3. Incorrect. The answer
is true not false
Take the natural log of PVg=C
ln P+glnV=lnC or lnP=lnC-glnV
Setting lnP=y and lnV=x
Y=a+bx
The following table contains the results of the required calculations:
|
x |
y |
x2 |
xy |
|
3.99 |
4.11 |
15.92 |
16.40 |
|
4.12 |
3.89 |
16.97 |
16.03 |
|
4.28 |
3.62 |
18.32 |
15.49 |
|
4.48 |
3.34 |
20.07 |
14.96 |
|
4.77 |
2.94 |
22.75 |
14.02 |
|
5.26 |
2.30 |
27.67 |
12.10 |
|
Sx=26.9 |
Sy=20.2 |
Sx2=121.7 |
Sxy=89.00 |
From the normal equations:
A=(ΣyΣx2-ΣxΣxy)/(nΣx2- (Σx)2)=9.75
B=(nΣxy-ΣxΣy)/(nΣx2-(Σx)2)=-1.42
Then y=9.75-1.42x
a. C=e9.75=17,154.22 and g=-b=1.42.
b. PV1.42=17,154.22
c. If V=150, x=ln V=5.01 and y=lnP=9.75 - 1.42*5.01=2.6358. Then P=e2.6358=13.95 lb/in3.