A 8.30 L container holds a mixture of two gases at 29 °C. The partial pressures of gas A and gas B, respectively, are 0.189 atm and 0.661 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

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**Problem Statement:**

A 8.30 L container holds a mixture of two gases at 29 °C. The partial pressures of gas A and gas B, respectively, are 0.189 atm and 0.661 atm. If 0.130 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

\[ \text{P}_{\text{total}} = \text{________} \, \text{atm} \]

**Explanation:**

To find the total pressure after adding a third gas without changing the volume or temperature, use the Ideal Gas Law in the form of Dalton's Law of Partial Pressures. According to Dalton’s Law, the total pressure exerted by a mixture of gases is the sum of the partial pressures of each individual gas.

1. **Initial Total Pressure**: 
   - The initial total pressure is the sum of the partial pressures of gas A and gas B:
   \[
   \text{P}_{\text{initial}} = \text{P}_A + \text{P}_B = 0.189 \, \text{atm} + 0.661 \, \text{atm} = 0.850 \, \text{atm}
   \]

2. **Pressure Contribution of the Third Gas**:
   - Use the Ideal Gas Law \((PV = nRT)\) to find the pressure exerted by the additional 0.130 mol of gas. 
   - Assume \(R = 0.0821 \, \text{L atm/mol K}\) and convert temperature to Kelvin:
   \[
   T = 29 + 273.15 = 302.15 \, \text{K}
   \]
   - Solve for \(P\) where \(V = 8.30 \, \text{L}\), \(n = 0.130 \, \text{mol}\):
   \[
   P = \frac{nRT}{V} = \frac{(0.130 \, \text{mol})(0.0821 \, \text{L atm/mol K})(302.15 \, \text{K})}{8.30 \, \text{L}}
   \]

3. **Calculate the Total Pressure**:
   - Add the pressure exerted by the third gas to the initial total pressure to find \(\text
Transcribed Image Text:**Problem Statement:** A 8.30 L container holds a mixture of two gases at 29 °C. The partial pressures of gas A and gas B, respectively, are 0.189 atm and 0.661 atm. If 0.130 mol of a third gas is added with no change in volume or temperature, what will the total pressure become? \[ \text{P}_{\text{total}} = \text{________} \, \text{atm} \] **Explanation:** To find the total pressure after adding a third gas without changing the volume or temperature, use the Ideal Gas Law in the form of Dalton's Law of Partial Pressures. According to Dalton’s Law, the total pressure exerted by a mixture of gases is the sum of the partial pressures of each individual gas. 1. **Initial Total Pressure**: - The initial total pressure is the sum of the partial pressures of gas A and gas B: \[ \text{P}_{\text{initial}} = \text{P}_A + \text{P}_B = 0.189 \, \text{atm} + 0.661 \, \text{atm} = 0.850 \, \text{atm} \] 2. **Pressure Contribution of the Third Gas**: - Use the Ideal Gas Law \((PV = nRT)\) to find the pressure exerted by the additional 0.130 mol of gas. - Assume \(R = 0.0821 \, \text{L atm/mol K}\) and convert temperature to Kelvin: \[ T = 29 + 273.15 = 302.15 \, \text{K} \] - Solve for \(P\) where \(V = 8.30 \, \text{L}\), \(n = 0.130 \, \text{mol}\): \[ P = \frac{nRT}{V} = \frac{(0.130 \, \text{mol})(0.0821 \, \text{L atm/mol K})(302.15 \, \text{K})}{8.30 \, \text{L}} \] 3. **Calculate the Total Pressure**: - Add the pressure exerted by the third gas to the initial total pressure to find \(\text
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