3. 1.00L of a gas at 300K has a pressure of 155kPa. The temperature increases to 400K and the pressure to 605kPa. What is the final volume of the gas?
3. 1.00L of a gas at 300K has a pressure of 155kPa. The temperature increases to 400K and the pressure to 605kPa. What is the final volume of the gas?
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...
Related questions
Question
![**Gas Law Problem**
**Problem Statement:**
3. A gas occupying a volume of 1.00 L at a temperature of 300 K has a pressure of 155 kPa. The temperature increases to 400 K and the pressure rises to 605 kPa. What is the final volume of the gas?
**Explanation:**
To solve this problem, you can use the combined gas law, which is derived from Boyle’s Law, Charles’s Law, and Gay-Lussac’s Law. The combined gas law formula is:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Where:
- \( P_1 \) = initial pressure (155 kPa)
- \( V_1 \) = initial volume (1.00 L)
- \( T_1 \) = initial temperature (300 K)
- \( P_2 \) = final pressure (605 kPa)
- \( V_2 \) = final volume (unknown)
- \( T_2 \) = final temperature (400 K)
**Steps:**
1. Substitute the given values into the combined gas law equation:
\[ \frac{155 \, \text{kPa} \times 1.00 \, \text{L}}{300 \, \text{K}} = \frac{605 \, \text{kPa} \times V_2}{400 \, \text{K}} \]
2. By simplifying the equation, we solve for \( V_2 \):
\[ V_2 = \frac{(155 \, \text{kPa} \times 1.00 \, \text{L} \times 400 \, \text{K})}{(300 \, \text{K} \times 605 \, \text{kPa})} \]
3. Perform the calculation:
\[ V_2 = \frac{62000}{181500} \approx 0.3415 \, \text{L} \]
Therefore, the final volume of the gas is approximately 0.3415 liters.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3d3fcad5-0847-4348-902e-95e13b235eb3%2Fcdb308b0-d58f-4ab0-aee2-1c939131a69c%2Fvgzrnr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Gas Law Problem**
**Problem Statement:**
3. A gas occupying a volume of 1.00 L at a temperature of 300 K has a pressure of 155 kPa. The temperature increases to 400 K and the pressure rises to 605 kPa. What is the final volume of the gas?
**Explanation:**
To solve this problem, you can use the combined gas law, which is derived from Boyle’s Law, Charles’s Law, and Gay-Lussac’s Law. The combined gas law formula is:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Where:
- \( P_1 \) = initial pressure (155 kPa)
- \( V_1 \) = initial volume (1.00 L)
- \( T_1 \) = initial temperature (300 K)
- \( P_2 \) = final pressure (605 kPa)
- \( V_2 \) = final volume (unknown)
- \( T_2 \) = final temperature (400 K)
**Steps:**
1. Substitute the given values into the combined gas law equation:
\[ \frac{155 \, \text{kPa} \times 1.00 \, \text{L}}{300 \, \text{K}} = \frac{605 \, \text{kPa} \times V_2}{400 \, \text{K}} \]
2. By simplifying the equation, we solve for \( V_2 \):
\[ V_2 = \frac{(155 \, \text{kPa} \times 1.00 \, \text{L} \times 400 \, \text{K})}{(300 \, \text{K} \times 605 \, \text{kPa})} \]
3. Perform the calculation:
\[ V_2 = \frac{62000}{181500} \approx 0.3415 \, \text{L} \]
Therefore, the final volume of the gas is approximately 0.3415 liters.
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 2 steps with 1 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Recommended textbooks for you
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY