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eric99gt
guest

Oct1805, 09:06 PM (EST) 

"Rearranging Equation"

Alright I'm going absolutely nuts here. It's a simple matter of rearranging an equation and for the life of me I can't do it. I've been working at it for like an hour now to no avail. Here's the equation. P = RT/(vb)  a/v(v+b)T^0.5 I'm trying to solve for v. I've manipulated this equation every way I can think of and still can't get it. I'm sure it's something simple but I'm stuck. Please help. Thanks


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alexb
Charter Member
1670 posts 
Oct1805, 09:11 PM (EST) 

2. "RE: Rearranging Equation"
In response to message #0

>I'm trying to solve for v. In v you are going to get an equation of 3^{rd} degree. I do not know if you want to solve a 3^{rd} degree equation. It's not quite clear from the form in which you put your equation but I would guess you have two fractions one with (vb) and the other with v(v+b) in the denominator. All you have to do is multiply all terms of the equation (3 of them in all) by the product (vb)v(v+b) and subsequently simplify the result. A 3^{rd} degree equation it is going to be. 

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mpdlc
guest

Oct1905, 07:54 AM (EST) 

3. "RE: Rearranging Equation"
In response to message #0

As Alex said, without any more info you end in a third degree eequation. However by the names of variables you are using, my guessing is that you are dealing with gas equation or Van der Waals equation. In such case try to multiply both term by v then substitute Pv for RT, and if my quick calculation was right,  check it I did in a hurry you should end in the the following : v = b(1+bRT^3/2)/(abRT^3/2) 

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eric99gt
guest

Oct1905, 06:38 PM (EST) 

4. "RE: Rearranging Equation"
In response to message #3

and you would be right. It's a gas equation. What the real problem is, is that I'm trying to find the partial of (dv/dt)with constant pressure using that equation. P=<(RT/vb)><(a)/(v(v b)T^0.5)>. Hope that clears up any of the ambiguity of the equation. 

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mr_homm
Member since May2205

Oct2005, 05:55 PM (EST) 

5. "RE: Rearranging Equation"
In response to message #4

Hi eric99gt, >What the real >problem is, is that I'm trying to find the partial of >(dv/dt)with constant pressure using that equation. >P=<(RT/vb)><(a)/(v(v b)T^0.5)>. Hope that clears up any of >the ambiguity of the equation. In that case, there is a much easier way. In solving for partial derivatives, it is not necessary to solve for your variable first; in fact, sometimes it is impossible. Nevertheless, you can always get the partial derivative by the following method: 1) take the differential, not derivative, of both sides 2) set the differentials of any variables being held constant to zero 3) rearrange the remaining terms to get your derivative. This always works because the product rule and chain rule cause the equation for the derivative to always be linear in terms of the differentials. In your case, I would rewrite the equation as P = (RT)/(vb)  a/(v(v+b))( T^0.5) and take the differentials to get dP = (R dT)/(vb) (RT)dv/(vb)^2 + a(2v+b)dv/(v(v+b))^2(T^0.5) + 0.5a dT(v(v+b))(T^1.5). Then you set dP = 0 and collect the terms in dT and dv to get 0 = {R/(vb) + a/(2v(v+b)T^1.5}dT  {RT/(vb)^2  a(2v+b)/(v(v+b))^2(T^0.5)}dv From there, you can obviously rearrange terms to solve for the ratio dv/dT, which is what you want. Notice that although this method uses ordinary differentials (not "curly" d partial differentials), the explicit assumption that dP = 0 forces the ratio to represent the sensitivity of v to changes in T while P is constant, which is exactly the definition of the partial derivative at constant P. So although the notation looks like an ordinary derivative, you really are getting a partial derivative this way. Note: I have not carefully proofread my algebra here, since my purpose was not to get the answer for you, but to illustrate the method. Please check to see whether it is correct. Hope this helps. Stuart Anderson


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mpdlc
guest

Oct2205, 07:21 AM (EST) 

6. "RE: Rearranging Equation"
In response to message #4

I think you should double check your original equation again. The first term after the equal is allright but in the second term the dimension of the constant a it'sounds strange, R times (T)^3/2 (L)^3 it does not make sense from an usual thermodynamic point of view. Indeed Van der Waals equation is : P = RT/(Vb)  a/V^2 and the dimension for a means presure times (volume/mol)^2, so I believe the T^0.5 is wrong. 

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mr_homm
Member since May2205

Oct2205, 10:39 AM (EST) 

7. "RE: Rearranging Equation"
In response to message #3

Hi mpdic, I don't think this advice is right: >However by the names of variables you are using, my guessing >is that you are dealing with gas equation or Van der Waals >equation. In such case try to multiply both term by v then >substitute Pv for RT, If you do this, you are assuming that Pv = RT; but that is the ideal gas equation, and is a different physical equation of state from the one that he is trying to solve. In other words, you are applying two different equations of state simultaneously to the same gas, which is a physically inconsistent set of assupmtions about the behavior of the gas. Algebraically, you can of course do this, but the result will not be physically meaningful. Stuart Anderson 

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mpdlc
guest

Oct2205, 02:35 PM (EST) 

8. "RE: Rearranging Equation"
In response to message #7

Hi, Stuart What I tried to state in my quick first remark is that equation was related to the gas equation or Van de Waals Equation, the last one improves the perfect gas equation and even it can be applied successfully in vapour phase, indeed as a retired Naval Architech and Marine Engineer, I have used it for steam when not being the water vapour tables handy. In my second comment, after a closer look to the new equation and the old one, in my opinion both are dimensionally inconsistent. Since b means a Volume,  indeed b is called, if I remember, covolume and take care of the correction for the limited packing of the gas molecules at high extremely pressures , so I conclude the first term of the second member is all right. However the second term I insist is wrong since the square root of T either in the numerator or denominator will force the constant a to be the dimension inverse or direct of square root of temperature what is rather uncommon, when it is assume to be a pressure times a volume times an specific volume so taking care of the so called internal attraction. Furthermore the Van der Waals equation is written and tabulated using a and b traditionally in this way, in the same fashion that R is reserved for perfect gas constant. As I stated in my last remark the Van der Waals equation in term of pressure is : P= (RT/(Vb))  (a/V^2) so if his intend was to correct the second term using V(V+b) instead V^2 it will be dimensionally correct but the temperature T in that term is out of place. Regarding to your procedure to get the derivative is a good approach to start to learn differentials with several variables. My approach is somewhat different, since in the given equation is easy to separate T as a funtion of v and p, I would do it first and then calculate dT/dv, treating p as a constant what it is easier and then write the inverse of obtained expression for the derivative which will give us dv/dT. We will get the result in explicit form for v (variable) and p as a constant, T will not appear. However if T is the given data, we will not avoid either to caculate v from the originally given equation or to eliminate v among the derivative and the original one, in the same way you are forced to do in your solution in anyway. Finally even I ocasionally concur this excelente website as an entertaining brainteaser, I do have noticed your mostly smart remarks, mathematical proofs and above all I can envision on you a profound dedication on teaching as a vital enterprise.
Best Regards Mariano Perez de la Cruz 

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mr_homm
Member since May2205

Oct2205, 07:26 PM (EST) 

9. "RE: Rearranging Equation"
In response to message #8

Hi Mariano, >What I tried to state in my quick first remark is that >equation was related to the gas equation or Van de Waals >Equation, the last one improves the perfect gas equation and >even it can be applied successfully in vapour phase, indeed >as a retired Naval Architech and Marine Engineer, I have >used it for steam when not being the water vapour tables >handy. I agree completely, this equation is some kind of gas equation. As you say, the van der Waals equation gives a much better approximation to the true behavior of a gas than the ideal gas law does, escpecially when the gas is near to a phase boundary. > >In my second comment, after a closer look to the new >equation and the old one, in my opinion both are >dimensionally inconsistent. Since b means a Volume,  indeed > b is called, if I remember, covolume and take care of the >correction for the limited packing of the gas molecules at >high extremely pressures , so I conclude the first term of >the second member is all right. Yes, this is also how I understand the "b" constant in van der Waals. >However the second term I insist is wrong since the square >root of T either in the numerator or denominator will force >the constant a to be the dimension inverse or direct of >square root of temperature what is rather uncommon, when it >is assume to be a pressure times a volume times an specific >volume so taking care of the so called internal >attraction. I agree that the dimension of "a" is strange, but this equation may not be exactly the van der Waals equation. If eric99gt really meant to use the ordinary van der Waals equation, it would be a very strange mistake to put T^.5 where did not exist. Also, replacing v^2 with v(v+b) shows that some changes have been made, so that this is not the true van der Waals equation. I can see two possibilities: first, this may be a problem assigned by a college professor to give students practice in working out partial derivatives. In this case, the professor would make changes in the equation so that it is hard to solve for v or T, just to make the students think about how to compute dv/dT. This would of course change the meaning of the a and b constants. Second, it is possible that using some empirical data, eric99gt found that he should modify the van der Waals equation to better fit his data, and that the new terms are placed in the equation to improve the fit. In this case, a and b are new empirical constants that are not the same as the a and b in the van der Waals equation. In either case, I agree that the T^.5 term looks very strange, but it may nevertheless be correct for eric99gt's equation. > >Regarding to your procedure to get the derivative is a good >approach to start to learn differentials with several >variables. > >My approach is somewhat different, since in the given >equation is easy to separate T as a funtion of v and p, I >would do it first and then calculate dT/dv, treating p as >a constant what it is easier and then write the inverse of >obtained expression for the derivative which will give us >dv/dT. This will work if the factor of T^.5 is removed, and in that case, it is a better (quicker, easier) method than the one I suggested. But if the factor of T^.5 is really supposed to be there, then there is a trouble: since T^.5 is on the denominator in the second term and T is on the numerator in the first term, solving for T will give a cubic equation in the variable T^.5, which is just as hard as solving for v. This is what makes me suspect that this is a college homework problem: the changes to the equation are exactly right to take away the easy method of finding the partial derivative and force the students to find another way. I could be wrong of course; this is just a guess. > >We will get the result in explicit form for v (variable) and >p as a constant, T will not appear. However if T is the >given data, we will not avoid either to caculate v from the >originally given equation or to eliminate v among the >derivative and the original one, in the same way you are >forced to do in your solution in anyway. Yes, I agree. Your method is nicer because it gives a formula which only uses two variables. In fact, I would advise anyone always to try your way first, and use my way in case the algebra is too hard to solve for the variable that you want. >Finally even I ocasionally concur this excelente website as >an entertaining brainteaser, I do have noticed your mostly >smart remarks, mathematical proofs and above all I can >envision on you a profound dedication on teaching as a vital >enterprise.
Thank you, I try my best to help and to be clear. I have enjoyed reading your interesting and helpful remarks as well. Stuart Anderson 

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