Benzene has a heat of vaporization of 30.72kJ/mol and a normal boiling point of 80.1 degrees C.
At what temperature does benzene boil when the external pressure is 405 torr?
how do i go about answering this question? i don't know where to start. am i supposed to use the clausius-clayperon equation?
333K
To determine the boiling temperature of benzene at a given external pressure, we can use the Clausius-Clapeyron equation:
ln(P1/P2) = (ΔH_vap/R) * ((1/T2) - (1/T1))
Where:
P1 = initial pressure (in this case, atmospheric pressure)
P2 = final pressure (external pressure)
ΔH_vap = heat of vaporization
R = gas constant (8.314 J/(mol·K))
T1 = initial temperature (normal boiling point)
T2 = final temperature (unknown, boiling temperature at given pressure)
We have:
P1 = atmospheric pressure = 1 atm
P2 = external pressure = 405 torr = 405/760 atm (since 1 atm = 760 torr)
ΔH_vap = 30.72 kJ/mol = 30.72 * 1000 J/mol
R = 8.314 J/(mol·K)
T1 = normal boiling point = 80.1 °C
After substituting the known values into the equation, we can solve for T2:
ln(1 / (405/760)) = (30.72 * 1000 / 8.314) * ((1/T2) - (1/353.25))
Simplifying the equation yields:
ln(760/405) = (30.72 * 1000 / 8.314) * ((1/T2) - (1/353.25))
Now, we can rearrange the equation and solve for T2:
T2 = 1 / ((1/353.25) + (8.314/(30.72 * 1000)) * ln(760/405))
By substituting the value of ln(760/405) and evaluating the expression, we can find the boiling temperature of benzene at the given external pressure.
Yes, to answer this question, you can use the Clausius-Clapeyron equation. This equation relates the boiling point of a substance to its heat of vaporization and the external pressure. The equation is:
ln(P2/P1) = ΔHvap / R * (1/T1 - 1/T2)
Where:
P1 and P2 are the initial and final pressures (in this case, P1 is the normal boiling point pressure and P2 is the given pressure of 405 torr)
ΔHvap is the heat of vaporization (given as 30.72 kJ/mol)
R is the ideal gas constant (8.314 J/mol·K)
T1 and T2 are the initial and final temperatures
To solve for the boiling point temperature when the pressure is 405 torr, you need to rearrange the equation as follows:
T2 = 1 / [(ΔHvap / R) * (1/T1 - ln(P2/P1))]
Now let's plug in the values:
P2 = 405 torr
P1 = pressure at the normal boiling point
ΔHvap = 30.72 kJ/mol
R = 8.314 J/mol·K
T1 = normal boiling point temperature in Kelvin (80.1 °C + 273.15 °C)
Since we need to convert temperature to Kelvin, let's calculate T1:
T1 = 80.1 °C + 273.15 °C = 353.25 K
Now, before plugging the values into the equation, we need to convert the units to match:
ΔHvap = 30.72 kJ/mol × 1000 J/kJ = 30,720 J/mol
R = 8.314 J/mol·K
Now we can plug in the numbers and calculate T2:
T2 = 1 / [(30,720 J/mol / 8.314 J/mol·K) * (1/353.25 K - ln(405 torr/P1))]
Please note that to determine the pressure at the normal boiling point (P1), you may need to consult a reference source or calculate it based on other given information about benzene.
Solving this equation will give you the boiling point temperature when the external pressure is 405 torr.