Question

What is the vapour pressure of a `0.44*mol` quantity of bromine at a temperature of `298.15*K`?

Answer 1

Well, what is the vapour pressure of bromine at `25` `""^@C`...?

Explanation:

It is not something I want to find out for myself, bromine is one of the most corrosive substances you can handle, and can cause horrendous burns. You got a pot of bromine well below its normal boiling point of `58.8` `""^@C`... It will express a vapour pressure (dependent on temperature), but most of the stuff will be in the liquid phase.

If the question wished to test your knowledge of the gas laws they should have specified dioxygen, or dinitrogen, or helium....

Answer 2

Bromine will exert a pressure of 0.280 atm.

Explanation:

If bromine were an ideal gas, the Ideal Gas Law would predict

`p = (nRT)/V = (0.44 color(red)(cancel(color(black)("mol"))) × "0.082 06" color(red)(cancel(color(black)("L")))·"atm"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 298.15 color(red)(cancel(color(black)("K"))))/(10.2 color(red)(cancel(color(black)("L"))))= "1.1 atm"`

However, bromine is not an ideal gas. It condenses to a liquid at 58.8 °C.

A pressure of 1.1 atm is greater than the vapour pressure of bromine at 25 °C (0.280 atm).

Thus, at 25 °C, bromine will condense to a liquid until the pressure equals its vapour pressure of 0.280 atm.