Question

Acetone must be kept in a​ closed, pressure tank due to evaporation. Acetone evaporates at a rate of 44 grams per minute left bracket StartFraction g Over min EndFraction right bracket g min. If the tank holds 5454 ​pound-mass ​[lbm​] of acetone and is allowed to remain open to the​ atmosphere, how many days​ [d] will it take for half of the acetone to​ evaporate?

Answer 1

The vapor pressure of acetone at 25.0°C under 1 atm of pressure is approximately 2.05 atm.

Step-by-step explanation:

The molar entropy of vaporization of acetone under 1 atm of pressure can be calculated using the Clausius-Clapeyron equation. The equation is given as:

ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)

Where P1 and P2 are the initial and final vapor pressures, ΔHvap is the molar enthalpy of vaporization, R is the ideal gas constant, T1 and T2 are the initial and final temperatures respectively. By rearranging the equation, we can solve for the final vapor pressure:

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

Substituting the given values, we have:

P1 = 1 atm

T1 = 56°C = 329 K

T2 = 25°C = 298 K

ΔHvap = 31.3 kJ/mol = 31,300 J/mol

R = 8.314 J/(mol·K)

Plugging these values into the equation:

P2 = 1 atm * exp(31300 J/mol / (8.314 J/(mol·K)) * (1/329 K - 1/298 K))

Simplifying the equation gives:

P2 ≈ 2.05 atm

Therefore, the vapor pressure of acetone at 25.0°C is approximately 2.05 atm.