GAMSAT Forum - Gold Standard GAMSAT Prep

The Only Prep You Need

Chapter 4: Phases and Phase Equilibria

Chapter 4: Phases and Phase Equilibria

Postby admin » Fri Aug 27, 2010 12:36 am

Be the first to discuss this chapter!
admin
 
Posts: 1065
Joined: Tue Jun 29, 2010 7:47 pm

Re: Chapter 4: Phases and Phase Equilibria

Postby ms.sarah » Fri Nov 16, 2012 3:40 pm

Hi,

In the worked question in 4.1.6 the ideal gas law, is the answer in K-moles or just moles?

Thanks
ms.sarah
 
Posts: 1
Joined: Fri Nov 16, 2012 3:33 pm

Re: Chapter 4: Phases and Phase Equilibria

Postby goldstanda3269 » Sun Dec 30, 2012 8:59 am

The answer is in ONLY moles.

If you look carefully at the step before the final step, you will see K twice in the denominator in such a way that K cancels.
goldstanda3269
 
Posts: 1673
Joined: Wed Aug 25, 2010 10:59 pm

Re: Chapter 4: Phases and Phase Equilibria

Postby Legend786 » Wed May 14, 2014 9:07 pm

Pv=NRT

Hi there, I am slightly sonfused as to what value to use for the constant R... Having done A-Level Chemistry I am in a habit of using 8.30..however when doing the practice questions, the worked explanations seems to use the value 0.08.. can anyone explain how and why it is used, also in the gold standard book, the value 8.31 is in use

Thanks
Legend786
 
Posts: 6
Joined: Tue Apr 01, 2014 12:14 am

Re: Chapter 4: Phases and Phase Equilibria

Postby Legend786 » Wed May 14, 2014 9:25 pm

*confused
Legend786
 
Posts: 6
Joined: Tue Apr 01, 2014 12:14 am

Re: Chapter 4: Phases and Phase Equilibria

Postby goldstanda3269 » Thu May 15, 2014 3:10 am

Yes, R is a constant but its value depends entirely on the units. There are many different values of R and you are not expected to have memorized any of them. You are, however, definitely expected to evaluate the units of any numbers provided and to ensure that all values are consistent in terms of units (this is called 'dimensional analysis' and it is usually the source of 5-10 marks on the real exam).
goldstanda3269
 
Posts: 1673
Joined: Wed Aug 25, 2010 10:59 pm

Re: Chapter 4: Phases and Phase Equilibria

Postby contactmax2682 » Tue Jul 08, 2014 5:43 am

Yes, in the book on page: CHM-38 Phases and Phase Equilibria - it states:

R=8.31 kPa-dm^3/K-Mole

However, everywhere I have checked it states:

R=8.31 Pa-m^3/K-Mole

Truth is as long as you convert the units you are given, it doesn't matter which one you use.
contactmax2682
 
Posts: 1
Joined: Fri May 30, 2014 3:28 am

Re: Chapter 4: Phases and Phase Equilibria

Postby goldstanda3269 » Wed Jul 09, 2014 6:49 am

"Truth is as long as you convert the units you are given, it doesn't matter which one you use."
- That is correct. Of course, it is still important that you know how to interconvert SI units for the GAMSAT.

The units found in the GS book are commonly used. http://crescentok.com/staff/jaskew/isr/ ... lvin10.htm

Nonetheless, let's derive the reason for the units presented in the book.

We know from Archimedes Principle (i.e. standard molar volume CHM 4.1.1) that 1 mol of a gas occupies 22.4 L at STP (STP being a temperature of 273.15 K and a pressure of 101.325 kPa).

PV = nRT

R = PV/nR = (101.325kPa)(22.4L)/(273.15K)(1mol)

R = 8.31 L·kPa/mol·K

Before reading, try to convert the above to the units 8.31 Pa-m^3/K-Mole and 8.31 J/Kmol (since this type of conversion is definitely an expected GAMSAT skill and these 3 sets of units for 8.31 can be commonly found).

1 L = 1 dm^3
1 m^3 = 1000 dm^3

The above means that L·kPa = (10^-3)m^3 x (10^3) Pa = Pa-m^3

Now to convert to Joules. Keep in mind the pressure in pascals in force per unit area meaning Pa = N/m^2.

So Pa-m^3 = (N/m^2)(m^3) = Nm = joule

Had you forgotten that a Nm = a joule, recall that a joule is a unit of energy and that work (joules) is force (newtons) times distance (meters).
goldstanda3269
 
Posts: 1673
Joined: Wed Aug 25, 2010 10:59 pm

Re: Chapter 4: Phases and Phase Equilibria

Postby goldstanda3269 » Tue Aug 04, 2015 2:19 am

"I am not able to understand W=p[Vf-Vi].
With my stack knowledge, Work = Fs, P=F/A -> F=PA thus Work = PAs. isn't it?
BUT why W=p[Vf-Vi]?"

You got it exactly right, you just need to interpret your conclusion:

Work = PAs

Area times distance = volume!

So Work = PAs = PV

Of course it means that PV is a form of energy and the SI units would be in joules.

This type of reasoning (manipulating very basic equations) and/or their units is a common question type on the real GAMSAT.


"give me reference in the textbook."

I'll give you at least 2!

1) The first is the last paragraph of General Chemistry Chapter 7 and it underlines a very important point given that W = PV, it means that work is the area under a P vs V graph. This point is also listed as one of the important points to "Understand" on the cover page for the next chapter: CHM-99.

2) CHM 8.1 goes through the step by step reasoning.



"Which of the following processes represents a phase change from least ordered molecular structure?"
Really good question! Be careful with the wording!

FROM least ordered means that it used to be disordered but now it is more ordered. So this only describes answer choice D which is the only answer choice which starts with the least ordered structure (gas) and then becomes ordered (makes solid).
goldstanda3269
 
Posts: 1673
Joined: Wed Aug 25, 2010 10:59 pm

Re: Chapter 4: Phases and Phase Equilibria

Postby ElleRT » Fri Aug 21, 2015 2:04 am

Hello,

For Q.15 I'm unsure where these calculations have come from:

Explanation: Isothermal means that the temperature is held constant. In an isothermal expansion, the work done by gas is given by W=p[Vf-Vi]. W = p(A x 0.10m - A x 0.05m)=(3)1.013 x l05N/m2 [0.5 L - 0 . 25 L]. W=(3)(1.013 x l05 N/m2)(0.25 L) = 3.039 x l05 N/m2 [ 0.25 (1 x10-3 m3)] = 75 N x m = 75 J.

I understand everything up to W = p(A x 0.10m - A x 0.05m). But then it goes a bit haywire.

Many Thanks
ElleRT
 
Posts: 55
Joined: Thu Jul 09, 2015 4:50 am

Re: Chapter 4: Phases and Phase Equilibria

Postby goldstanda3269 » Fri Aug 21, 2015 3:39 pm

Important first concept for the GAMSAT: Work is in units of joules (SI unit) and it is equal to pressure times volume.

"I understand everything up to W = p(A x 0.10m - A x 0.05m)"

So at that point, we have pressure times the difference in volume. I will try to lay out everything step by step.

The pressure p is given in the passage and is equal to 3 atm (not SI unit). BUT the conversion to SI is given in the passage.

3 atm x (1.013 x 10^5 N/m^2 )/(1 atm) = (3)1.013 x 10^5 N/m^2

The area A is given in the passage as 50 cm^2 which is not SI unit so it must be converted to m^2 (I will do this differently than the explanation, avoiding the use of liters L).

(50 cm^2) x (1 m)^2/(100 cm)^2 = (50 cm^2) x (1 m^2)/(10000 cm^2) = 5 x 10^-3 m^2

W = p(A x 0.10m - A x 0.05m) = (3)1.013 x 10^5(5 x 10^-4 - 2.5 x 10^-4) = (3)1.013 x 10^5(2.5 x 10^-4) = 7.5 x 10^1 joules

Note: 1.013 is easily estimated as the number 1, then you have 3 x 2.5 = 7.5, and 10^5 x 10^-4 = 10^1.

That's good practice for dimensional analysis!
goldstanda3269
 
Posts: 1673
Joined: Wed Aug 25, 2010 10:59 pm

Re: Chapter 4: Phases and Phase Equilibria

Postby goldstanda3269 » Thu Sep 17, 2015 7:17 pm

A. ACER will almost never ask about sulfur for H-bonds because it is complicated. Remembering FON is enough.

B. Yes, 4 H bonds, 2 received and 2 donated as shown in CHM 4.2 and in the H bond video.
goldstanda3269
 
Posts: 1673
Joined: Wed Aug 25, 2010 10:59 pm

Re: Chapter 4: Phases and Phase Equilibria

Postby laura.cp897117 » Fri Oct 30, 2015 9:53 am

Could you help clarify Q13?

13) Assume 5 moles of an ideal gas S are placed in the apparatus. During the course of the experiment, the cylinder cap gets stuck at the x = 10 cm point, at which time the gas temperature is 300.15 K. The gas is then heated to a temperature of 450.15 K. What pressure will the gas be under at that point?

A) 3.5 atm
B) 4.2 atm
C) 4.5 atm
D) 5.7 atm

Explanation: In this problem, pressure becomes a variable; however, volume is now a constant. From pV = nRT, we obtain p/T nR/V = C, for some constant C. Thus P1/T1 = P2/T2. Substituting for variables results in (3 atm / 300 K)= (P2 450 K). P2 = (450/300) x (3 atm) = 9/6 x (3 atm) = 9/2 x (1 atm) = 4.5 atm.


I'm not sure how you got 3atm for the first part. I did:
PV=nRT
P(10cm x 50cmsquared) = 5(8.31)(300)
P = 12000/500 = 24atm???

Also, why wouldn't just doing PV=nRT for the final scenario work?
i.e. V = 10cm x 50cmsquared (place at which the lid got stuck)
n = 5 moles (amount of gas has not changed)
R = 8.31 (constant - does not change)
T = 450K (given value)
P = ?
therefore PV=nRT
P(500) = 5(8.31)(450)
P = 37 atm... which is still wrong
laura.cp897117
 
Posts: 27
Joined: Mon Mar 02, 2015 7:42 pm

Re: Chapter 4: Phases and Phase Equilibria

Postby laura.cp897117 » Fri Oct 30, 2015 10:09 am

Actually I just figured it out - I didn't realise the Pressure was given in the passage (3 atm)!.

But, I would still like to know why PV = nRT doesn't work for this example? I.e. slotting in V = 500cmsquared, n = 5 mols, R = 8.31 and T = 450K?
laura.cp897117
 
Posts: 27
Joined: Mon Mar 02, 2015 7:42 pm

Re: Chapter 4: Phases and Phase Equilibria

Postby goldstanda3269 » Fri Oct 30, 2015 1:02 pm

R can be in many, many different units. If it is given as 8.31 kPa•L/K mol then all other values must be in kPa for pressure (not atm), L for volume (not cm cubed), etc. Otherwise, conversions must be done.

[Note: Ensuring that all values share the same units is a common aspect to real GAMSAT problems. Also note that area is "cm squared" and volume is cm cubed.]
goldstanda3269
 
Posts: 1673
Joined: Wed Aug 25, 2010 10:59 pm

Next

Return to Masters Series GAMSAT Section 3 General Chemistry

Who is online

Users browsing this forum: No registered users and 1 guest

cron