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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![**Boyle’s Law: Pressure-Volume Relationship**
**Problem Statement:**
If a gas at 25°C occupies 3.5L at a pressure of 0.99 atm, what would be the volume if the pressure is increased to 2.04 atm?
**Solution:**
This problem can be approached using Boyle's Law, which states that for a given mass of gas at constant temperature, the product of its pressure and volume is constant. Mathematically, it can be represented as:
\[ P_1V_1 = P_2V_2 \]
Where:
- \( P_1 \) = initial pressure = 0.99 atm
- \( V_1 \) = initial volume = 3.5 L
- \( P_2 \) = final pressure = 2.04 atm
- \( V_2 \) = final volume
From Boyle’s Law, we can solve for \( V_2 \):
\[ V_2 = \frac{P_1V_1}{P_2} \]
Substituting the values:
\[ V_2 = \frac{(0.99 \text{ atm})(3.5 \text{ L})}{2.04 \text{ atm}} = \frac{3.465}{2.04} \approx 1.70 \text{ L} \]
Thus, the volume of the gas when the pressure is increased to 2.04 atm is approximately 1.70 L.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcf4fd61a-46d9-45b2-8502-88fe4eac45cd%2F3f6052e5-824e-46d1-9024-3311f07b0ea3%2F7aqn81f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Boyle’s Law: Pressure-Volume Relationship**
**Problem Statement:**
If a gas at 25°C occupies 3.5L at a pressure of 0.99 atm, what would be the volume if the pressure is increased to 2.04 atm?
**Solution:**
This problem can be approached using Boyle's Law, which states that for a given mass of gas at constant temperature, the product of its pressure and volume is constant. Mathematically, it can be represented as:
\[ P_1V_1 = P_2V_2 \]
Where:
- \( P_1 \) = initial pressure = 0.99 atm
- \( V_1 \) = initial volume = 3.5 L
- \( P_2 \) = final pressure = 2.04 atm
- \( V_2 \) = final volume
From Boyle’s Law, we can solve for \( V_2 \):
\[ V_2 = \frac{P_1V_1}{P_2} \]
Substituting the values:
\[ V_2 = \frac{(0.99 \text{ atm})(3.5 \text{ L})}{2.04 \text{ atm}} = \frac{3.465}{2.04} \approx 1.70 \text{ L} \]
Thus, the volume of the gas when the pressure is increased to 2.04 atm is approximately 1.70 L.
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