Question

Liquid ethyl ether is introduced drop by drop into a 3.0 L closed container at 18 degrees Celsius. After 122 drops of ethyl ether are added. No more evaporates and one drop is present in the container. What is the pressure in the container. The volume of one drop is 0.05 ml and the density of ethyl ether is 0.713 g/mL. The molar mass of ethyl ether is 74.12 g/ml. R=0.0821

The answer is 0.46 atm. Can you please show me how to do this problem?

Answer 1

since 122 drops are added, one drop of liquid remains, no of drops evaporate= 122-1= 121 drops

no of moles of evaporated, n= volume of single drop* no of drops* density ( mass of total drops)/ molar mass of ethyl ether = (121drops* 0.05 ml/drop*0.732 g/ml ) /74.12 =0.0597 moles

The pressure in the container is due to vaporized drops. The pressure can be calculated from gas law equation, P= nRT/V

V= volume of container =3 L, R= gas law constant =0.0821 L.atm/mole.K, T= 18 deg.c= 18+273= 291K

P= 0.0597*0.0821*291/3 =0.475 atm