STPM Physics Question Help!!!

[ChristopherCWL]

Hey, im currently had no idea about this question...can any seniors or Pro here help me out?? pls...The question is:

The behaviour of real gases deviates from PVm = RT. However, at a particular range of temperature and pressure, their behaviour can be represented by the van der waal's equation: (P+a/V^2m)(Vm-b) = RT, in which the values of a and b are characteristic of that particular gas. Determine the units for a and b. [Units for P:Pa,Vm:m^3mol^-1, R: Jmol^-1K^-1, T:K]

[nickvl]

From the equation find a and b in terms of the dimensions given. Then only substitute with units....

[ChristopherCWL]

From the equation find a and b in terms of the dimensions given. Then only substitute with units....

But where are the dimensions given?? im sry im still a noob since i had just started to learn physics for form 6 and i dont even have a book with me right now...can u mind to explain in more detail??

[nickvl]

Erm...haha I'm a noob too as i just started my form 6 too...so i guess it's the blind leading the blind...LOL

I think just find a and b in terms of the P,Vm, R,T(whatever given) and then replace the P with m^3mol^-1 thing and so on for Vm, R, T....

[withoutwax]

Hint:
a/V^2m has the same unit as pressure;
b has the same unit as molar volume.

The equation is originated from PV=nRT. Comparing them will give you the idea how to get answer.

Hope that helps.
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[chongkeat]

Hmm, I think I've seen this question before in Ace Ahead Phy Vol. 1....

*checking*

Yup, it's at page 10, Quick Check 1.2!

Well, you can flip to the answer at the back, but just in case you don't have the book...


Okay, it's all about being dimensionally consistent.

In order for an equation to be correct, all the parts of the equation must at least have the same "final/resulting" dimensions.

So, for "p + a/V^2", "a+V^2" must have the same dimensions or units as "p", which is "Pa".

So, as a result, "a" must have the unit "Pa m^6 mol^-2", as this, when divided by "(m^3 mol^-1)^2", give the unit "Pa".

It's the same reasoning for "b". In this case, "b" simply must have the same units as "V^m", which is "m^3 mol^-1".

This is because you can't really subtract, say, "Length" from "Force". You can only subtract "Force" from "Force".

Or something like that.....


Not really sure about my answer. Can anyone please go through it or something?

[ChristopherCWL]

Hey, may i know what is the dimension of Joule and Mole?? Any tips for me on how to find the dimension correctly??

[Johnivan]

Hey, may i know what is the dimension of Joule and Mole?? Any tips for me on how to find the dimension correctly??

Joules is used for work done or Energy.
[work] = [F][s] = (MLT^-2)(L) = ML^2T^-1

Mol is for number of substance.
[Number of substance] = N
It is one of the basic quantities, by the way...

[henry_yew]

Joules is used for work done or Energy.
[work] = [F][s] = (MLT^-2)(L) = ML^2T^-1


I believe the dimension should have been M*(L^2)*(T^-2), not (T^-1).

[Johnivan]

I believe the dimension should have been M*(L^2)*(T^-2), not (T^-1).

Yeah, thanks for the corrections.:)