0-6 a 2m²³ OF an Ideal gas is compressed from lookig to 200kPa.. As a result of the process, the internal energy of the gas increases by loks, and Koks is lost. bu the surrounding What the is the work- done by gas during bhis process is the work by the gas negative or positive ?
0-6 a 2m²³ OF an Ideal gas is compressed from lookig to 200kPa.. As a result of the process, the internal energy of the gas increases by loks, and Koks is lost. bu the surrounding What the is the work- done by gas during bhis process is the work by the gas negative or positive ?
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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![### Thermodynamics Problem: Compression of an Ideal Gas
**Question 1**
- **Scenario:**
- \(2 \, m^3\) of an ideal gas is compressed from \(100 \, kPa\) to \(200 \, kPa\).
- As a result of the process, the internal energy of the gas increases by \(10 \, kJ\) and \(4 \, kJ\) is lost to the surroundings.
- **Questions:**
1. **(a)** What is the work done by the gas during this process?
2. **(b)** Is the work done by the gas negative or positive?
**Solution Steps:**
1. To solve for the work done by the gas (W) and determine if it is negative or positive, use the first law of thermodynamics:
\[
\Delta U = Q - W
\]
where:
- \(\Delta U\) is the change in internal energy
- \(Q\) is the heat added to the system
- \(W\) is the work done by the gas
2. Given:
\[
\Delta U = 10 \, kJ\quad\text{(increase in internal energy)}
\]
\[
Q = -4 \, kJ\quad\text{(heat lost to surroundings is negative)}
\]
3. Rearrange the formula to solve for \(W\):
\[
W = Q - \Delta U
\]
\[
W = (-4 \, kJ) - (10 \, kJ)
\]
\[
W = -14 \, kJ
\]
4. **Interpretation of the Result:**
- The work done by the gas, \(W = -14 \, kJ\), is negative.
So:
- **Answer to 1(a):** The work done by the gas during this process is \(-14 \, kJ\).
- **Answer to 1(b):** The work done by the gas is negative.
These computations help in understanding the sign convention used in the first law of thermodynamics, where negative work indicates that work is done on the gas rather than by the gas.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7e36a6f6-5c57-4c3e-b4bb-5bd0d2f1f6a7%2F552426ab-df76-47cb-8788-e2c32ab406c1%2Fw89tda4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Thermodynamics Problem: Compression of an Ideal Gas
**Question 1**
- **Scenario:**
- \(2 \, m^3\) of an ideal gas is compressed from \(100 \, kPa\) to \(200 \, kPa\).
- As a result of the process, the internal energy of the gas increases by \(10 \, kJ\) and \(4 \, kJ\) is lost to the surroundings.
- **Questions:**
1. **(a)** What is the work done by the gas during this process?
2. **(b)** Is the work done by the gas negative or positive?
**Solution Steps:**
1. To solve for the work done by the gas (W) and determine if it is negative or positive, use the first law of thermodynamics:
\[
\Delta U = Q - W
\]
where:
- \(\Delta U\) is the change in internal energy
- \(Q\) is the heat added to the system
- \(W\) is the work done by the gas
2. Given:
\[
\Delta U = 10 \, kJ\quad\text{(increase in internal energy)}
\]
\[
Q = -4 \, kJ\quad\text{(heat lost to surroundings is negative)}
\]
3. Rearrange the formula to solve for \(W\):
\[
W = Q - \Delta U
\]
\[
W = (-4 \, kJ) - (10 \, kJ)
\]
\[
W = -14 \, kJ
\]
4. **Interpretation of the Result:**
- The work done by the gas, \(W = -14 \, kJ\), is negative.
So:
- **Answer to 1(a):** The work done by the gas during this process is \(-14 \, kJ\).
- **Answer to 1(b):** The work done by the gas is negative.
These computations help in understanding the sign convention used in the first law of thermodynamics, where negative work indicates that work is done on the gas rather than by the gas.
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