At what temperature does benzene boil when the external pressure is 440 torr ?
ln (k2/k1) = (delta H/R)(1/T1 - 1/T2)
You will need to look up delta H. I dont have those numbers at my finger tips. Post your work if you get stuck.
k1 and k2 are the vapor pressures @ T1 and T2
T1 will be 80.1 C. T2 is the unknown. k1 is the vapor pressure at T1 which is 760 torr. k2 is the vapor pressure of the problem or 440 torr.
To determine the boiling point of benzene at a pressure of 440 torr, we can consult a phase diagram or use the Clausius-Clapeyron equation.
However, since the phase diagram for benzene might not be readily available, we can use the Clausius-Clapeyron equation to approximate the boiling point.
The Clausius-Clapeyron equation is written as:
ln(P1/P2) = ΔHvap/R * (1/T2 - 1/T1)
Where:
P1 and P2 are the initial and final pressures respectively
ΔHvap is the heat of vaporization
R is the ideal gas constant
T1 and T2 are the initial and final temperatures respectively
In this case, we know the initial pressure (P1) is the atmospheric pressure, which is typically 760 torr. The boiling point of benzene at this pressure is 80.1 °C.
Plugging in the values:
ln(440/760) = ΔHvap/R * (1/T2 - 1/80.1)
We need to rearrange the equation and solve for T2 (the boiling point at 440 torr).
1/T2 = [ln(440/760) * R/ΔHvap] + 1/80.1
Now we can substitute the values:
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
ΔHvap = 31.4 kJ/mol (heat of vaporization of benzene)
Converting the units, the equation becomes:
1/T2 = [ln(440/760) * 0.0821 L·atm/(mol·K) / (31.4 kJ/mol)] + 1/80.1
Calculating this equation will give us the reciprocal of T2 (1/T2). By taking the reciprocal of that value, we can find the boiling point at 440 torr.