Help with rearranging equations for gas laws

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SUMMARY

The discussion focuses on the dimensional analysis of the gas law equation (p + a/v²)(V - b) = ZT, where Z, a, and b are constants. Participants determined the dimensions of b as L³ and a as ML⁵T⁻², while Z was concluded to be expressed in Joules per Kelvin (J/K). The analysis emphasized the importance of recognizing variable notation and dimensional consistency in equations. Key insights included the necessity of using square brackets to differentiate between dimensions and variable labels.

PREREQUISITES
  • Understanding of dimensional analysis in physics
  • Familiarity with the ideal gas law and real gas equations
  • Knowledge of fundamental quantities: mass (M), length (L), and time (T)
  • Basic algebraic manipulation skills for equations
NEXT STEPS
  • Study dimensional analysis techniques in physics
  • Learn about the ideal gas law and its applications
  • Explore the concept of state equations for real gases
  • Investigate the significance of units in scientific equations
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Students studying physics, particularly those focusing on thermodynamics and gas laws, as well as educators seeking to enhance their teaching of dimensional analysis concepts.

talebone
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Homework Statement



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 and b are constants, and P represents pressure, V represents volume, and T represents temperature. 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 -1T-2.
D M-1 L T2 .Q2
The dimensions of a are
A L6.
B M L5 T-2.
C M L-I 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-l.
C N JK-1.
D J K-1

Homework Equations

V = cm3
d = m/v
P = N/m2
PV = nRT?
W = FD

The Attempt at a Solution



Ok. I'll be honest these questions are beyond me. At first glance I am tempted to skip it. But then with what little knowledge I had, I gave it a shot.

Q1
For one I know V = L3. Ok, we'll start there. If I'm finding b, that means L is going to be on top. I can't show it in working, but just looking at the equation, I'm guessing it. So it's either A or B.

Q2
Again it looks like L will be on top. So that leaves A or B.

Q3
Immediately it looks like K will be under. So that leaves C or D. I look at the answer I see Joules which is work (=FD), which has Newtons in it, so I don't think the ans is C. Therefore it must be D.

As you can see, I have no idea how to do these type of questions. I did a google search and found a term 'Dimensional Analysis', and apparently P = M/LT2. However, this still does not allow me to answer the question. Please if a kind soul can help me out, or is there a website you can direct me which would explain these kind of questions?

Thanks for your time, appreciate it.
 
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You can only sum terms with the same dimensions.
You can check your results by completing the dimensional analysis to see if the dimensions on each side match up.

Notation is case-sensitive so "v" is a different variable to "V". You have only defined one of them.
I'm guessing they are both supposed to be upper case.

example: if we have a relation ##y=a(x+b)## then,
if [x] = L then =L so that [x+b]=L ...
if [y] = [force] = MLT-2 then [a] = MT-2
 
Last edited:
thanks Simon, got it! (yes you are right the two V's are meant to be capital)

Q1
b has to be L3 according to the rule of sums.

Q2
pressure = F/A = N/m2 = kg m/s2 x 1/m2 = kg/ms2 = M/LT2

so M/LT2 + a/(L3)2 then a = ML5/T2

Q3
(M/LT2)(L3) = ZT
ML2/T2 = ZT

work = FD = Nm = kgm/s2 x m kgm2/s2 = ML2/T2 = J

Z = J/K
 
talebone said:
thanks Simon, got it! (yes you are right the two V's are meant to be capital)

Q1
b has to be L3 according to the rule of sums.

Q2
pressure = F/A = N/m2 = kg m/s2 x 1/m2 = kg/ms2 = M/LT2

so M/LT2 + a/(L3)2 then a = ML5/T2

Q3
(M/LT2)(L3) = ZT
ML2/T2 = ZT

work = FD = Nm = kgm/s2 x m kgm2/s2 = ML2/T2 = J

Z = J/K

Very nice job. The only thing I would change would be to recognize in your Q3 analysis, the rhs's of the first two equations should be ZK, not ZT. You have been using the symbol T for time, not temperature.
 
(M/LT2)(L3) = [ZT] to distingush between T := "dimensions of time" and [T] := "dimensions of temperature". Without the square brackets, the statement is false. :)

As Chestermillar points out, it is good practice to temporarily change a quantity label when it is the same as a dimension label... just while you are doing dimensional analysis. The unit label is handy so [K] would mean "dimensions of Kelvins".

One way to figure if the math you have written makes sense is to translate it into words.

That said - good work!
 

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