i think i have posted this before except i forgot to put the decimal in the temperature. i was hoping you could help me now that it might make more sense.
The total pressure in a flask containing air and ethanol at 25.7 C is 878 mm Hg. The pressure of the air in the flask at 25.7 C is 762 mm Hg. If the flask is immersed in a water bath at 45.0 C, the total pressure is 980 mm Hg. The vapor pressure of ethanol at the new temperature is____ mm Hg.
I have already answered this question with the corrected decimal point. Please review previous posts
To find the vapor pressure of ethanol at the new temperature, we need to use the Clausius-Clapeyron equation. This equation relates the vapor pressure of a substance at one temperature to its vapor pressure at a different temperature.
The Clausius-Clapeyron equation is given by:
ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 = vapor pressure at temperature T1
P2 = vapor pressure at temperature T2
ΔHvap = heat of vaporization
R = ideal gas constant
T1 = initial temperature
T2 = final temperature
In this case, we know the initial temperature (25.7 C) and the final temperature (45.0 C). We want to find the vapor pressure of ethanol at the final temperature.
Before we can use the equation, we need to convert the temperatures from Celsius to Kelvin. The Kelvin scale starts at absolute zero (0 K) and is the same size as the Celsius scale, so we can convert by simply adding 273.15 to the Celsius temperature.
T1 = 25.7 C + 273.15 = 298.85 K
T2 = 45.0 C + 273.15 = 318.15 K
Now, we have the initial and final temperatures in Kelvin. We also need to know the value of the heat of vaporization (ΔHvap) and the ideal gas constant (R) for ethanol.
The heat of vaporization and the ideal gas constant will depend on the substance, so we need to look up their values. For ethanol, the heat of vaporization is approximately 38.56 kJ/mol, and the ideal gas constant is 0.0821 L·atm/(K·mol).
Now we can plug in all the known values into the Clausius-Clapeyron equation and solve for P2, which represents the vapor pressure of ethanol at the final temperature:
ln(P2/762 mm Hg) = -(38.56 kJ/mol / 0.0821 L·atm/(K·mol)) * (1/318.15 K - 1/298.85 K)
Simplifying the equation:
ln(P2/762) = -470.54 * (0.003146 - 0.003349)
ln(P2/762) = -470.54 * (-0.000203)
ln(P2/762) = 0.09553
To get the value of P2, take the exponential of both sides of the equation:
P2/762 = e^(0.09553)
P2 = 762 * e^(0.09553)
Using a calculator, we can evaluate this expression:
P2 ≈ 792.77 mm Hg
Therefore, the vapor pressure of ethanol at the new temperature of 45.0 C is approximately 792.77 mm Hg.