The normal boiling point of acetone is 56.2 (degrees)C and the molar heat of vaporization is 32.0 kj/mol. At what temperature will acetone boil under a pressure of 50.0 mmHg?
Use the Clausius-Clapeyron equation.
To determine the boiling point of acetone under a pressure of 50.0 mmHg, we can use the Clausius-Clapeyron equation:
ln(P1/P2) = ΔHvap/R * (1/T2 - 1/T1)
Where:
P1 = initial pressure (normal boiling point)
P2 = final pressure (50.0 mmHg)
ΔHvap = molar heat of vaporization (32.0 kJ/mol)
R = gas constant (8.314 J/(mol*K))
T1 = initial temperature (normal boiling point)
T2 = final temperature (unknown)
Let's substitute the known values into the equation:
ln(P1/P2) = ΔHvap/R * (1/T2 - 1/T1)
ln(760 mmHg / 50.0 mmHg) = (32.0 kJ/mol) / (8.314 J/(mol*K)) * (1/T2 - 1/329.2 K)
Now we can solve for T2. First, let's simplify the equation:
ln(15.2) = 3.8558 / T2 - 0.0118
Next, rearrange the equation to solve for 1/T2:
3.8558 / T2 = ln(15.2) + 0.0118
Now, solve for 1/T2:
1/T2 = 3.8558 / (ln(15.2) + 0.0118)
Finally, calculate T2:
T2 = 1 / (3.8558 / (ln(15.2) + 0.0118))
T2 ≈ 47.4°C
Therefore, acetone will boil at approximately 47.4°C under a pressure of 50.0 mmHg.
To find the boiling temperature of a substance under a given pressure, we can use the Clausius-Clapeyron equation:
ln(P1/P2) = (ΔH_vap/R) * (1/T2 - 1/T1)
Where:
P1 is the initial pressure (normal boiling point pressure)
P2 is the final pressure (50.0 mmHg)
ΔH_vap is the molar heat of vaporization (32.0 kj/mol)
R is the ideal gas constant (8.314 J/(mol·K))
T1 is the initial temperature (normal boiling point temperature)
T2 is the final temperature (the one we want to find)
First, let's convert the pressure from mmHg to kPa:
1 mmHg = 0.13332 kPa
50.0 mmHg * 0.13332 kPa/mmHg = 6.666 kPa
Now we can rearrange the equation to solve for T2:
ln(P1/P2) = (ΔH_vap/R) * (1/T2 - 1/T1)
ln(P1/P2) / (ΔH_vap/R) = 1/T2 - 1/T1
1/T2 = ln(P1/P2) / (ΔH_vap/R) + 1/T1
T2 = 1 / (ln(P1/P2) / (ΔH_vap/R) + 1/T1)
Substituting the known values:
P1 = pressure at normal boiling point = 1 atm = 101.3 kPa
T1 = normal boiling point temperature = 56.2°C = 329.35 K
T2 = 1 / (ln(P1/P2) / (ΔH_vap/R) + 1/T1)
T2 = 1 / (ln(101.3 kPa / 6.666 kPa) / (32.0 * 10^3 J/mol / (8.314 J/(mol·K))) + 1/329.35 K)
Now we can calculate T2.