Question

A mixture of three gases has a total pressure of 786 mmHg. The mixture is 27% gas A and 53% gas B. What is the partial pressure due to the third gas?

212.2 mmHg

417 mmHg

157.2 mmHg

629.8 mmHg

Answer 1

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, \( P_B = 53\% \)

1. Calculate the partial pressure of gas A:
\[
P_A = \frac{27}{100} \times P_{\text{total}} = 0.27 \times 786 \, \text{mmHg} = 212.22 \, \text{mmHg}
\]

2. Calculate the partial pressure of gas B:
\[
P_B = \frac{53}{100} \times P_{\text{total}} = 0.53 \times 786 \, \text{mmHg} = 416.58 \, \text{mmHg}
\]

3. 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:
\[
P_C = P_{\text{total}} - P_A - P_B
\]

4. 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**.