Ethanol has a heat of vaporization of 38.56 KJ/Mol. And a normal boiling point of 78.4 degrees celcius.
WHAT is the vapor pressure of ethanol at 18 degrees celcius??
Note the correct spelling of celsius.
Use the Clausius-Clapeyron equation. One of the points is the normal boiling point (where the vapor pressure is 760 torr) and the other point is vapor pressure at 18 C. Don't forget to use Kelvin for temperature.
Thanks. But I'm still getting it wrong, cause maybe the sig figs are off by a little. What did you get for this problem?
Show your work (best in a new post together with the problem) so it will get to the top of the list again) and I will help you find the error.
To determine the vapor pressure of ethanol at 18 degrees Celsius, you can use the Clausius-Clapeyron equation. The equation is as follows:
ln(P2/P1) = (∆Hvap/R) * (1/T1 - 1/T2)
Where:
P1 = Initial vapor pressure (at the boiling point)
P2 = Vapor pressure at the given temperature (18 degrees Celsius)
∆Hvap = Heat of vaporization (38.56 kJ/mol)
R = Ideal Gas Constant (8.314 J/(mol*K))
T1 = Initial temperature (boiling point in Kelvin)
T2 = Given temperature (18 degrees Celsius)
First, convert the given temperature to Kelvin by adding 273.15:
T2 = 18 + 273.15 = 291.15 K
The boiling point of ethanol is given as 78.4 degrees Celsius. Convert it to Kelvin as well:
T1 = 78.4 + 273.15 = 351.55 K
Substituting the values into the equation:
ln(P2/P1) = (38.56 * 10^3 J/mol) / (8.314 J/(mol*K)) * (1/351.55 K - 1/291.15 K)
Simplifying the equation, we are left with:
ln(P2/P1) = (38.56 * 10^3 J/mol) * (1/8.314 J/(mol*K)) * (1/351.55 K - 1/291.15 K)
ln(P2/P1) = 47.23 * (1/351.55 K - 1/291.15 K)
Now, we can solve for P2/P1 by taking the exponent of both sides of the equation:
P2/P1 = e^(47.23 * (1/351.55 K - 1/291.15 K))
Finally, multiply the vapor pressure at the boiling point (P1) by P2/P1 to find the vapor pressure at 18 degrees Celsius:
P2 = P1 * (P2/P1)
Please provide the actual boiling point (in Celsius) of ethanol for an accurate calculation.