problem Statement

A sample of an ideal gas goes through the process shown in the figure below. 

1. From A to B, the process is adiabatic.
2. From B to C, it is isobaric with 345 kJ of energy entering the system by heat.
3. From C to D, the process is isothermal.
4. From D to A, it is isobaric with 371 kJ of energy leaving the system by heat.

Determine the difference in internal energy \( E_{\text{Int,B}} - E_{\text{Int,A}} \).

#### Given Data
- Energy entering from B to C: \(345\ \text{kJ}\)
- Energy leaving from D to A: \(371\ \text{kJ}\)

#### Diagram Description

The diagram displays a Pressure (P) vs. Volume (V) graph with the following key points:
- **Point A:** \( P = 1\ \text{atm}, V = 0.09\ \text{m}^3 \)
- **Point B:** \( P = 3\ \text{atm}, V = 0.2\ \text{m}^3 \)
- **Point C:** \( P = 3\ \text{atm}, V = 0.4\ \text{m}^3 \)
- **Point D:** \( P = 1\ \text{atm}, V = 1.2\ \text{m}^3 \)

The graph shows the transitions among these points:
- **A to B:** An upward curved line indicating an adiabatic process.
- **B to C:** A horizontal line indicating an isobaric process.
- **C to D:** A downward curved line indicating an isothermal process.
- **D to A:** A horizontal line indicating an isobaric process.

\( \Delta E_{\text{Int}} = E_{\text{Int,B}} - E_{\text{Int,A}} \)

By examining these processes and the energy exchanges, we aim to calculate the change in internal energy  \(\Delta E_{\text{Int}} \).

#### Calculation

We apply the first law of thermodynamics to each process segment and then integrate the results to find the change in internal energy. 

``` 
First law of thermodynamics:
ΔU = Q - W

Where
Transcribed Image Text:### Understanding the Thermodynamic Process of an Ideal Gas #### Problem Statement A sample of an ideal gas goes through the process shown in the figure below. 1. From A to B, the process is adiabatic. 2. From B to C, it is isobaric with 345 kJ of energy entering the system by heat. 3. From C to D, the process is isothermal. 4. From D to A, it is isobaric with 371 kJ of energy leaving the system by heat. Determine the difference in internal energy \( E_{\text{Int,B}} - E_{\text{Int,A}} \). #### Given Data - Energy entering from B to C: \(345\ \text{kJ}\) - Energy leaving from D to A: \(371\ \text{kJ}\) #### Diagram Description The diagram displays a Pressure (P) vs. Volume (V) graph with the following key points: - **Point A:** \( P = 1\ \text{atm}, V = 0.09\ \text{m}^3 \) - **Point B:** \( P = 3\ \text{atm}, V = 0.2\ \text{m}^3 \) - **Point C:** \( P = 3\ \text{atm}, V = 0.4\ \text{m}^3 \) - **Point D:** \( P = 1\ \text{atm}, V = 1.2\ \text{m}^3 \) The graph shows the transitions among these points: - **A to B:** An upward curved line indicating an adiabatic process. - **B to C:** A horizontal line indicating an isobaric process. - **C to D:** A downward curved line indicating an isothermal process. - **D to A:** A horizontal line indicating an isobaric process. \( \Delta E_{\text{Int}} = E_{\text{Int,B}} - E_{\text{Int,A}} \) By examining these processes and the energy exchanges, we aim to calculate the change in internal energy \(\Delta E_{\text{Int}} \). #### Calculation We apply the first law of thermodynamics to each process segment and then integrate the results to find the change in internal energy. ``` First law of thermodynamics: ΔU = Q - W Where
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