Section III question

derekmvr · Rookie Joined: 10 Dec 2007 Posts: 5

Hi all

I am struggling to solve one of the GAMSAT practice question units (Prac test book, section III, unit 32). Question is as follows:

An equation of state that has been used to model the behaviour of a fixed amount of a real gas is:
(P+a/v2)(v-b)=ZT
In the equation Z, a, b are constants, and P represents pressure, V = volume, T = temp. The constant Z depends on the amount of gas.
Consider what the dimensions of a and b are in terms of the fundamental quantities of mass (M), length (l), and time (t).

Q1: The dimensions of b are:
a) L3
b) L6
c) M.L-1.T-2
d) M-1.L.T2

Q2: The dimensions of a are:
a) L6
b) M.L5.T-2
c) M.L-1.T-2
d) M.L-5.T-1

Q3: The value of the constant Z could be expressed in terms of the unit:
a) W
b) W.N.s-1
c) N.J.K-1
d) J.K-1

If any one has any advice it would be greatly appreciated.

Thanks in advance.

Cheers

ozzyaaron · Rookie Joined: 16 Mar 2010 Posts: 6

Hi there, I've just worked through this myself and got the answers they were looking for. It's all to do with matching units in an equation, for instance :

s = ut + 1/2(at^2)
in units :

m = (m/t)t + 1/2(m/t^2)(t^2)
m = m + m (we remove constants)

Essentially when you multiply terms, their units multiply, but each added/subtracted component of the equation should have matching units, I'm probably not explaining it well but I'm sure you'll see :

For b :

We have b's units having to be the same as V as they're adding. V => L^3 so b must have the units of L^3.

for a :

We have (a/V^2) units having to be the same as P. Pressure is equivalent to M/TL^2 or kg per metre second squared

So we have M/LT^2 = a/V^2
a/V^2 => a/L^6
which gives :
M/LT^2 = a/L^6
a = ML^5/ T^2

So far, so good.

Now for Z.

In this case we have the units multiplying between the left bracket and right bracket to give us a term of :

M/LT^2 * L^3 = ZT
(ML^2/T^2)/T = Z

What we have there is a Joule (kg m^2 / s ^2) and then divided by T

So the units of Z are JT^-1

bingo Smile

Hope this is both correct and helpful.

derekmvr · Rookie Joined: 10 Dec 2007 Posts: 5

Hey ozzyaaron

Thanks a ton for the solution. Makes perfect sense. Congrats on figuring that out - should be a help to others as well.

Cheers