Based in the given PV diagram, what is the work done on the gas in the process shown, in Joules? Your answer needs to have 2 significant figures, including the negative sign in your answer if needed. Do not include the positive sign if the answer is positive. No unit is needed in your answer, it is already given in the question statement. p (kPa)
Based in the given PV diagram, what is the work done on the gas in the process shown, in Joules? Your answer needs to have 2 significant figures, including the negative sign in your answer if needed. Do not include the positive sign if the answer is positive. No unit is needed in your answer, it is already given in the question statement. p (kPa)
Physics for Scientists and Engineers
10th Edition
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter18: Temperature
Section: Chapter Questions
Problem 1P
Related questions
Question
answer both or none
1. Isnt 30
2. Isnt 4

Transcribed Image Text:### Educational Content: Understanding Work Done on a Gas
**Problem Statement:**
Based on the given pV diagram, what is the work done on the gas in the process shown, in Joules?
**Instructions:**
Your answer needs to have 2 significant figures, including the negative sign in your answer if needed. Do not include the positive sign if the answer is positive. No unit is needed in your answer, as it is already given in the question statement.
**pV Diagram Explanation:**
- **Axes:**
- The horizontal axis represents Volume (V) in cubic centimeters (cm³).
- The vertical axis represents Pressure (p) in kilopascals (kPa).
- **Data Points:**
- The process starts at point **i** with a volume of approximately 100 cm³ and a pressure of 200 kPa.
- The process ends at point **f** with a volume of approximately 300 cm³ and a pressure of 400 kPa.
- **Graph Description:**
- The diagram shows a straight line increasing linearly from point **i** to point **f**.
- This linear path indicates a direct proportional relationship between pressure and volume during the process.
**Educational Note:**
The work done on or by the gas in this process can be calculated by finding the area under the curve in the pV diagram between the initial and final state. For a straight-line process, this can typically be calculated as the area of a trapezoid formed between these two points on the pV plane.
### Tips for Solving:
- Recognize the shape under the curve.
- Use the formula for the area of a trapezoid when applicable.
- Remember that work done on the system is typically indicated by the area under the curve in the positive volume direction. A clockwise path indicates work done by the system, while a counterclockwise path indicates work done on the system.
![**Title: Understanding the Ideal-Gas Process in a Pressure-Volume Diagram**
**Introduction to the Problem:**
The question posed is: "What is the ratio \( T_D/T_B \) for this ideal-gas process D→B?"
**Description of the Graph:**
The graph is a Pressure-Volume (p-V) diagram with:
- The vertical axis labeled as \( p \) (pressure in atm).
- The horizontal axis labeled as \( V \) (volume in m³).
**Key Points on the Graph:**
1. **Point D:**
- Located at \( V = 1 \, \text{m}^3 \) and \( p = 1 \, \text{atm} \).
2. **Point A:**
- Shown for reference at \( V = 1 \, \text{m}^3 \) and \( p = 4 \, \text{atm} \).
3. **Point B:**
- Located at \( V = 5 \, \text{m}^3 \) and \( p = 1 \, \text{atm} \).
**Process Description:**
- The process D→B is a horizontal line on the graph, indicating an isobaric process (constant pressure).
- The process involves an increase in volume from 1 m³ to 5 m³, with the pressure remaining constant at 1 atm.
**Explanation:**
In an isobaric process for an ideal gas, the temperature ratio \( T_D/T_B \) can be found using the ideal gas law (\( PV = nRT \)). Since pressure is constant, the temperature ratio is proportional to the volume ratio:
\[
\frac{T_D}{T_B} = \frac{V_D}{V_B}
\]
Given \( V_D = 1 \, \text{m}^3 \) and \( V_B = 5 \, \text{m}^3 \), the ratio is:
\[
\frac{T_D}{T_B} = \frac{1}{5}
\]
**Conclusion:**
The temperature ratio \( T_D/T_B \) for the ideal-gas process D→B is \( \frac{1}{5} \). This demonstrates how the volume change affects the temperature during an isobaric process.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf73d5ff-2378-4701-8e60-6def9ee03b70%2Fe816a9ee-45c1-4ec6-b0f2-e9dc69e4f2ad%2Fakrkmbk_processed.png&w=3840&q=75)
Transcribed Image Text:**Title: Understanding the Ideal-Gas Process in a Pressure-Volume Diagram**
**Introduction to the Problem:**
The question posed is: "What is the ratio \( T_D/T_B \) for this ideal-gas process D→B?"
**Description of the Graph:**
The graph is a Pressure-Volume (p-V) diagram with:
- The vertical axis labeled as \( p \) (pressure in atm).
- The horizontal axis labeled as \( V \) (volume in m³).
**Key Points on the Graph:**
1. **Point D:**
- Located at \( V = 1 \, \text{m}^3 \) and \( p = 1 \, \text{atm} \).
2. **Point A:**
- Shown for reference at \( V = 1 \, \text{m}^3 \) and \( p = 4 \, \text{atm} \).
3. **Point B:**
- Located at \( V = 5 \, \text{m}^3 \) and \( p = 1 \, \text{atm} \).
**Process Description:**
- The process D→B is a horizontal line on the graph, indicating an isobaric process (constant pressure).
- The process involves an increase in volume from 1 m³ to 5 m³, with the pressure remaining constant at 1 atm.
**Explanation:**
In an isobaric process for an ideal gas, the temperature ratio \( T_D/T_B \) can be found using the ideal gas law (\( PV = nRT \)). Since pressure is constant, the temperature ratio is proportional to the volume ratio:
\[
\frac{T_D}{T_B} = \frac{V_D}{V_B}
\]
Given \( V_D = 1 \, \text{m}^3 \) and \( V_B = 5 \, \text{m}^3 \), the ratio is:
\[
\frac{T_D}{T_B} = \frac{1}{5}
\]
**Conclusion:**
The temperature ratio \( T_D/T_B \) for the ideal-gas process D→B is \( \frac{1}{5} \). This demonstrates how the volume change affects the temperature during an isobaric process.
Expert Solution

This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 3 images

Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Recommended textbooks for you

Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning

Physics for Scientists and Engineers with Modern …
Physics
ISBN:
9781337553292
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning

College Physics
Physics
ISBN:
9781938168000
Author:
Paul Peter Urone, Roger Hinrichs
Publisher:
OpenStax College

Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning

Physics for Scientists and Engineers with Modern …
Physics
ISBN:
9781337553292
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning

College Physics
Physics
ISBN:
9781938168000
Author:
Paul Peter Urone, Roger Hinrichs
Publisher:
OpenStax College

Physics for Scientists and Engineers: Foundations…
Physics
ISBN:
9781133939146
Author:
Katz, Debora M.
Publisher:
Cengage Learning

Glencoe Physics: Principles and Problems, Student…
Physics
ISBN:
9780078807213
Author:
Paul W. Zitzewitz
Publisher:
Glencoe/McGraw-Hill

Physics for Scientists and Engineers, Technology …
Physics
ISBN:
9781305116399
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning