A certain substance has a heat of vaporation 56.21 kj/mol. At what kelvin temperature will the vapor pressure be 6.50 times higher than it was at 361k?
Thank you
To determine the Kelvin temperature at which the vapor pressure will be 6.50 times higher than it was at 361K, we need to use the Clausius-Clapeyron equation:
ln(P2/P1) = ΔHvap/R * (1/T1 - 1/T2)
Where:
P1 = initial vapor pressure (at 361K)
P2 = final vapor pressure (6.50 times higher than P1)
ΔHvap = heat of vaporization (56.21 kJ/mol)
R = ideal gas constant (8.314 J/(mol·K))
T1 = initial temperature (361K)
T2 = final temperature (unknown, to be determined)
First, let's rewrite the equation to isolate T2:
ln(P2/P1) * R/ΔHvap = 1/T1 - 1/T2
Now we substitute the known values into the equation:
ln(6.50) * (8.314 J/(mol·K)) / (56.21 kJ/mol) = 1/361K - 1/T2
Evaluating the left side of the equation:
ln(6.50) * (8.314 J/(mol·K)) / (56.21 kJ/mol) ≈ 0.25135 K
Now we rearrange the equation to solve for T2:
1/T2 = 1/361K - 0.25135 K
Combining terms:
1/T2 ≈ 1/361K - 0.25135 K
Let's calculate the right-hand side of the equation:
1/T2 ≈ 0.00277 - 0.25135 K
Now, solve for T2 by taking the reciprocal of both sides:
T2 ≈ 1 / (0.00277 - 0.25135 K)
Evaluating the expression:
T2 ≈ 1 / -0.24858 K
T2 ≈ -4.02 K
The calculated value for T2 is -4.02 K, which suggests an error may have occurred during the calculation. Please double-check the values and calculations to ensure accuracy.