42 posts
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But is R not also 8.31cm^3 kPa /Kmol

Therefore:

PV = nRT

P(500cm^3) = (5mol)(8.31cm^3 kPa/Kmol) (450K)

P = 37.395kPa

if 1 atm = 101.325 kPa, then P = 0.37atm which is still not the right answer!

In any case, the answer should only differ by where the decimal point is, if it is only the Gas Constant R which is wrong in this equation? But the answer is 4.5atm not 3.7atm?

Therefore:

PV = nRT

P(500cm^3) = (5mol)(8.31cm^3 kPa/Kmol) (450K)

P = 37.395kPa

if 1 atm = 101.325 kPa, then P = 0.37atm which is still not the right answer!

In any case, the answer should only differ by where the decimal point is, if it is only the Gas Constant R which is wrong in this equation? But the answer is 4.5atm not 3.7atm?

- laura.cp897117
**Posts:**27**Joined:**Mon Mar 02, 2015 7:42 pm

laura.cp897117 wrote:But is R not also 8.31cm^3 kPa /Kmol

That is not quite accurate.

dm^3 = L

So 8.31 kPa•L/K mol = 8.31 kPa•dm^3/K mol but not cm^3.

Also, I should have re-read the question earlier but it is clear that PV=nRT does not apply to this question because the volume V is constant because "the cylinder cap gets stuck at the x = 10 cm point". This is why the explanation describes how P/T = constant applies, and not PV=nRT. You may consider rewatching the video on the Equation of State, specifically the different ways that it can be used depending on what variables are constant.

- goldstanda3269
**Posts:**1673**Joined:**Wed Aug 25, 2010 10:59 pm

ilbe wrote:Queries with regards to Q10.

1. all options are the assumption for ideal gas. Right?

2. Plow, Thigh. is requirement for ideal gas. But in the reality it may be Phigh Tlow like this question.

If I change D into right sentence, 'The volume of the gas molecule become 'significant' in comparison with the volume occupied by the gas. Right?

3. In the reality, mostly D is not valid?

1. Yes.

2. Yes.

3. Yes.

- goldstanda3269
**Posts:**1673**Joined:**Wed Aug 25, 2010 10:59 pm

PV = nRT does not apply if P, V or T is constant.

It would be helpful to watch the Equation of State video again. If either P, V or T is constant then the Equation of State (PV = nRT) changes and so an equation related to the Equation of State must be used.

It would be helpful to watch the Equation of State video again. If either P, V or T is constant then the Equation of State (PV = nRT) changes and so an equation related to the Equation of State must be used.

- goldstanda3269
**Posts:**1673**Joined:**Wed Aug 25, 2010 10:59 pm

ilbe wrote:Confused with Raoult's law and Dalton's law.

These two laws looks very similar and easy to confuse.

Are they saying same things??

Please let me know.

Just to be clear: GAMSAT does not require that you memorise the names of these laws. But of course, it is expected that you understand the concepts involved. The concepts behind the 2 laws are very different.

Raoult's Law is most important for GAMSAT (ACER's Blue Booklet Unit 3 has 2 questions) and basically says that as the solute in solution increases, the vapour pressure reduces (you should watch the video on Raoult's Law again because it is clearly explained). Unlike Raoult's law, Dalton's law is NOT related to solutions, solvents or solutes. Dalton's law basically, and simply, states that the total pressure exerted on a mixture of gases is equal to the sum of the partial pressures of the individual gases (this is also discussed in the videos).

- goldstanda3269
**Posts:**1673**Joined:**Wed Aug 25, 2010 10:59 pm

ilbe wrote:I also add others..

Please answer these

https://www.youtube.com/watch?v=CmtRj-z ... 5YtYgb714b

https://www.youtube.com/watch?v=JTqWAdd ... 5YtYgb714b

We must apologise but we are not able to reliably answer questions regarding our off site videos. We leave the comment box open for our YouTube videos mostly so that students can interact, and so we can respond when time permits. We are the only company that has ever created detailed worked solutions to ACER's materials and then we made them freely available. With over 100 videos and questions coming in every day, we would have to hire full-time staff in order to answer questions about freely available videos. Unfortunately, we simply do not have those resources. We hope, however, that you find the videos helpful for your GAMSAT preparation.

NB: We are happy to announce that in the last week, we added about 4 hours of helpful science review videos to our already extensive collection (if you click on Videos in the top Menu when you are logged in, you will notice new videos in bold). We hope these new videos will further enhance your preparation. Good luck!

- goldstanda3269
**Posts:**1673**Joined:**Wed Aug 25, 2010 10:59 pm

Hi,

For question 30 it asks you find the unit of the gas constant R.

I got to (joules)/(mole)(k) but the answer was (joules)/(mole)/(k) and now I'm totally lost.

before I did this question would have assumed (joules)/(mole)/(k) would be expressed as

(joules)(k)/(mole)

As the reciprocal of (mole)/(k) is (k)/(mole) and I was under the impression that dividing by a fraction was the same as multiplying by it's reciprocal.

Now after messing about with a calculator I found this is only the case some of the time.

Example:

5/ 3/4 = 5/0.75 = (5)(4)/3 however it does not = 5/3/4 which instead = 5/12

how can I tell the difference when working with units?

Please help I can't find any explanation in the book that makes sense to me and this just keeps tripping me up when manipulating formulae and it's driving me nuts because I'm sure it's something simple I'm missing.

Regards,

Jesse.

For question 30 it asks you find the unit of the gas constant R.

I got to (joules)/(mole)(k) but the answer was (joules)/(mole)/(k) and now I'm totally lost.

before I did this question would have assumed (joules)/(mole)/(k) would be expressed as

(joules)(k)/(mole)

As the reciprocal of (mole)/(k) is (k)/(mole) and I was under the impression that dividing by a fraction was the same as multiplying by it's reciprocal.

Now after messing about with a calculator I found this is only the case some of the time.

Example:

5/ 3/4 = 5/0.75 = (5)(4)/3 however it does not = 5/3/4 which instead = 5/12

how can I tell the difference when working with units?

Please help I can't find any explanation in the book that makes sense to me and this just keeps tripping me up when manipulating formulae and it's driving me nuts because I'm sure it's something simple I'm missing.

Regards,

Jesse.

- JesseG
**Posts:**4**Joined:**Tue Jan 24, 2017 3:54 am

Don't worry, it is very unlikely on the real exam that this will trip you up because you already know that it is an issue. Also, in this particular question, only one answer was even possible which would force you into the correct interpretation.

Nonetheless, I will start by showing you an example that you should be already familiar with since you are studying Chemistry (most students study Physics before Chemistry):

Acceleration is known as the rate of velocity, or, velocity over time:

a = v/t = m/s/s = m/s^2

With that in mind, here is a longer response to your question: https://www.quora.com/How-is-m-s-s-math ... t-to-m-s-2

Nonetheless, I will start by showing you an example that you should be already familiar with since you are studying Chemistry (most students study Physics before Chemistry):

Acceleration is known as the rate of velocity, or, velocity over time:

a = v/t = m/s/s = m/s^2

With that in mind, here is a longer response to your question: https://www.quora.com/How-is-m-s-s-math ... t-to-m-s-2

- goldstanda3269
**Posts:**1673**Joined:**Wed Aug 25, 2010 10:59 pm

In Q.15 of gamsat mcq's, the answer is this "Isothermal means that the temperature is held constant. In an isothermal expansion, the passage provided the equation that 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."

My question is why are we using A when the formula doesn't state to use it?

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."

My question is why are we using A when the formula doesn't state to use it?

- ahmedijaz91235
**Posts:**2**Joined:**Sun Feb 18, 2018 12:14 am

ahmedijaz91235 wrote:...

My question is why are we using A when the formula doesn't state to use it?

The passage states:

" the work done by a gas is given by W = p[Vf-Vi], where p is the pressure, Vf is the final volume and Vi is the initial volume."

It is assumed that you recognize that volume V is the length cubed (i.e. length x length x length) or otherwise stated as area times length since the area of something is length x length.

V = A x L.

- goldstanda3269
**Posts:**1673**Joined:**Wed Aug 25, 2010 10:59 pm

Hello there!

Q5 states:

Water is less dense in its solid phase than in its liquid phase. Which of the following best explains the unusual density of water?

A) Water in the liquid phase evaporates which decreases its volume.

B) Water molecules exist in an ordered structure in the solid phase which prevents them from decreasing the volume of the solid.

C) Intermolecular bonding between water molecules in the liquid phase decreases the overall volume of the liquid.

D) The density of water decreases at temperatures above 4°C.

Aren't B and C both essentially two sides of the same answer?

Thanks!!!

Q5 states:

Water is less dense in its solid phase than in its liquid phase. Which of the following best explains the unusual density of water?

A) Water in the liquid phase evaporates which decreases its volume.

B) Water molecules exist in an ordered structure in the solid phase which prevents them from decreasing the volume of the solid.

C) Intermolecular bonding between water molecules in the liquid phase decreases the overall volume of the liquid.

D) The density of water decreases at temperatures above 4°C.

Aren't B and C both essentially two sides of the same answer?

Thanks!!!

- Jaylin
**Posts:**1**Joined:**Fri Jul 12, 2019 12:20 pm

Jaylin wrote:Hello there!

Q5 states:

Water is less dense in its solid phase than in its liquid phase. Which of the following best explains the unusual density of water?

A) Water in the liquid phase evaporates which decreases its volume.

B) Water molecules exist in an ordered structure in the solid phase which prevents them from decreasing the volume of the solid.

C) Intermolecular bonding between water molecules in the liquid phase decreases the overall volume of the liquid.

D) The density of water decreases at temperatures above 4°C.

Aren't B and C both essentially two sides of the same answer?

Thanks!!!

You are correct, answer choice C has been repaired (made to be more obviously incorrect):

C) Intermolecular bonding between water molecules in the liquid phase increases the overall volume of the liquid.

The next free webinar, Feb 9!

https://www.gamsat-prep.com/GAMSAT-free-online-seminar/

Good luck with your studies, keep up the good work!

- goldstanda3269
**Posts:**1673**Joined:**Wed Aug 25, 2010 10:59 pm

This is regarding GEN CHM Chapter 4 Worked Solutions Question 20 (CHM-110)

Since we always use SI units I have used Kelvin not "degree" for my unit of temperature...

I ended up with R = Joules/(mol)(K)

Why would you use degree over K and why mol times degree = mol/K?

Regards

Ania

Since we always use SI units I have used Kelvin not "degree" for my unit of temperature...

I ended up with R = Joules/(mol)(K)

Why would you use degree over K and why mol times degree = mol/K?

Regards

Ania

- horbowiec.7527
**Posts:**71**Joined:**Mon Mar 08, 2021 11:35 am

This is regarding GEN CHM Chapter 4 Worked solution question 27 (CHM-113)

I am a bit confused by this question. I have chosen B as being not consistent with the graph and because of that I didn't even consider answer D.

Explanation for B feels wrong to me... it appears that X rays end just before hitting 0.001 g/m^3 level of atmospheric density but choice B in question states atmospheric density of 10^-4 g/m^3 by X rays = 0.0001 g/m^3 and X rays are reaching that level so likely to affect it? Am I not understanding something here?

Regards,

Ania

I am a bit confused by this question. I have chosen B as being not consistent with the graph and because of that I didn't even consider answer D.

Explanation for B feels wrong to me... it appears that X rays end just before hitting 0.001 g/m^3 level of atmospheric density but choice B in question states atmospheric density of 10^-4 g/m^3 by X rays = 0.0001 g/m^3 and X rays are reaching that level so likely to affect it? Am I not understanding something here?

Regards,

Ania

- horbowiec.7527
**Posts:**71**Joined:**Mon Mar 08, 2021 11:35 am

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