The normal boiling point of benzene is 80 degree C and its heat of vaporizations is 30.8 kj/mol. What is its boiling point at the top of Pikes Peak where P=260 mmHg?
Use the Clausius-Clapeyron equation.
Thanks Dr. B, I know to use that equation but the math is the part that I'm stuck on.
ln(P1/P2) = deltaHvap/R(1/T2 - 1/T1)
ln(760/260) = 30,800/8.314(1/T2 - 1/353)
Solve for T2. You can check you numbers with this. If those numbers are ok you can post your work and I'll try to find the error.
Yeah that's the same setup I have but I'm not sure once I divide 30800/8.314 if I take that number and multiply by the other side of the equation or if I'm supposed to FOIL it out with the temps in parenthesis? Thanks!
To determine the boiling point of benzene at the top of Pikes Peak, we need to use the Clausius-Clapeyron equation, which relates the boiling point of a substance to its vapor pressure at different temperatures. The equation is as follows:
ln(P1/P2) = (ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 and P2 are the initial and final vapor pressures respectively,
ΔHvap is the heat of vaporization,
R is the ideal gas constant,
T1 is the initial temperature (normal boiling point), and
T2 is the final temperature (boiling point at Pikes Peak).
We are given:
P1 = 1 atm (since the normal boiling point is given at standard atmospheric pressure)
T1 = 80°C ( conversion to Kelvin: T1 = 80 + 273 = 353 K)
ΔHvap = 30.8 kJ/mol (conversion to J/mol: ΔHvap = 30.8 * 10^3 J/mol)
P2 = 260 mmHg (conversion to atm: P2 = 260 / 760 atm)
Now we can plug these values into the Clausius-Clapeyron equation and solve for T2:
ln(1/P2) = (ΔHvap/R) * (1/T2 - 1/T1)
ln(1/(260/760)) = (30.8 * 10^3 J/mol) / (8.314 J/(mol·K)) * (1/T2 - 1/353)
ln(760/260) = (30.8 * 10^3) / (8.314) * (1/T2 - 1/353)
Let's simplify the equation:
ln(38/13) = 3714.12 * (1/T2 - 1/353)
Now, we can rearrange the equation to solve for T2:
(1/T2 - 1/353) = ln(38/13) / 3714.12
1/T2 = (ln(38/13) / 3714.12) + 1/353
Now, solve for T2:
T2 = 1 / [(ln(38/13) / 3714.12) + 1/353]
Using a calculator, you can find T2 ≈ 74.7°C.
Therefore, the boiling point of benzene at the top of Pikes Peak, where the pressure is 260 mmHg, is approximately 74.7°C.