The vapor pressure of methanol is 03 atm at 367 degree C Liq

The vapor pressure of methanol is 0.3 atm at 36.7 degree C. Liquid methanol is kept in a sealed flask @ 36.7 degree C; the gas above the liquid contains air and methanol vapor at a pressure of 1.28 atm. What is the partial pressure of methanol in the gas. the partial pressure of N_2, the mole fraction of acetone in the gas? How would you determine the boiling point of acetone, assuming a constant P_r of 1.28 atm?

Solution

a) In the flask, above the liquid methanol there is certain volume which is occupied by the mixture of methanol and air in gaseous form. Further the air is assumed as a mixture of nitrogen and oxygen.

Let us assume,

p_m = partial pressure of methanol

p_o = partial pressure of oxygen

p_n = partial pressure of nitrogen

So, the total pressure, p = p_m + p_o + p_n

Vapour pressure of methanol is 0.3 atm which means that at equilibrium state, the methanol vapour formed above the liquid methanol will exert a pressure of 0.3 atm in absence of any other gaseous substance. This is the partial pressure of methanol gas.

Hence, partial pressure of methanol in the gas is 0.3 atm. at 36.7°C.

b) The partial pressure of air is 1.28 - 0.3 = 0.98 atm

p_o + p_{n =} 0.98

The partial pressure of N₂ can be calculated from the formula, p_n*v = n*R*T ; where n = number of moles and R = universal gas constant = 8314.3 Nm/kg mol K.

Since the values of v and n are unknown and air contains about 78% of N₂, the partial pressure of N₂ acn be approximated to the partial pressure of air.

Hence, partial pressure of methanol in the gas is 0.98 atm. at 36.7°C.

c) Since the mixture does not contain acetone, the mole fraction of acetone in the gas = 0