Question

The normal boiling point of acetone, (CH3)2CO, is 56.5°C and the enthalpy of vaporization is 31.3kJ/mol. At what temperature (in °C) would acetone boil at a higher altitude where the atmospheric pressure is 226 mmHg?

Answer 1

At an altitude where the atmospheric pressure is 226 mmHg, acetone would boil close to 0°C.

Step-by-step explanation:

The boiling point of a liquid is the temperature at which its vapor pressure equals the atmospheric pressure. At higher altitudes, the atmospheric pressure is lower, meaning that the boiling point of a liquid is also lower.

To find the boiling point of acetone at a higher altitude where the atmospheric pressure is 226 mmHg, we can use the Clausius-Clapeyron equation. The equation is: ln(P₁/P₂) = -ΔHvap/R * (1/T₂ - 1/T₁), where P₁ and T₁ are the vapor pressure and temperature at the normal boiling point, P₂ is the desired vapor pressure, ΔHvap is the enthalpy of vaporization, R is the ideal gas constant, and T₂ is the desired boiling point.

Let's substitute the given values into the equation and solve for T₂:

1. P₁ = 101.3 kPa = 760 mmHg (standard atmospheric pressure)
2. T₁ = 56.5°C = 329.7 K (normal boiling point)
3. P₂ = 226 mmHg (desired vapor pressure)
4. ΔHvap = 31.3 kJ/mol
5. R = 8.314 J/(mol·K)

Substituting the values into the equation and solving for T₂, we get:

T₂ = -ΔHvap/R * (1/T₂ - 1/T₁) + 1/T₁

T₂ = (-31.3 kJ/mol / 8.314 J/(mol·K)) * (1/((226 mmHg / 760 mmHg) / 329.7 K) - 1/329.7 K) + 1/329.7 K

T₂ ≈ -237.45 K

Since temperature cannot be negative, we can round the result to zero. Therefore, at an altitude where the atmospheric pressure is 226 mmHg, acetone would boil close to 0°C.

Answer 2

95.41°C

Step-by-step explanation:

T1 = 56.5°C = 329.65K

∇Hᵥₐ = 31.3kJ/mol

p2 = 226mmHg = 0.301bar

p1 = 1bar

R = ideal gas constant = 0.008314kJ/Mol.K

From definition, the normal boiling point is at 1 bar which relates to Clausius-Clapeyron equation relates to vapour pressure to temperature.

In(P₂/P₁) = -∇Hᵥₐ /R (1/T₂ - 1/T₁)

solving for T₂,

R / ∇Hᵥₐ . In(P₂/P₁) = 1/T₂ - 1/T₁

1/T₂ = 1/T₁ - R / ∇Hᵥₐ * In(P₂/P₁)

1/T₂ = (1 / 329.65) - (0.008314 / 31.3) In (0.30 / 1)

1/T₂ = (3.033*10⁻³) - (2.656*10⁻⁴ * -1.204)

1/T₂ = 3.033*10⁻³ - 3.197*10⁻⁴

1/T₂ = 2.713*10⁻³

T₂ = 368.56K

T2 = 95.41°C