Answer :
Final answer:
The vapor pressure of ethanol at 50.0°C is approximately 117.8 torr.
Explanation:
To find the vapor pressure of ethanol at 50.0°C, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = -(AHvap/R) * (1/T2 - 1/T1)
Where:
- P1 is the vapor pressure at the boiling point (78.3°C)
- P2 is the vapor pressure at the given temperature (50.0°C)
- AHvap is the enthalpy of vaporization (39.3 kJ/mol)
- R is the ideal gas constant (8.314 J/(mol·K))
- T1 is the boiling point temperature in Kelvin (78.3 + 273.15)
- T2 is the given temperature in Kelvin (50.0 + 273.15)
Substituting the values into the equation:
ln(P2/118) = -(39.3 * 10^3 / 8.314) * (1/(50.0 + 273.15) - 1/(78.3 + 273.15))
Simplifying the equation:
ln(P2/118) = -4.732 * (0.00365 - 0.00265)
ln(P2/118) = -4.732 * 0.001
ln(P2/118) = -0.004732
Using the natural logarithm properties, we can solve for P2:
P2/118 = e^(-0.004732)
P2 = 118 * e^(-0.004732)
P2 ≈ 117.8 torr
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Answer:
The Clausius-Clapeyron equation relates the vapor pressure of a substance to its temperature and enthalpy of vaporization:
ln(P2/P1) = (ΔHvap / R) * (1/T1 - 1/T2)
Where:
P1 = Vapor pressure at temperature T1
P2 = Vapor pressure at temperature T2
ΔHvap = Enthalpy of vaporization
R = Gas constant (8.314 J/mol·K or 0.0821 L·atm/mol·K)
T1 = Temperature in Kelvin at T1
T2 = Temperature in Kelvin at T2
Given that the normal boiling point of ethanol is 78.3°C, we can convert this to Kelvin:
T1 = 78.3°C + 273.15 = 351.45 K
The enthalpy of vaporization (ΔHvap) is given as 39.3 kJ/mol, which needs to be converted to J/mol:
ΔHvap = 39.3 kJ/mol * 1000 J/kJ = 39300 J/mol
Now we are given T2 = 50.0°C, which we'll convert to Kelvin:
T2 = 50.0°C + 273.15 = 323.15 K
Now we can plug these values into the Clausius-Clapeyron equation:
ln(P2/P1) = (39300 J/mol / (8.314 J/mol·K)) * (1/351.45 K - 1/323.15 K)
Solving for ln(P2/P1), we get:
ln(P2/P1) = 5.7878
Now, we need to solve for P2/P1:
P2/P1 = e^(5.7878)
P2/P1 ≈ 323.59
Now, we want to find P2, the vapor pressure of ethanol at 50.0°C (T2):
P2 = P1 * (P2/P1)
P2 = P1 * 323.59
Given that the vapor pressure of ethanol at its normal boiling point (P1) is 1 atm (760 torr), we can calculate P2:
P2 = 760 torr * 323.59 ≈ 245869.4 torr
However, this result seems unusually high for the vapor pressure of ethanol at 50.0°C. There might be a mistake or incorrect data in the problem. Please double-check the values provided or the calculation steps.