A 8.35 L container holds a mixture of two gases at 21 °C. The partial pressures of gas A and gas B, respectively, are 0.273 atm and 0.566 atm. If 0.120 mol of a third gas is added with no change in volume or temperature, what will the total pressure become? Ptotal = atm
A 8.35 L container holds a mixture of two gases at 21 °C. The partial pressures of gas A and gas B, respectively, are 0.273 atm and 0.566 atm. If 0.120 mol of a third gas is added with no change in volume or temperature, what will the total pressure become? Ptotal = atm
Introductory Chemistry: A Foundation
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ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Donald J. DeCoste
Chapter15: Solutions
Section: Chapter Questions
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![**Problem Statement:**
A 8.35 L container holds a mixture of two gases at 21 °C. The partial pressures of gas A and gas B, respectively, are 0.273 atm and 0.566 atm. If 0.120 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?
**Solution:**
To find the total pressure \((P_{\text{total}})\), follow these steps:
1. **Identify the given data:**
- Volume of container \((V) = 8.35 \, L\)
- Temperature \((T) = 21 °C\) (must be converted to Kelvin for calculations: \(21 + 273.15 = 294.15 \, K\))
- Partial pressure of gas A \((P_{A}) = 0.273 \, atm\)
- Partial pressure of gas B \((P_{B}) = 0.566 \, atm\)
- Amount of third gas added \((n_{\text{C}}) = 0.120 \, mol\)
2. **Use the ideal gas law to determine the partial pressure of the added gas \(P_{\text{third gas}}\):**
- The ideal gas law is \( PV = nRT \)
- Rearrange to solve for pressure: \( P = \frac{nRT}{V} \)
- Use the values:
- \( n = 0.120 \, mol \)
- \( R = 0.0821 \, \text{atm} \cdot \text{L} / (\text{mol} \cdot \text{K}) \)
- \( T = 294.15 \, K \)
- \( V = 8.35 \, L \)
- Calculation:
\[
P_{\text{third gas}} = \frac{(0.120 \, \text{mol}) \cdot (0.0821 \, \text{atm} \cdot \text{L} / (\text{mol} \cdot \text{K})) \cdot (294.15 \, K)}{8.35 \, L}
\]
3. **Calculate \(P_{\text{third gas}}\):**
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2531879c-adde-4e1b-ab06-85ed93219951%2F6ac744bf-b0a2-4d88-a456-86671d849c99%2Fqfnmamo.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A 8.35 L container holds a mixture of two gases at 21 °C. The partial pressures of gas A and gas B, respectively, are 0.273 atm and 0.566 atm. If 0.120 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?
**Solution:**
To find the total pressure \((P_{\text{total}})\), follow these steps:
1. **Identify the given data:**
- Volume of container \((V) = 8.35 \, L\)
- Temperature \((T) = 21 °C\) (must be converted to Kelvin for calculations: \(21 + 273.15 = 294.15 \, K\))
- Partial pressure of gas A \((P_{A}) = 0.273 \, atm\)
- Partial pressure of gas B \((P_{B}) = 0.566 \, atm\)
- Amount of third gas added \((n_{\text{C}}) = 0.120 \, mol\)
2. **Use the ideal gas law to determine the partial pressure of the added gas \(P_{\text{third gas}}\):**
- The ideal gas law is \( PV = nRT \)
- Rearrange to solve for pressure: \( P = \frac{nRT}{V} \)
- Use the values:
- \( n = 0.120 \, mol \)
- \( R = 0.0821 \, \text{atm} \cdot \text{L} / (\text{mol} \cdot \text{K}) \)
- \( T = 294.15 \, K \)
- \( V = 8.35 \, L \)
- Calculation:
\[
P_{\text{third gas}} = \frac{(0.120 \, \text{mol}) \cdot (0.0821 \, \text{atm} \cdot \text{L} / (\text{mol} \cdot \text{K})) \cdot (294.15 \, K)}{8.35 \, L}
\]
3. **Calculate \(P_{\text{third gas}}\):**
\[
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