
#1
Jan2313, 03:00 AM

P: 9

1. The problem statement, all variables and given/known data
An equation of state that has been used to model the behaviour of a fixed amount of a real gas is (p + a/v^{2}) (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 L^{3} B L^{6 } C M L ^{1}T^{2}. D M^{1} L T^{2} . Q2 The dimensions of a are A L^{6}. B M L^{5} 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} 2. Relevant equations V = cm^{3} d = m/v P = N/m^{2} PV = nRT? W = FD 3. 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 = L^{3}. 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/LT^{2}. 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. 



#2
Jan2313, 06:32 AM

Homework
Sci Advisor
HW Helper
Thanks ∞
PF Gold
P: 10,918

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 casesensitive 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 [b]=L so that [x+b]=L ... if [y] = [force] = MLT^{2} then [a] = MT^{2} 



#3
Jan2313, 02:58 PM

P: 9

thanks Simon, got it! (yes you are right the two V's are meant to be capital)
Q1 b has to be L^{3} according to the rule of sums. Q2 pressure = F/A = N/m^{2} = kg m/s^{2} x 1/m^{2} = kg/ms^{2} = M/LT^{2} so M/LT^{2} + a/(L^{3})^{2} then a = ML^{5}/T^{2} Q3 (M/LT^{2})(L^{3}) = ZT ML^{2}/T^{2} = ZT work = FD = Nm = kgm/s^{2} x m kgm^{2}/s^{2} = ML^{2}/T^{2} = J Z = J/K 



#4
Jan2313, 04:04 PM

Sci Advisor
HW Helper
Thanks
PF Gold
P: 4,396

Help with rearranging equations for gas laws 



#5
Jan2313, 09:30 PM

Homework
Sci Advisor
HW Helper
Thanks ∞
PF Gold
P: 10,918

(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! 


Register to reply 
Related Discussions  
Rearranging Equations  Linear & Abstract Algebra  2  
Rearranging equations  Precalculus Mathematics Homework  4  
Rearranging Equations  Calculus & Beyond Homework  3  
rearranging equations  Introductory Physics Homework  3  
Rearranging equations.  Introductory Physics Homework  28 