benzene has a heat of vaporization of 30.72kj/mol and a normal boiling point of 80.1c. at what temperature does benzene boil when the external pressure is 445 torr

To determine the boiling temperature of benzene at a given external pressure, we need to use the Clausius-Clapeyron equation, which relates the boiling point to the heat of vaporization, the gas constant, and the external pressure.

The Clausius-Clapeyron equation is as follows:

ln(P1/P2) = (ΔHvap/R) * (1/T2 - 1/T1)

where:
P1 = initial pressure (normal boiling point, in this case)
P2 = final pressure (given as 445 torr)
ΔHvap = heat of vaporization
R = gas constant (8.314 J/(mol*K))
T1 = initial temperature (normal boiling point, in this case)
T2 = final temperature (the boiling temperature we are trying to find)

We need to convert all units to SI units before plugging them into the equation.

ΔHvap = 30.72 kJ/mol = 30.72 * 1000 J/mol
T1 = 80.1°C = 80.1 + 273.15 K
P1 = 1 atm (since it is the normal boiling point)
P2 = 445 torr = 445 / 760 atm
R = 8.314 J/(mol*K)

Now let's plug all the values into the Clausius-Clapeyron equation:

ln(1/(445/760)) = (30.72 * 1000 J/mol) / (8.314 J/(mol*K)) * (1/T2 - 1/(80.1 + 273.15))

Simplifying:

ln(760/445) = (30.72 * 1000) / 8.314 * (1/T2 - 1/(80.1 + 273.15))

ln(760/445) = 3704.928 / (1/T2 - 1/353.25)

Now let's solve for T2 by rearranging the equation:

1/T2 - 1/353.25 = (3704.928 / ln(760/445))

Taking the reciprocal of both sides:

T2 = 1 / (1/353.25 + (3704.928 / ln(760/445)))

Calculating T2 using the equation:

T2 ≈ 80.61°C

Therefore, at an external pressure of 445 torr, benzene boils at approximately 80.61°C.

To determine the boiling point of benzene at a given pressure, we can use the Clausius-Clapeyron equation:

ln(P1/P2) = (ΔHvap/R) * (1/T2 - 1/T1)

Where:
P1 = initial pressure (usually taken as 1 atm or 760 torr)
P2 = final pressure
ΔHvap = heat of vaporization (in this case, for benzene it is 30.72 kJ/mol)
R = gas constant (8.314 J/(mol·K) or 0.0821 L·atm/(mol·K))
T1 = initial temperature
T2 = final temperature (boiling point)

First, convert the pressure from torr to atm (1 atm = 760 torr):
P2 = 445 torr / 760 torr/atm
P2 ≈ 0.585 atm

Now, rearrange the Clausius-Clapeyron equation to solve for T2:

ln(P1/P2) = (ΔHvap/R) * (1/T2 - 1/T1)

Since we want to find the boiling point (T2) at a given external pressure, we know P1, P2, and ΔHvap. We can rearrange the equation to solve for T2:

1/T2 = (ln(P1/P2) * R / ΔHvap) + 1/T1

Now, substitute the known values into the equation:

1/T2 = (ln(1 atm/0.585 atm) * (0.0821 L·atm/(mol·K)) / (30.72 kJ/mol) + 1/(80.1 + 273.15) K

Now, solve for T2:

T2 = 1 / [(ln(1 atm/0.585 atm) * (0.0821 L·atm/(mol·K)) / (30.72 kJ/mol) + 1/(80.1 + 273.15)] K

T2 ≈ 353.6 K

Finally, convert the temperature from Kelvin to Celsius:

T2 ≈ 353.6 K - 273.15 K
T2 ≈ 80.5°C

Therefore, benzene boils at approximately 80.5°C when the external pressure is 445 torr.

The Clausius–Clapeyron relation is to be sued.

Tb=(R ln (Po)/deltaH - 1/To)

Po is 445torr (convert that)
To is what you are looking for
deltaHv given
Tb=80.1C convert that to K

R is universal gas constant. Watch units.