Question

6. The data below correspond to solubility of oxygen in water (gOy/100 gH:0) T°C S) 0 0.006945 SER FLER 0.006072 10 SENE 0.003368 15 EEEEEEEEEE 0.00480 20 0.004339 10.003931 30 W 0.003588 0.003315 40 0.003082 45 3 85558 0.002858 3022 PE0.002657 35 PET 0.002274 160 a. Plot solubility (Yaxis) vs I (X axis) b. Fit the graph to a polynomial fit. Show the equation of the fit. 1 0 . Calculate T and

Answer 1

a)

x1=T

x2=solubility

在 《 —0.007/ -0.006 0.006945 0.006072 0.005368 -0.005 10.00480 0.004339 -0.0044 0.003931 0.003588 0.003315 -0.003 0.003082 0.002858 0.002657 -0.002 0.002274 0.001856 -0.0011 o | 0 10 | 20 30 |

b)

Cubic Polynomial is best fit for given data.

mv 0:007 0.004339 0.003931 -0.006 0.003588 0.003315 -0.005 0.003082 0.002858 0.002657 -0.004 0.002274 0.001856 -0.003 Y; ~ax + bxį + cx+d -0.002 STATISTICS RESIDUALS R2 = 0.9996 e plot PARAMETERS a=-3.0103 x 10-8 b = 0.0000035993 C= -0.000192047 d= 0.00695291 -0.001 10 20 304050 nowered by

The Cubic equation is:

y = -3.0103 · 10-87 + 0.0000035995.62 – 0.000192047.0 + 0.00695291

or

Solubility = -3.0103-10-873 +0.000003599512-0.000192047T +0.00695291