If the mole fraction of n-pentane in a liquid mixture is 0.43, and its partial pressure (vapor pressure) above the liquid is 0.24 bars, what is the vapor pressure of pure n-pentane at the same temperature (in bars)? Assume it is an ideal solution. 0.0 0.24 1.79 0.56 0.103
If the mole fraction of n-pentane in a liquid mixture is 0.43, and its partial pressure (vapor pressure) above the liquid is 0.24 bars, what is the vapor pressure of pure n-pentane at the same temperature (in bars)? Assume it is an ideal solution. 0.0 0.24 1.79 0.56 0.103
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![### Problem Statement:
If the mole fraction of n-pentane in a liquid mixture is 0.43, and its partial pressure (vapor pressure) above the liquid is 0.24 bars, what is the vapor pressure of pure n-pentane at the same temperature (in bars)? Assume it is an ideal solution.
### Options:
- ○ 0.0
- ○ 0.24
- ○ 1.79
- ○ 0.56
- ○ 0.103
### Solution:
To find the vapor pressure of pure n-pentane at the same temperature, you can use Raoult's Law, which is applicable to ideal solutions. Raoult's Law states:
\[ P_i = X_i \cdot P_i^0 \]
Where:
- \( P_i \) = partial pressure of component i in the mixture.
- \( X_i \) = mole fraction of component i in the liquid phase.
- \( P_i^0 \) = vapor pressure of pure component i.
Given:
- \( X_{\text{n-pentane}} = 0.43 \)
- \( P_{\text{n-pentane}} = 0.24 \) bar
Rearranging Raoult’s Law to solve for \( P_{\text{n-pentane}}^0 \):
\[ P_{\text{n-pentane}}^0 = \frac{P_{\text{n-pentane}}}{X_{\text{n-pentane}}} \]
\[ P_{\text{n-pentane}}^0 = \frac{0.24 \, \text{bars}}{0.43} \]
\[ P_{\text{n-pentane}}^0 \approx 0.56 \, \text{bars} \]
Therefore, the vapor pressure of pure n-pentane at the same temperature is approximately 0.56 bars.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3c924432-5356-4d2a-9799-f03f3a53c4ab%2Fe0962ee8-4dfc-4ffd-aa67-3b3d1b47ed1a%2Fmy31e8c_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
If the mole fraction of n-pentane in a liquid mixture is 0.43, and its partial pressure (vapor pressure) above the liquid is 0.24 bars, what is the vapor pressure of pure n-pentane at the same temperature (in bars)? Assume it is an ideal solution.
### Options:
- ○ 0.0
- ○ 0.24
- ○ 1.79
- ○ 0.56
- ○ 0.103
### Solution:
To find the vapor pressure of pure n-pentane at the same temperature, you can use Raoult's Law, which is applicable to ideal solutions. Raoult's Law states:
\[ P_i = X_i \cdot P_i^0 \]
Where:
- \( P_i \) = partial pressure of component i in the mixture.
- \( X_i \) = mole fraction of component i in the liquid phase.
- \( P_i^0 \) = vapor pressure of pure component i.
Given:
- \( X_{\text{n-pentane}} = 0.43 \)
- \( P_{\text{n-pentane}} = 0.24 \) bar
Rearranging Raoult’s Law to solve for \( P_{\text{n-pentane}}^0 \):
\[ P_{\text{n-pentane}}^0 = \frac{P_{\text{n-pentane}}}{X_{\text{n-pentane}}} \]
\[ P_{\text{n-pentane}}^0 = \frac{0.24 \, \text{bars}}{0.43} \]
\[ P_{\text{n-pentane}}^0 \approx 0.56 \, \text{bars} \]
Therefore, the vapor pressure of pure n-pentane at the same temperature is approximately 0.56 bars.
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