Acetone has a delta Hvap of 29.1 kj/mol and a normal boiling point of 56.2 degrees Celsius. At what temperature (in degrees Celsius) does acetone have a Pvap = 118 mm Hg?

Question

Acetone has a delta Hvap of 29.1 kj/mol and a normal boiling point of 56.2 degrees Celsius. At what temperature (in degrees Celsius) does acetone have a Pvap = 118 mm Hg?

Answer 1

Use the Clausius-Clapeyron equation.

ln(p2/p1) = (dHvap/R)(1/T1 - 1/T2)
Substitute 29,100 for dHvap, 56.2 for T1 and 760 mm for p1. Then solve for T2 when p2 = 118. I've used celsius for both temperatures; don't forget to change them to kelvin before substituting.

Answer 2

To determine the temperature at which acetone has a vapor pressure (Pvap) of 118 mm Hg, we can use the Clausius-Clapeyron equation. The equation relates the vapor pressure of a substance to its enthalpy of vaporization (delta Hvap), the gas constant (R), and the temperature (T).

Here is the Clausius-Clapeyron equation:

ln(P2/P1) = (-delta Hvap/R) * (1/T2 - 1/T1)

Where:
P1 = initial pressure
P2 = final pressure
delta Hvap = enthalpy of vaporization
R = gas constant (8.31 J/molK)
T1 = initial temperature
T2 = final temperature

In this case, we are given:
Pvap = final pressure = 118 mm Hg
delta Hvap = 29.1 kJ/mol (Note: We need to convert this to J/mol)
T1 = initial temperature = normal boiling point of acetone = 56.2 degrees Celsius = 329.35 K
R = gas constant = 8.31 J/molK

First, we need to convert delta Hvap from kJ/mol to J/mol:
delta Hvap = 29.1 kJ/mol = 29,100 J/mol

Now, we can substitute the given values into the equation:

ln(118/760) = (-29,100/8.31) * (1/T2 - 1/329.35)

Simplifying further:

ln(118/760) = (-3,500.3) * (1/T2 - 0.00304)

Next, we need to rearrange the equation to solve for T2:

1/T2 - 0.00304 = (ln(118/760))/(-3,500.3)

1/T2 = (ln(118/760))/(-3,500.3) + 0.00304

Now, solve for T2:

T2 = 1 / [(ln(118/760))/(-3,500.3) + 0.00304]

T2 ≈ 299.47 K

Converting this temperature to degrees Celsius:

T2 ≈ 299.47 - 273.15 ≈ 26.32 degrees Celsius

Therefore, acetone has a vapor pressure of 118 mm Hg at approximately 26.32 degrees Celsius.

Answer 3

To determine the temperature at which acetone has a vapor pressure (Pvap) of 118 mm Hg, we can use the Clausius-Clapeyron equation:

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

where:
P1 = initial pressure = normal boiling point (in this case, 56.2 degrees Celsius)
P2 = final pressure = 118 mm Hg
ΔHvap = enthalpy of vaporization = 29.1 kJ/mol
R = ideal gas constant = 8.314 J/(mol·K)
T1 = initial temperature (boiling point in kelvin)
T2 = final temperature (unknown)

First, convert the boiling point from Celsius to Kelvin:
T1 = 56.2 + 273.15 = 329.35 K

Rearrange the equation to solve for T2:

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

Now, substitute the given values into the equation:

(1/T2 - 1/329.35) = (ln(118/760) / (-29100 / 8.314))

ln(118/760) is the natural logarithm of the ratio of the final pressure to the normal boiling point pressure. In this case, the atmospheric pressure of 760 mm Hg is chosen as the reference point.

Now, simplify the equation:

1/T2 - 1/329.35 = 0.2563

To isolate 1/T2, add 1/329.35 to both sides:

1/T2 = 0.2563 + 1/329.35

Combine the terms on the right side:

1/T2 = 1.283 + 0.0030

1/T2 = 1.286

Now, take the reciprocal of both sides:

T2 = 1 / 1.286 ≈ 0.777

Finally, convert T2 from Kelvin to Celsius by subtracting 273.15:

T2 = (0.777 - 273.15) ≈ -272.37°C

Therefore, acetone has a vapor pressure of 118 mm Hg at approximately -272.37 degrees Celsius.