Benzene has a heat of vaporization of 30.72kj/mol and a normal boiling point of 80.1 C . At what temperature does benzene boil when the external pressure is 410 ? This is how I am solving this problem,

Question

Benzene has a heat of vaporization of 30.72kj/mol and a normal boiling point of 80.1 C . At what temperature does benzene boil when the external pressure is 410 ? This is how I am solving this problem,

ln(415/760)= (30720/8.314)*(1/80.1 - 1/T2)

It is asking to answer using two significant figures and it keeps telling me I'm incorrect. Any help would be appreciated.

Thanks.

Answer 1

ops I meant to put 410 in the equation.

Answer 2

To solve for the boiling temperature of benzene (T2) when the external pressure is 410 mmHg, you can rearrange the equation you provided using logarithmic properties:

ln(415/760) = (30720/8.314) * (1/80.1 - 1/T2)

Convert the external pressure from mmHg to atm by dividing it by 760:

ln(0.545) = (30720/8.314) * (1/80.1 - 1/T2)

Simplify the equation:

-0.605 = 3711.81 * (1/80.1 - 1/T2)

Now solve for T2. Start by isolating the term (1/80.1 - 1/T2):

1/80.1 - 1/T2 = -0.605 / 3711.81

Next, simplify and take the reciprocal of both sides:

T2/80.1 - 1/80.1 = 3711.81 / -0.605

Common denominators:

(T2 - 80.1)/80.1 = 3711.81 / -0.605

Cross-multiply:

-0.605 * (T2 - 80.1) = 3711.81 * 80.1

Simplify:

-0.605T2 + 48.5055 = 297151.681

Rearrange:

-0.605T2 = 297151.681 - 48.5055

-0.605T2 = 297103.1755

Divide both sides by -0.605:

T2 = 297103.1755 / -0.605

T2 ≈ -491,112.2

Since a negative temperature value doesn't make physical sense, it seems there may be an error in the calculation or the given information. Double-check the given values and your calculations to ensure accuracy.

Answer 3

To solve this problem correctly, you need to use the Clausius-Clapeyron equation, which relates the boiling point of a substance to the pressure and heat of vaporization. The equation is as follows:

ln(P1/P2) = (ΔHvap/R)*(1/T2 - 1/T1)

Where:
P1 is the initial pressure (760 mmHg or 1 atm)
P2 is the final pressure (410 mmHg or 0.54 atm)
ΔHvap is the heat of vaporization of benzene (30.72 kJ/mol or 30,720 J/mol)
R is the ideal gas constant (8.314 J/mol∙K)
T1 is the initial temperature (given as the normal boiling point, 80.1 °C or 353.25 K)
T2 is the final temperature (what you need to solve for)

Rearranging the equation to solve for T2, we have:

1/T2 = (ln(P1/P2) * R / ΔHvap) + (1/T1)

Substituting the values given:

1/T2 = (ln(1/0.54) * 8.314 J/mol∙K / 30,720 J/mol) + (1/353.25 K)

1/T2 ≈ (0.617 * 8.314 / 30,720) + (1/353.25)

1/T2 ≈ 0.0001686 + 0.0028329

1/T2 ≈ 0.0030015

T2 ≈ 333 K

To answer the question with two significant figures, the boiling point of benzene when the external pressure is 410 mmHg is approximately 333 K.

Answer 4

Did you and I work on this problem yesterday. If not, then let me tell you quickly where your problem(s) is/are. First you MUST use Kelvin for T1. That is 80.1 + 273.16. I think the problem asks for T in C BUT you must solve the problem first, using Kelvin, then convert back to C. Second, I see you typed the problem as 410 but substituted 415. Take care that you are solving the problem with the right numbers.