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Phase diagram
rleung:
Hi,
I am a little confused about the following. An unknown chemical compound, X, has a molar mass of 74 g/mol, a liquid density of 1.14 g/mL, and obeys the phase diagram illustrated below. Suppose you execute the following two steps:
Step 1: 6.32-g of X(l) is injected into an initially evacuated 3.0-L container at 200-K.
Step 2: The equilibrated sample from step 1 is compressed to half its original volume while the temperature remains constant at 200-K.
(a) Describe what happens to the system immediately following each of the two steps.
1) I am slightly confused on this. I used PV=nRT to find the pressure conditions in the container immediately after step 1. So after step 1, I have P=nRT/V
P=(6.32-g/74g/mol)(0.0821)(200-K) / 3-L
P=0.467 atm
On the phase diagram for that pressure and temperature, it is in the gas phase. Am I correct in saying that the sample exists as a gas and will STAY AS as a gas under this same condition of pressure and temperature provided the container is sealed and not disturbed?
2) After step 2, since the V is halved, the pressure INITIALLY will be doubled, so 0.467atm * 2= 0.934atm. According to the phase diagram, that combination of pressure and temperature exists in the liquid phase, so does all of the sample condense to a liquid INITIALLY?
3) However, since the temperature is still the same, the system will eventually equilibrate, right? My question is, where does it equlibrate to? Does it equilibrate to T=200-K and P=0.75 atm (the place at T=200-K where liquid and gas exist in equilibrium)? If it does equilibrate there, I am confused as to how the pressure just "comes to" be 0.75 atm. Do systems usually just equilibrate to the equilibrium lines in phase diagrams?
Sorry for the simplicity of the questions, This was on a worksheet I found, but I can't find anywhere in my textbook that really relates phase diagrams to these types of problems. Thanks so much! :)
Ryan
Donaldson Tan:
I agree with what you did for part(a)
for part(b), the sample must be at equilibrium. This is because during the isothermal compression, the pressure will increase until it reaches the equilibrium point, ie. 0.7atm, 200K. Any more work done during the equilibrium will be used to convert the gas phase to the liquid phase, until all the gas phase is consumed, then the pressure of the system will continue to increase until the volume is half the original.
let X be molar fraction of liquid phase.
molar composition of vapour phase is thus 1 - X
total number of moles in the system N = 6.32/74 = 0.0854054 moles
Since the final point requires the volume to be halved, so the final volume is 1.5L
assuming perfect gas, volume of vapour phase = nRTeqbm/Peqbm where n = (1-X)N
volume of vapour phase = (1-X)NRTeqbm/Peqbm
volume of liquid phase = XN.M/DL where M is the molar mass and DL is the liquid density.
Volume of System = Volume of Vapour Phase + Volume of Liquid Phase
1.5L = XN.M/DL + (1-X)NRTeqbm/Peqbm
Donaldson Tan:
Solve for X
rleung:
Thanks so much geodome.
I am still confused, though, on a few points.
1) Does the pressure inside the container ever REACH 0.934 atm, even for a moment? I obtained this value from PV=nRT, using V=1.5-L since the volume was halved from the original 3-L. The way I understand it, the pressure should INCREASE at 200-K when the volume is halved all the way to 0.934 atm INITIALLY. THEN, after some time, more gas molecules condense into the liquid state to reduce the pressure, until equilibrium is reached at P=0.75 atm (the equilibrium line between liquid and gas at 200-K on the phase diagram below).
2) Would that also mean that the partial pressure of the gas at equilibrium is 0.75-atm?
3) To find the moles of liquid and gas, I used another way. Could you please check to see if it is right if you have time? Since the system ultimately equilibrates to 0.75-atm (obtained visually from the phase diagram below), I used PV=nRT to find the moles of gas present at equilibrium:
n=PV/RT
n=(0.75-atm)(1.5-L)
______________
(0.0821)(200-K)
n= 0.685 mol * 70g/mol = 5.07-g of gas.
To find the mass of liquid condensed, we would subtract the grams of gas from the grams of total compound, so...
6.32 - 5.07=1.25-g of liquid at equilibrium
From there, we can easily find the moles of liquid in equilibrium.
Is that correct? Thanks so much!
Ryan
Donaldson Tan:
--- Quote from: rleung on May 28, 2006, 02:15:23 PM ---1) Does the pressure inside the container ever REACH 0.934 atm, even for a moment? I obtained this value from PV=nRT, using V=1.5-L since the volume was halved from the original 3-L. The way I understand it, the pressure should INCREASE at 200-K when the volume is halved all the way to 0.934 atm INITIALLY. THEN, after some time, more gas molecules condense into the liquid state to reduce the pressure, until equilibrium is reached at P=0.75 atm (the equilibrium line between liquid and gas at 200-K on the phase diagram below).
--- End quote ---
Read the quote below:
--- Quote from: geodome on May 28, 2006, 01:34:14 PM ---for part(b), the sample must be at equilibrium. This is because during the isothermal compression, the pressure will increase until it reaches the equilibrium point, ie. 0.7atm, 200K. Any more work done during the equilibrium will be used to convert the gas phase to the liquid phase, until all the gas phase is consumed, then the pressure of the system will continue to increase until the volume is half the original.
--- End quote ---
The answer is NO. This is because compression is not an instantaneous process. The pressure of the system gradually increases as the volume of the system is increasingly constricted. The pressure of the system will increase until it hit the equilibrium point, ie. saturated conditions.
I also want to point out even if complete condensation occurs, the resultant volume is 6.32/1.14 = 5.54ml. This is much less than 1.5L, so complete condensation cannot happen. This means the resultant fluid consists of both vapour and liquid phases.
--- Quote from: rleung on May 28, 2006, 02:15:23 PM ---2) Would that also mean that the partial pressure of the gas at equilibrium is 0.75-atm?
--- End quote ---
Yes, because at equilibrium, the partial pressure of the gas corresponds to the saturated vapour pressure of 0.75atm at 200K.
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