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
- Calculate the volume of 85 g of chlorine gas at 73 oC and 1.54 atm.
![**Problem Statement:**
Calculate the volume of 85 g of chlorine gas at 73 °C and 1.54 atm.
**Explanation:**
To solve this problem, we will use the Ideal Gas Law, which is given by the formula:
\[ PV = nRT \]
Where:
- \( P \) is the pressure (in atm),
- \( V \) is the volume (in liters),
- \( n \) is the number of moles,
- \( R \) is the ideal gas constant (0.0821 L·atm/K·mol),
- \( T \) is the temperature (in Kelvin).
**Steps:**
1. **Convert the temperature to Kelvin:**
\[ T(K) = 73 + 273.15 = 346.15 \, K \]
2. **Calculate the number of moles of chlorine gas:**
The molar mass of chlorine (Cl₂) is approximately 70.9 g/mol.
\[ n = \frac{85 \, g}{70.9 \, g/mol} = 1.20 \, mol \]
3. **Use the Ideal Gas Law to find the volume:**
\[ V = \frac{nRT}{P} \]
\[ V = \frac{(1.20 \, mol)(0.0821 \, L·atm/K·mol)(346.15 \, K)}{1.54 \, atm} \]
4. **Calculate the volume:**
\[ V \approx \frac{1.20 \times 0.0821 \times 346.15}{1.54} \]
\[ V \approx 22.1 \, L \]
Thus, the volume of 85 g of chlorine gas at 73 °C and 1.54 atm is approximately **22.1 liters**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbcea870b-a558-4e25-bd85-f50b869e1110%2Fb2096e51-9d44-4660-911f-2a9c918d51d2%2Flityie9_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Calculate the volume of 85 g of chlorine gas at 73 °C and 1.54 atm.
**Explanation:**
To solve this problem, we will use the Ideal Gas Law, which is given by the formula:
\[ PV = nRT \]
Where:
- \( P \) is the pressure (in atm),
- \( V \) is the volume (in liters),
- \( n \) is the number of moles,
- \( R \) is the ideal gas constant (0.0821 L·atm/K·mol),
- \( T \) is the temperature (in Kelvin).
**Steps:**
1. **Convert the temperature to Kelvin:**
\[ T(K) = 73 + 273.15 = 346.15 \, K \]
2. **Calculate the number of moles of chlorine gas:**
The molar mass of chlorine (Cl₂) is approximately 70.9 g/mol.
\[ n = \frac{85 \, g}{70.9 \, g/mol} = 1.20 \, mol \]
3. **Use the Ideal Gas Law to find the volume:**
\[ V = \frac{nRT}{P} \]
\[ V = \frac{(1.20 \, mol)(0.0821 \, L·atm/K·mol)(346.15 \, K)}{1.54 \, atm} \]
4. **Calculate the volume:**
\[ V \approx \frac{1.20 \times 0.0821 \times 346.15}{1.54} \]
\[ V \approx 22.1 \, L \]
Thus, the volume of 85 g of chlorine gas at 73 °C and 1.54 atm is approximately **22.1 liters**.
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