A certain substance has a heat of vaporization of 55.30 kJ/mol. At what Kelvin temperature will the vapor pressure be 3.50 times higher than it was at 361 K?
Use the Clausius-Clapeyron equation.
ln(p2/p1) = (dHvap/R)(1/T1 - 1/T2)
make p2 = 3.5p1.Remember T must be in kelvin.
To solve this problem, you need to use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its molar heat of vaporization and temperature. The equation is as follows:
ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 is the initial vapor pressure (361 K in this case)
P2 is the final vapor pressure (3.5 times P1)
ΔHvap is the molar heat of vaporization (55.30 kJ/mol)
R is the ideal gas constant (8.314 J/(mol·K))
T1 is the initial temperature (361 K)
T2 is the final temperature (what we need to find)
Now, let's plug in the values into the equation:
ln(3.5P1/P1) = -(55.30 kJ/mol)/(8.314 J/(mol·K)) * (1/T2 - 1/361 K)
ln(3.5) = -(55.30 kJ/mol)/(8.314 J/(mol·K)) * (1 - 1/361 K)
Now we can solve the equation for T2:
ln(3.5) = -(55.30 kJ/mol)/(8.314 J/(mol·K)) * (360/361)
ln(3.5) = -6.6636 * (360/361)
To isolate T2, we divide both sides by -6.6636:
T2 = (-(55.30 kJ/mol)/(8.314 J/(mol·K)) * (360/361)) / ln(3.5)
T2 ≈ 342 K
So, at approximately 342 Kelvin temperature, the vapor pressure will be 3.50 times higher than it was at 361 K.