# Using Raoult's law... and knowing that #A# and #B# form an ideal binary mixture but #R# and #S# have negative deviation...?

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#a)# Show that #P_(AB) = P_B^@ + chi_A(P_A^@ - P_B^@)# .

#b)# If #P_A^@ > P_B^@# , #P_A^@ = P_R^@# and #P_B^@ = P_S^@# , sketch a diagram against the mol fraction of #B# for solution #AB# and of #S# for solution #RS# , showing how #P_A# , #P_B# , #P_R# , and #P_S# vary, and how #P_(AB)# and #P_(RS)# look.

##### 1 Answer

Okay, so the idea with this is to compare an **ideal binary mixture** with a **nonideal binary mixture** wherein *negative deviation* occurs upon creation of the mixture.

**"SATURATED" VAPOR PRESSURE**

If you remember Raoult's law, it dealt with the following relationship:

#\mathbf(P_j = chi_jP_j^@)# where:

#P_j# is thevapor pressureof the solutioncontainingsolute (i.e. non-pure).#chi_j = (n_j)/(n_i + n_j)# is themol fractionof solvent#j# in solution. When#chi_j darr# ,#chi_j < 1# , and#chi_i# , the mol fraction of solute#i# , is increasing.#P_j^@# is the vapor pressure of the solution containingNO solute at all(i.e. purely solvent). This is known as thevapor pressure of the pure solvent, or the "saturated" vapor pressure.

Let's just do the derivation part of the problem right here. Starting from Dalton's law of partial pressures, we just have, for the ideal binary mixture

#P_"tot" = color(blue)(P_(AB)^@) = P_A + P_B#

#= chi_AP_A^@ + chi_BP_B^@#

#= chi_AP_A^@ + (1-chi_A)P_B^@#

#= chi_AP_A^@ + P_B^@ - chi_AP_B^@#

#= color(blue)(P_B^@ + chi_A(P_A^@ - P_B^@))#

The physical interpretation of this result is that the total pressure can be computed for the mixture of any two substances, assuming ideality, by using their pure vapor pressures and by knowing how much

If you look back at how I did this derivation, you should notice that

*Thus, you should remember that*

**ENERGIES OF INTERACTION IN AN IDEAL BINARY MIXTURE**

For the *ideal* binary mixture **equal** to those between

#2epsilon_(AB) = epsilon_(A A) + epsilon_(BB)#

This says that the energy of the system *after* mixing **equal** to the energy of the system *before* mixing

**POSITIVE AND NEGATIVE DEVIATIONS FROM IDEAL MOLAR VOLUMES**

Now, we can have two variations on this.

#\mathbf(2epsilon_(AB) < epsilon_(A A) + epsilon_(BB))# (1)

In **(1)**, the energy

That means the particles, after mixing, have a lower overall energy, and thus, mixing is **favorable**. Thus, the average distance between **negative deviation**---the volume of the solution is **smaller** than for the ideal solution.

*That leads to a dip relative to Raoult's law behavior when the mol fraction of the solvent is less than #1# but that of the solute is not yet #1#.*

#\mathbf(2epsilon_(AB) > epsilon_(A A) + epsilon_(BB))# (2)

Similarly, in **(2)**, since the energy for the **unfavorable**, and the average distance between particles is larger; thus, the volume increases and we have **positive deviation**.

*That leads to a "hump" relative to Raoult's law behavior when the mol fraction of the solvent is less than #1# but that of the solute is not yet #1#.*

**THE AB VS THE RS SOLUTION**

**Based on your problem setup, what you're dealing with is ideality for the AB solution and negative deviation for the RS solution.**

For the ideal mixture, the *total vapor pressure*, ** horizontal** with the addition of solute

**only**

**at first**.

This is because there is *no* favorability in either the positive or negative direction when it comes to mixing *doesn't make a difference*.

Note that since we defined *decreases from*

For the **a dip downwards** in the graph of *total vapor pressure*,

When there is

When there is

Thus,