### Understanding the Thermodynamic Process of an Ideal Gas#### Problem StatementA 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 DescriptionThe 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 \(\Delta E_{\text{Int}} \).#### CalculationWe 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 - WWhere

Introduction: The gases which follows the Boyle's law, Charles's law, Avogadro's law and…

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)

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)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

### 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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