To find the partial pressure of the third gas in the mixture, we first need to determine the partial pressures of gases A and B using their respective percentages.
Given:
- Total pressure, \( P_{\text{total}} = 786 , \text{mmHg} \)
- Percentage of gas A, \( P_A = 27% \)
- Percentage of gas B, \( P_B = 53% \)
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Calculate the partial pressure of gas A: \[ P_A = \frac{27}{100} \times P_{\text{total}} = 0.27 \times 786 , \text{mmHg} = 212.22 , \text{mmHg} \]
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Calculate the partial pressure of gas B: \[ P_B = \frac{53}{100} \times P_{\text{total}} = 0.53 \times 786 , \text{mmHg} = 416.58 , \text{mmHg} \]
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Use Dalton's law of partial pressures, which states that the total pressure is the sum of the partial pressures: \[ P_{\text{total}} = P_A + P_B + P_C \] Where \( P_C \) is the partial pressure of the third gas. Rearranging the equation gives: \[ P_C = P_{\text{total}} - P_A - P_B \]
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Substitute the values to find \( P_C \): \[ P_C = 786 , \text{mmHg} - 212.22 , \text{mmHg} - 416.58 , \text{mmHg} \] \[ P_C = 786 - 212.22 - 416.58 = 157.2 , \text{mmHg} \]
Therefore, the partial pressure due to the third gas is 157.2 mmHg.
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