Pressure Help!

Question

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

Symmeatra,

In order, the steps you need to follow to solve this problem are:

1) determine the molecular mass of CH₃I from its formula

2) determine how many moles of CH₃I the stated mass provided represents

3) plug this # moles into the general gas law pV = nRT ; use the appropriate form of R for the conditions you are given, or else convert convert those conditions first to more convenient expressions -- p in mm Hg to KPa, for example

You must leave one condition "open", pressure for example, and solve for it assuming the liquid all vaporizes to an ideal gas state

4) if this value is >100 mm Hg, then liquid will only vaporize to supply 100 mm Hg pressure that represents equilibrium between gas and liquid for CH₃I. If, on the other hand, the value is <100 mm Hg, then all the liquid will vaporize, and the vapor will be unsaturated with respect to condensation to the liquid.

Hope this sets you on your way,

-- S. de Riel

Answer 2

Hello Symmeatra,

 

Use PV = nRT. You know V, n, R, and T. If the pressure you calculate is greater than 100 mm Hg, then the liquid will not all evaporate since its vapor pressure maxes out at 100 mm Hg. (It doesn't boil away since iodomethane boils at 315 K.) If the calculated pressure is less than 100 mm Hg, all of the liquid will evaporate and the equilibrium pressure will be the pressure you calculated.

 

Try solving the problem now before reading further.

 

 

 

 

 

 

 

Still need more guidance? Use V = 0.49 L, n = 0.541 g * mol CH₃I/141.9 g = 3.81*10^-3 mol. Use R = 0.08206 L*atm/(K*mol), and convert atmospheres to mm Hg using 760 mm Hg = 1 atm.

 

Hope this helps...I've well paved the path you can now take.

 

 

Eric M.

McMinnville OR