Question

If 0.15 moles of solid carbon dioxide were added to a 6L rigid container at 25 degrees C, what was the final pressure in the container after the carbon dioxide sublimed? The container already had 740 toor of air and returned to 25 degrees c after sublimation.

Answer 1

moles of air initially present can be calculated from gas law, PV= nRT

whrere P= 740Torr, 760 Torr = 1 atm, P= 740/760 atm =0.973 atm, V= 6L, T= 25 deg.c= 25+273 K= 298K, R= gas constant = 0.0821 L.atm/mole.K

moles of air, n= PV/RT= 0.973*6/(0.0821*298)= 0.24 moles

when 0.15 moles of CO2 is added, total moles of air +Co2= 0.25+0.15= 0.39 moles, this is the new n.

again applying gas law, n= 0.39moles, T= 25 deg.c= 298K, R=0.0821 L.atm/mole.K

V= 6 L( it is a rigid container), P= nRT/V= 0.39*0.0821*298/6 atm =1.59 atm =1.59*760 Torr =1208.4 Torr