Answer :
Sure! Let's go through the problem step-by-step using the Clausius-Clapeyron equation:
### Given Data
1. Normal boiling point of acetone: \( T_2 = 56.5^\circ C \)
2. Enthalpy of vaporization of acetone: \( \Delta H_{vap} = 31.3 \frac{kJ}{mol} \)
3. Temperature at which vapor pressure is to be found: \( T_1 = 25.0^\circ C \)
4. Universal gas constant: \( R = 8.3145 \frac{J}{mol\cdot K} \)
### Step 1: Convert Temperatures to Kelvin
- Normal boiling point in Kelvin:
[tex]\[ T_2 = 56.5^\circ C + 273.15 = 329.65 K \][/tex]
- Temperature at which to find vapor pressure in Kelvin:
[tex]\[ T_1 = 25.0^\circ C + 273.15 = 298.15 K \][/tex]
### Step 2: Convert Enthalpy of Vaporization to \( \frac{J}{mol} \)
[tex]\[ \Delta H_{vap} = 31.3 \frac{kJ}{mol} \times 1000 = 31300 \frac{J}{mol} \][/tex]
### Step 3: Use the Clausius-Clapeyron Equation
[tex]\[ \ln \left(\frac{P_1}{P_2}\right) = \frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \][/tex]
We are asked to find the vapor pressure at 25.0°C. We know at the normal boiling point, the vapor pressure \( P_2 \) is 1 atm.
### Step 4: Substitute the Known Values
[tex]\[ \ln \left(\frac{P_1}{1}\right) = \frac{31300}{8.3145} \left(\frac{1}{329.65} - \frac{1}{298.15}\right) \][/tex]
### Step 5: Calculate the Right-Hand Side
1. Calculate the temperature reciprocal difference:
[tex]\[ \frac{1}{329.65} - \frac{1}{298.15} = 0.003034 - 0.003354 = -0.000320 \][/tex]
2. Calculate the numerator:
[tex]\[ \frac{31300}{8.3145} = 3764.5 \][/tex]
3. Multiply the results:
[tex]\[ 3764.5 \times (-0.00032) = -1.20464 \][/tex]
### Step 6: Solve for \( P_1 \)
[tex]\[ \ln P_1 = -1.20464 \][/tex]
[tex]\[ P_1 = e^{-1.20464} \][/tex]
### Step 7: Calculate the Vapor Pressure
[tex]\[ P_1 \approx 0.299 \, \text{atm} \][/tex]
Therefore, the vapor pressure of acetone at [tex]\( 25.0^\circ C \)[/tex] is approximately [tex]\( 0.299 \, \text{atm} \)[/tex].
### Given Data
1. Normal boiling point of acetone: \( T_2 = 56.5^\circ C \)
2. Enthalpy of vaporization of acetone: \( \Delta H_{vap} = 31.3 \frac{kJ}{mol} \)
3. Temperature at which vapor pressure is to be found: \( T_1 = 25.0^\circ C \)
4. Universal gas constant: \( R = 8.3145 \frac{J}{mol\cdot K} \)
### Step 1: Convert Temperatures to Kelvin
- Normal boiling point in Kelvin:
[tex]\[ T_2 = 56.5^\circ C + 273.15 = 329.65 K \][/tex]
- Temperature at which to find vapor pressure in Kelvin:
[tex]\[ T_1 = 25.0^\circ C + 273.15 = 298.15 K \][/tex]
### Step 2: Convert Enthalpy of Vaporization to \( \frac{J}{mol} \)
[tex]\[ \Delta H_{vap} = 31.3 \frac{kJ}{mol} \times 1000 = 31300 \frac{J}{mol} \][/tex]
### Step 3: Use the Clausius-Clapeyron Equation
[tex]\[ \ln \left(\frac{P_1}{P_2}\right) = \frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \][/tex]
We are asked to find the vapor pressure at 25.0°C. We know at the normal boiling point, the vapor pressure \( P_2 \) is 1 atm.
### Step 4: Substitute the Known Values
[tex]\[ \ln \left(\frac{P_1}{1}\right) = \frac{31300}{8.3145} \left(\frac{1}{329.65} - \frac{1}{298.15}\right) \][/tex]
### Step 5: Calculate the Right-Hand Side
1. Calculate the temperature reciprocal difference:
[tex]\[ \frac{1}{329.65} - \frac{1}{298.15} = 0.003034 - 0.003354 = -0.000320 \][/tex]
2. Calculate the numerator:
[tex]\[ \frac{31300}{8.3145} = 3764.5 \][/tex]
3. Multiply the results:
[tex]\[ 3764.5 \times (-0.00032) = -1.20464 \][/tex]
### Step 6: Solve for \( P_1 \)
[tex]\[ \ln P_1 = -1.20464 \][/tex]
[tex]\[ P_1 = e^{-1.20464} \][/tex]
### Step 7: Calculate the Vapor Pressure
[tex]\[ P_1 \approx 0.299 \, \text{atm} \][/tex]
Therefore, the vapor pressure of acetone at [tex]\( 25.0^\circ C \)[/tex] is approximately [tex]\( 0.299 \, \text{atm} \)[/tex].
