What will be the mole fraction of at equilibrium at 1000

Signup with Email. Gender Male Female. Create Account. Already Have an Account? A sparingly soluble gas solute is in equilibrium with a solvent at 10 bar. The mole fraction of the solvent in the gas phase is 0. Answer to Question. Jaswant Kaur Mar 17, This discussion on A sparingly soluble gas solute is in equilibrium with a solvent at 10 bar. For each mole of sodium chloride dissolved, you get 1 mole of sodium ions and 1 mole of chloride ions - in other words, you get twice the number of moles of particles as of original salt.

So, if you added 0. Unless you think carefully about it, Raoult's Law only works for solutes which do not change their nature when they dissolve. For example, they must not ionize or associate e. If it does either of these things, you have to treat Raoult's law with great care.

The effect of Raoult's Law is that the saturated vapor pressure of a solution is going to be lower than that of the pure solvent at any particular temperature. That has important effects on the phase diagram of the solvent.

The next diagram shows the phase diagram for pure water in the region around its normal melting and boiling points. The 1 atmosphere line shows the conditions for measuring the normal melting and boiling points. The line separating the liquid and vapor regions is the set of conditions where liquid and vapor are in equilibrium.

It can be thought of as the effect of pressure on the boiling point of the water, but it is also the curve showing the effect of temperature on the saturated vapor pressure of the water. These two ways of looking at the same line are discussed briefly in a note about half-way down the page about phase diagrams follow the last link above.

If you draw the saturated vapor pressure curve for a solution of a non-volatile solute in water, it will always be lower than the curve for the pure water. If you look closely at the last diagram, you will see that the point at which the liquid-vapor equilibrium curve meets the solid-vapor curve has moved. That point is the triple point of the system - a unique set of temperature and pressure conditions at which it is possible to get solid, liquid and vapor all in equilibrium with each other at the same time.

Since the triple point has solid-liquid equilibrium present amongst other equilibria , it is also a melting point of the system - although not the normal melting point because the pressure is not one atmosphere.

The curves for the pure water and for the solution are often drawn parallel to each other. That has got to be wrong! Suppose you have a solution where the mole fraction of the water is 0. The vapor pressure of the solution will be 99 kPa - a fall of 1 kPa. At a lower temperature, where the vapor pressure of the pure water is 10 kPa, the fall will only be 0. For the curves to be parallel the falls would have to be the same over the whole temperature range. They aren't! That must mean that the phase diagram needs a new melting point line a solid-liquid equilibrium line passing through the new triple point.

That is shown in the next diagram. Now we are finally in a position to see what effect a non-volatile solute has on the melting and freezing points of the solution. Look at what happens when you draw in the 1 atmosphere pressure line which lets you measure the melting and boiling points.

The diagram also includes the melting and boiling points of the pure water from the original phase diagram for pure water black lines.

We have looked at this with water as the solvent, but using a different solvent would make no difference to the argument or the conclusions. The only difference is in the slope of the solid-liquid equilibrium lines. For most solvents, these slope forwards whereas the water line slopes backwards. You could prove to yourself that that does not affect what we have been looking at by re-drawing all these diagrams with the slope of that particular line changed. You will find it makes no difference whatsoever.

We can calculate the vapor pressure of the solution in two ways, depending on the volatility of the solute. If the solute is volatile, it will exert its own vapor pressure and this amount is a significant contribution to the overall vapor pressure of the solution, and thus needs to be included in the calculations. On the other hand, if it is nonvolatile, the solute will not produce vapor pressure in solution at that temperature. These calculations are fairly straightforward if you are comfortable with stoichiometric conversions.

The standard conditions are 70 F and cfm. The diameter of the stack is found to be 1. Solution : Actual volumetric flow rates are always used to calculate stack discharge velocity. Step 2 : Calculate crossectional area of the stack. Discharge vel. Note : The velocity calculated in the above problem is average velocity. It provides information about the flow behavior of the fluid, i. The density of the gas is 0.