Question

The normal boiling point for acetone is 56.5°C. At an elevation of 5300 ft the atmospheric pressure is 630. torr. What would be the boiling point of acetone (ΔHvap = 32.0 kJ/mol) at this elevation? WebAssign will check your answer for the correct number of significant figures. °C What would be the vapor pressure of acetone at 24.0°C at this elevation?

Answer 1

Answer: Boiling point is 51.3 *C

Step-by-step explanation:

The variation of the vapor pressure p with temperature is given by the Clausius Clapeyron equation:

d(ln p)/dT = ?Hv / (RT²)

(?Hv heat of vaporization, R universal gas constant, T absolute temperature in Kelvins)

integration with initial condition p(T0) = p0 leads to:

ln(p/p0) = -?Hv/R · (1/T - 1/T0)

Thus the boiling temperature as function of ambient pressure p is:

T = 1 / [1/T0 - ln(p/p0) ·R / ?Hv]

For acetone at 630torr:

with T0 = (56.5 +273.15)K =329.65K

and p0 = 1atm = 760torr

T = 1 / [1/329.65 - ln(630/760) · 8.314J/(molK) / 32000J/mol]

= 324.45K = 51.3°C