xA and xB are mole fractions of each components - Vapour-liquid phase ? why is AI saying it's liquid phase composition ?

Henry's Law : p = KH x “partial pressure of as in vapour phase (p) is proportional to ole fraction of gas (x) in solution” and is expressed as:

Raoult’s Law: A=xA pA∘

pi = yi P total - this is the ratio of 2 components in vapour phase

AI Solved -

Given:
p°A = 450 mm Hg
p°B = 700 mm Hg
Total vapour pressure, P = 600 mm Hg

Let xA be the mole fraction of A in the liquid phase.
Then xB = 1 − xA

Using Raoult’s law:
P = xA·p°A + xB·p°B

600 = 450xA + 700(1 − xA)
600 = 700 − 250xA
xA = 0.40

Therefore:
xA = 0.40
xB = 0.60

Partial vapour pressures:
pA = xA·p°A = 0.40 × 450 = 180 mm Hg
pB = xB·p°B = 0.60 × 700 = 420 mm Hg

Vapour phase composition:
yA = pA / P = 180 / 600 = 0.30
yB = pB / P = 420 / 600 = 0.70

Final Answer:

Liquid phase composition:
xA = 0.40
xB = 0.60

Vapour phase composition:
yA = 0.30
yB = 0.70

8 hours ago, HbWhi5F said:

xA and xB are mole fractions of each components - Vapour-liquid phase ? why is AI saying it's liquid phase composition ?

Henry's Law : p = KH x “partial pressure of as in vapour phase (p) is proportional to ole fraction of gas (x) in solution” and is expressed as:

Raoult’s Law: A=xA pA∘

pi = yi P total - this is the ratio of 2 components in vapour phase

AI Solved -

Given:
p°A = 450 mm Hg
p°B = 700 mm Hg
Total vapour pressure, P = 600 mm Hg

Let xA be the mole fraction of A in the liquid phase.
Then xB = 1 − xA

Using Raoult’s law:
P = xA·p°A + xB·p°B

600 = 450xA + 700(1 − xA)
600 = 700 − 250xA
xA = 0.40

Therefore:
xA = 0.40
xB = 0.60

Partial vapour pressures:
pA = xA·p°A = 0.40 × 450 = 180 mm Hg
pB = xB·p°B = 0.60 × 700 = 420 mm Hg

Vapour phase composition:
yA = pA / P = 180 / 600 = 0.30
yB = pB / P = 420 / 600 = 0.70

Final Answer:

Liquid phase composition:
xA = 0.40
xB = 0.60

Vapour phase composition:
yA = 0.30
yB = 0.70

I am not interested in debating AI output.

How do you work it out?

Edited 23 hours ago23 hr by exchemist

I worked same solution but why is AI saying it's liquid phase composition ? It's -xA and xB are mole fractions of each components - Vapour-liquid phase ?

1 hour ago, HbWhi5F said:

I worked same solution but why is AI saying it's liquid phase composition ? It's -xA and xB are mole fractions of each components - Vapour-liquid phase ?

What liquid composition did you get?

@exchemist Given the context of the question AI is right but If I got the definattion right, it is not possible to be get the value (from the infomation in the book). Either I got the question wrong or the definetion.

Ok the conceptual problem is -

We got ratio amongst vapours

and ratios of the components amonst their phases

How are they connected ?

---

It should be something like -

Let V and L be phase of Vapour and LIquid in the solution

V/L = [1/Q1] / [(xA/y1) + (xB/y2)]/Q2

Q is some ratio to normalize the value from 1 to a fraction of the solution - it should be y1+y2/xa+xb

V/L=[1/(y1+y2/xa+xb)] / [(xA/y1) + (xB/y2)]/(y1+y2/xa+xb)

V/L=[1*xa+xb/y1+y2] / [(xA*y2+xB+y1/y1y2) / (y1+y2/xa+xb)]

0.30+0.70/L=xA + xB

1/L = 0.40+0.60

1/L=1

L=1/1

V/L=ya/xa

Edited 7 hours ago7 hr by HbWhi5F

14 minutes ago, HbWhi5F said:

@exchemist Given the context of the question AI is right but If I got the definattion right, it is not possible to be get the value (from the infomation in the book). Either I got the question wrong or the definetion.

Ok the conceptual problem is -

We got ratio amongst vapours

and ratios of the components amonst their phases

How are they connected ?

---

It should be something like -

Let V and L be phase of Vapour and LIquid in the solution

V/L = 0.30+0.70/L=xA + xB

1/L = 0.40+0.60

1/L=1

L=1/1

V/L=ya/xa

If you want to learn, you need to forget AI and work the problem out yourself. It sounds to me as if you have not done that. If you had, you would not be asking this question. Look up Raoult’s Law - then calculate for yourself the first part of the problem and show me how you did that.

Then we can talk about the second part.

Edited 7 hours ago7 hr by exchemist

@exchemist updated my answer

Also did i get the definations right

and the question ?

Edited 7 hours ago7 hr by HbWhi5F

17 minutes ago, HbWhi5F said:

@exchemist updated my answer

Also did i get the definations right

and the question ?

Just do what I asked you, and make a new post showing your reasoning.

I’m going to get up and have my breakfast now (it is 0730 in the UK) and then I can come back to this for you. Let’s take it step by step without the distraction of AI and get the logic clear.

Edited 6 hours ago6 hr by exchemist

Given:

p°A = 450 mm Hg (pure component's vapour pressure), °B = 700 mm Hg, = 600 mm Hg

Let xA and xB are mole ractions, So xB = 1 − xA

Using Raoult’s law: Total Vapor pressure = mole fraction of component A in it's vapour and liquid phase times the Vapour pressure of pure component A + same thing for B

P = xA·p°A + xB·p°B

600 = 450xA + 700(1 − xA)

600 = 700 − 250xA

xA = 0.40

xB = 1-.4 = 0.60

Partial vapour pressures:

pA=pA^0 * xA (Parital pressure of a component = vapoyr pressure of pre component times it's mole fraction)

pA = xA·p°A = 0.40 × 450 = 180 mm Hg

pB = xB·p°B = 0.60 × 700 = 420 mm Hg

pA and pB are vapour partial pressure, ie the pressure exerted by them on solution

Vapour phase composition:

pi = yi * P (partial pressure of a component = the ratio of it's vapour phase the vapour phase of the other component) times the total vapour pressure on solution

yi=pi/P

yA = pA / P = 180 / 600 = 0.30

yB = pB / P = 420 / 600 = 0.70

1 hour ago, HbWhi5F said:

Given:

p°A = 450 mm Hg (pure component's vapour pressure), °B = 700 mm Hg, = 600 mm Hg

Let xA and xB are mole ractions, So xB = 1 − xA

Using Raoult’s law: Total Vapor pressure = mole fraction of component A in it's vapour and liquid phase times the Vapour pressure of pure component A + same thing for B

P = xA·p°A + xB·p°B

600 = 450xA + 700(1 − xA)

600 = 700 − 250xA

xA = 0.40

xB = 1-.4 = 0.60

Partial vapour pressures:

pA=pA^0 * xA (Parital pressure of a component = vapoyr pressure of pre component times it's mole fraction)

pA = xA·p°A = 0.40 × 450 = 180 mm Hg

pB = xB·p°B = 0.60 × 700 = 420 mm Hg

pA and pB are vapour partial pressure, ie the pressure exerted by them on solution

Vapour phase composition:

pi = yi * P (partial pressure of a component = the ratio of it's vapour phase the vapour phase of the other component) times the total vapour pressure on solution

yi=pi/P

yA = pA / P = 180 / 600 = 0.30

yB = pB / P = 420 / 600 = 0.70

Yes exactly. So there is a greater mole fraction of B in the vapour phase than there is in the liquid phase, because it has a greater vapour pressure than A. In other words, B is the more volatile component. And you can work out the mole fractions in the vapour phase by multiplying the vapour pressure of each component by its mole fraction in the liquid phase.

So I don't see why you think there is a problem.

For a physical picture of what is happening, you can consider the molecules on the liquid surface, as it is from this layer that molecules escape to create the vapour phase. 40% of them on the surface are A and 60% of them are B.

But in pure liquid form, molecules of A have less of a tendency to escape than the molceules of B. That is what a lower vapour pressure of A than B implies.

So in the mixture, not only are there only 40% of A molecules in the surface layer but also, they have less of an intrinsic tendency to escape. Hence we get 30% A and 70% B in the vapour phase.

This is the principle behind fractional distillation. In the fractionating column, the mixture is made to evaporate and condense repeatedly as it rises up the column, each time creating a mixture with a higher proportion of the more volatile component. So that at the top of the column you get the more volatile component in almost pure form. (I once worked at an oil refinery🙂)

5 hours ago, exchemist said:

If you want to learn, you need to forget AI and work the problem out yourself. It sounds to me as if you have not done that. If you had, you would not be asking this question. Look up Raoult’s Law - then calculate for yourself the first part of the problem and show me how you did that.

Then we can talk about the second part.

I take my hat off to this post and your subsequent persistence. +1