Question

The vapor pressure of ethanol is 400. mmhg at 63.5°c. its molar heat of vaporization is 39.3 kj/mol. what is the vapor pressure of ethanol, in mmhg, at 34.9°c? (r = 8.314 j/k • mol)

Answer 1

Answer: -

100 mm Hg

Explanation: -

P 1 =400 mm Hg

T 1 = 63.5 C + 273 = 336.5 K

T 2 = 34.9 C + 273 = 307.9 K

ΔHvap = 39.3 KJ/mol = 39.3 x 10³ J mol⁻¹

R = 8.314 J ⁻¹K mol⁻¹

Now using the Clausius Clapeyron equation

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

Plugging in the values

ln (400 mm/ P₂) = (39.3 x 10³ J mol⁻¹ / 8.314 J ⁻¹K mol⁻¹) x ($\frac{1}{307.9 K}$ - $\frac{1}{336.5 K}$

= 1.38

P₂ = 100 mm Hg