I need help in solving this problem:
A certain substance has a heat of vaporization of 42.46 kJ/mol. At what Kelvin temperature will the vapor pressure be 5.00 times higher than it was at 291 K?
I know that I have to use the Clausius-Clapeyron equation, but what would I be solving for essentially?
Thanks for the help!!!
You know p1 = P
and p2 = 5P1
You know T1 = 291
You solve for T2.
Awesome, thank you so much for the help!
To solve this problem, you would be solving for the temperature (in Kelvin) at which the vapor pressure of the substance is 5.00 times higher than it was at 291 K.
In the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its heat of vaporization and temperature, you have two variables: the temperature (T) and the vapor pressure (P). The equation is given by:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 is the initial vapor pressure (at T1)
P2 is the final vapor pressure (at T2)
ΔHvap is the heat of vaporization
R is the ideal gas constant
In this case, you are given the initial temperature T1 (291 K) and the factor by which the vapor pressure increases (5.00 times higher), which gives you the final vapor pressure P2. You need to solve for the final temperature T2.
To find T2, you can rearrange the equation:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
Since you know the value of ln(P2/P1), ΔHvap, R, and T1, you can substitute these values into the equation and solve algebraically for T2.
Note: Make sure to convert the heat of vaporization (ΔHvap) from kJ/mol to J/mol, and the value of R depends on the units used for ΔHvap (8.314 J/(mol*K) for ΔHvap in J/mol, or 0.008314 kJ/(mol*K) for ΔHvap in kJ/mol).