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??
Okay...I don't really get out to solve for it actually.
ln (P2/760 torr)= -38560/8.3145 * (1/291.15 - 1/351.15).
I got ln x/760 = 2.87 ( I must have made an error) and I got stuck right here.
Can someone find me the answer to this??? Answer has to be in 2 sig figs.
I solved it and got 42.94 torr, so by 2 sig figs i rounded it up to 43 torr. What did you get?
nevermind. i got it. thank you dr. bob for showing me how to do it tho
To find the vapor pressure of ethanol at 18 degrees Celsius, you can use the Clausius-Clapeyron equation:
ln(P2 / P1) = -(ΔHvap / R) * (1/T2 - 1/T1)
Where:
P1 = vapor pressure at the boiling point (78.4 degrees Celsius)
P2 = vapor pressure at the desired temperature (18 degrees Celsius)
ΔHvap = heat of vaporization (38.56 kJ/mol)
R = ideal gas constant (8.3145 J/(mol·K))
T1 = boiling point temperature in Kelvin (78.4 degrees Celsius + 273.15 = 351.15 K)
T2 = desired temperature in Kelvin (18 degrees Celsius + 273.15 = 291.15 K)
Now, let's plug in the values into the equation:
ln(P2 / 760 torr) = -(38.56 kJ/mol) / (8.3145 J/(mol·K)) * (1/291.15 K - 1/351.15 K)
First, convert the heat of vaporization to J/mol:
ΔHvap = 38.56 kJ/mol * (1000 J/1 kJ) = 38,560 J/mol
Next, calculate the temperature difference:
1/291.15 K - 1/351.15 K = 0.003434 - 0.002849 = 0.000585 K^-1
Now, substitute the values into the equation:
ln(P2 / 760 torr) = -(38,560 J/mol) / (8.3145 J/(mol·K)) * (0.000585 K^-1)
Simplifying further:
ln(P2 / 760 torr) = -16690.87
To solve for P2, we need to isolate it. Take the exponential of both sides of the equation:
P2 / 760 torr = e^(-16690.87)
P2 = 760 torr * e^(-16690.87)
Now you can calculate P2 using a scientific calculator or online calculator:
P2 ≈ 0.0098 torr
Rounding to 2 significant figures, the vapor pressure of ethanol at 18 degrees Celsius is approximately 0.0098 torr.