A 9.65 L container holds a mixture of two gases at 13 °C. The partial pressures of gas A and gas B, respectively, are 0.419 atm and 0.693 atm. If 0.130 mol of a third gas is added with no change in volume or temperature, what will the total pressure become? Ptotal = atm
A 9.65 L container holds a mixture of two gases at 13 °C. The partial pressures of gas A and gas B, respectively, are 0.419 atm and 0.693 atm. If 0.130 mol of a third gas is added with no change in volume or temperature, what will the total pressure become? Ptotal = atm
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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![**Problem Description:**
A 9.65 L container holds a mixture of two gases at 13 °C. The partial pressures of gas A and gas B, respectively, are 0.419 atm and 0.693 atm. If 0.130 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?
**Equation:**
\[ P_{\text{total}} = \]
**Unit for Total Pressure:**
atm
**Explainer:**
To find the total pressure, we can use Dalton's Law of Partial Pressures, which states that the total pressure of a gas mixture is equal to the sum of the partial pressures of the individual gases. Additionally, you can calculate the partial pressure of the third gas using the ideal gas law equation:
\[ P = \frac{nRT}{V} \]
Where:
- \( P \) is the pressure in atm,
- \( n \) is the number of moles,
- \( R \) is the ideal gas constant \( (0.0821 \, \text{L atm/mol K}) \),
- \( T \) is the temperature in Kelvin,
- \( V \) is the volume in liters.
Convert the temperature from Celsius to Kelvin by adding 273.15. Then, calculate the partial pressure of the third gas and add it to the existing partial pressures to find \( P_{\text{total}} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa563c0cc-a924-4c37-9f6b-c4caff62891a%2F2ff2ee67-dcff-4bc5-85b3-d38a8a214ce6%2Fztcl1f7_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Description:**
A 9.65 L container holds a mixture of two gases at 13 °C. The partial pressures of gas A and gas B, respectively, are 0.419 atm and 0.693 atm. If 0.130 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?
**Equation:**
\[ P_{\text{total}} = \]
**Unit for Total Pressure:**
atm
**Explainer:**
To find the total pressure, we can use Dalton's Law of Partial Pressures, which states that the total pressure of a gas mixture is equal to the sum of the partial pressures of the individual gases. Additionally, you can calculate the partial pressure of the third gas using the ideal gas law equation:
\[ P = \frac{nRT}{V} \]
Where:
- \( P \) is the pressure in atm,
- \( n \) is the number of moles,
- \( R \) is the ideal gas constant \( (0.0821 \, \text{L atm/mol K}) \),
- \( T \) is the temperature in Kelvin,
- \( V \) is the volume in liters.
Convert the temperature from Celsius to Kelvin by adding 273.15. Then, calculate the partial pressure of the third gas and add it to the existing partial pressures to find \( P_{\text{total}} \).
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