PRESSURE HELP!

Question

PRESSURE HELP!
The vapor pressure of liquid iodomethane, CH₃I, is 100. mm Hg at 266 K. A 0.541 g sample of liquid CH₃I is placed in a closed, evacuated 490. mL container at a temperature of 266 K.

Assuming that the temperature remains constant, will all of the liquid evaporate?

What will the pressure in the container be when equilibrium is reached? ______mm Hg

Answer 1

Not sure where David G was coming from, but try this...

PV = nRT and P = nRT/V

n = 0.541 g x 1 mol/142 g = 0.0038 moles

P = (0.0038 moles)(0.0821 L-atm/deg-mole)(266 deg) = 0.17 atm

0.17 atm x 760 mm Hg/atm = 129 mm Hg <--- new pressure

All the iodomethane will evaporate because the pressure is greater than that 100 mm Hg

Answer 2

So, to help..The liquid will stop boiling when the vapor pressure equals the pressure inside the container.

 

So use PV=nRT to figure out what n will equal P @ 100 torr

 

Alter the formula to solve for n...

 

n = PV/RT ...or (100 torr(same as mmHg) *  .490 L(units change) / (266 K * 62.36367)(gas constant in torr*L/Mol K) = 2.95 x 10^-3 mol to reach 100 torr

 

So - MM of Methyl Iodide = 141.39 g/mol

 

0.541g/141.39 g/mol = 3.81 x 10^-3 mol of Methyl Iodide

 

All of the liquid shouldn't evaporate.