Ethanol has an enthalpy of evaporation of 38.6kJ/mol, and a normal boiling point of 78.4oC. What is the vapor pressure of ethanol at 15oC?

Hint: the normal boiling point is when the vapor pressure reaches 1 atm. Use,
. Pay attention to use right units

equation: ln(P2/P1)=(delta H/R)((1/T1)-(1/T1)

1st:T1=78.4+273=354.4K
T2= 15+273=286K
R=8.3154J/mol
delta H=38.6*1000=38600J/mol
2nd: ln(P2/1atm)=((8600J/mol/8.3145J/mol)((1/354.4)-(1/286))= 3.0211
3rd: natural e^3.0211= 20.513
4th: (P2/1atm)=(1/20.513)
5th: P2=(1atm/20.513)=answer: 4.8750x10^-2

Je me comprends pas Dr. Neutron

To find the vapor pressure of ethanol at 15oC, we can use the Clausius-Clapeyron equation:

ln(P2/P1) = -(ΔHvap/R) * ((1/T2) - (1/T1))

Where:
P1 = vapor pressure at the boiling point (1 atm)
P2 = vapor pressure at the desired temperature (unknown)
ΔHvap = enthalpy of vaporization (38.6 kJ/mol)
R = gas constant (0.008314 kJ/(mol*K))
T1 = boiling point temperature (78.4oC + 273.15 = 351.35 K)
T2 = desired temperature (15oC + 273.15 = 288.15 K)

Now, let's solve the equation step by step:

ln(P2/1) = -(38.6 kJ/mol / 0.008314 kJ/(mol*K)) * ((1/288.15 K) - (1/351.35 K))

Simplifying the equation:

ln(P2) = -4659.92 * (0.00346767 - 0.00284555)

Calculating the difference inside the brackets:

ln(P2) = -4659.92 * 0.00062212

Multiplying the values:

ln(P2) = -2.89672944

To solve for P2, we can raise e to the power of both sides:

e^(-2.89672944) = P2

Solving this equation:

P2 ≈ 0.0542 atm

Therefore, the vapor pressure of ethanol at 15oC is approximately 0.0542 atm.

To find the vapor pressure of ethanol at 15°C, we can use the Clausius-Clapeyron equation. The equation is as follows:

ln(P1/P2) = (ΔHvap/R) * ((1/T2) - (1/T1))

Where:
P1 is the vapor pressure at one temperature (in this case, the boiling point)
P2 is the vapor pressure at the other temperature (in this case, 15°C)
ΔHvap is the enthalpy of evaporation (38.6 kJ/mol)
R is the ideal gas constant (8.314 J/(mol*K))
T1 is the temperature at P1 (78.4°C converted to Kelvin)
T2 is the temperature at P2 (15°C converted to Kelvin)

Let's calculate the vapor pressure using the given information:

First, let's convert the temperatures to Kelvin:
T1 = (78.4 + 273.15) K = 351.6 K
T2 = (15 + 273.15) K = 288.15 K

Now, let's plug the values into the Clausius-Clapeyron equation:
ln(P1/P2) = (38.6 * 10^3 J/mol) / (8.314 J/(mol*K)) * ((1/288.15 K) - (1/351.6 K))

Now, let's calculate the right side of the equation:
(38.6 * 10^3 J/mol) / (8.314 J/(mol*K)) * ((1/288.15 K) - (1/351.6 K)) ≈ 3.178

Now, let's solve for ln(P1/P2). We can exponentiate both sides to get rid of the natural logarithm:
e^(ln(P1/P2)) = e^(3.178)
P1/P2 = e^(3.178)
P1/P2 ≈ 24.151

Finally, let's solve for P2 (the vapor pressure at 15°C):
P2 = P1 / (P1/P2)
P2 = 1 atm / 24.151
P2 ≈ 0.0414 atm

Therefore, the vapor pressure of ethanol at 15°C is approximately 0.0414 atm.