Here is the question , i tried to solve it using lns but didnt work out for me
$\displaystyle Z = aV^bP^c$, 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 $\displaystyle a$.
$\displaystyle \ln(z) = \ln(a)+b\ln(V)+c\ln(P)$
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.