Question

A 4.6-L sealed bottle containing 0.39 g of liquid ethanol, C2H6O, is placed in a refrigerator and reaches equilibrium with its vapor at −11°C. The vapor pressure of ethanol is 10. torr at −2.3°C and 40. torr at 19°C. (a) What mass of ethanol is present in the vapor? (ΔHvap = 40.5 kJ/mol, boiling point = 78.5°C)

Answer 1

Vapor pressure of ethanol at -11 deg. c needs to be calculated which can be done using Classius- Clayperon equation

Ln (P2/P1= (delH/R)*(1/T1-1/T2)

R = 8.314 j/mole.K delH= 40.5 Kj/mol= 40.5*1000 j/mol=40500 j/mole.

P2= vapor pressure at -11 deg.c= -11+273.15=262.15K

P1= vapor pressure at -2.3 deg.c (T1=-2.3+273.15)=270.85 K} = 10 torr

Ln (P2/10)= (40.5*1000/8.314)*(1/270.85-1/262.15)=-0.59688

P2/10 = exp(-0.59688)=0.55 P2= 5.5 torr

At -11 deg.c , the vapor is at equilibrium with the liquid

Partial pressure of ethanol vapor = vapor pressure of liquid = 5.5 torr

Assuming a pressure of one atmosphere =760 torr

Partial pressure of ethanol/ total pressure = moles of ethanol/ total moles

Total moles = PV/RT where P= 1 atm V= 4.6 L T= -11deg.c = 262.15K and R=0.08206

Number of moles of mixture= 1*4.6/(0.08206*262.15)=0.2138 moles

Moles of ethanol/ 0.2138= 5.5/(760-5.5)=0.0073

Moles of ethanol= 0.2138*0.0073=0.001561 moles

Molecular weight of ethanol = 46, mass of ethanol vapor =0.001561*46= 0.072 gms