Here is the question , i tried to solve it using lns but didnt work out for me :(

http://i48.tinypic.com/2vc96wh.jpg

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- Jan 13th 2010, 03:46 AMkaboose786How to solve for unknowns ?
Here is the question , i tried to solve it using lns but didnt work out for me :(

http://i48.tinypic.com/2vc96wh.jpg - Jan 13th 2010, 04:14 AMdedust
$\displaystyle Z = aV^bP^c$, take the $\displaystyle \ln $ from both side,..

$\displaystyle \ln Z = \ln a + b \ln V + c \ln P$

for $\displaystyle P$ constant, we have

$\displaystyle \ln Z_{2} = \ln a + b \ln V_{2} + c \ln P$ $\displaystyle . . .eqn (1)$

$\displaystyle \ln Z_{3} = \ln a + b \ln V_{3} + c \ln P$ $\displaystyle . . .eqn (2)$

from $\displaystyle eqn (1) - eqn (2)$, we have

$\displaystyle \ln Z_{2} - \ln Z_{3}= b ( \ln V_{1} - \ln V_{2})$

solve it.

for $\displaystyle V$ constant,. use the same method to find $\displaystyle c$.

once you find $\displaystyle b$ and $\displaystyle c$,..you'll get $\displaystyle a$. - Jan 13th 2010, 04:20 AMbandedkrait
$\displaystyle \ln(z) = \ln(a)+b\ln(V)+c\ln(P)$

so, using the data obtained from the three trials, you'll get three equations in a, b, c, which you can then solve by using Cramer's rule, Matrix method, or using a calculator.

2) Obviously the most accurate would be Linear Regression, (I think) but there is another method I can spot here.

For trials 1,2,3 and 4, the Pressure remains constant. Therefore, you can assume the the above equation as being only a function of volume.

So plot a graph of the 4 readings Vs $\displaystyle ln(V)$ $\displaystyle \ln(z) = \ln(a)+b\ln(V) + c\ln(11.2)$

The slope of this graph gives you the value of b.

Now for trials 5,6,7 observe that the volume remains constant, hence plot the graph of $\displaystyle \ln(z) = \ln(a)+ c\ln(P) + b\ln(V)$

The slope of this graph would give you the value of c.

Now using c, and b obtained thus, you can estimate a. - Jan 14th 2010, 08:01 AMkaboose786