Lab 1: Introduction to Mathematica
Task 1: Determine \[CapitalDelta]Hvap and Boiling Point of Ammonia
a.)
temperatures = {197,200,203,209,215,221,224,228,232,236} vaporpressures = {47.8,60.5,95.0,157.1,159.4,237.4,406.9,436.1,438.7,551.4} inversetemperatures = 1/temperatures lnvaporpressures = Log[vaporpressures] graph1=Thread[{inversetemperatures,lnvaporpressures}] {197,200,203,209,215,221,224,228,232,236}
{47.8,60.5,95.,157.1,159.4,237.4,406.9,436.1,438.7,551.4}
{1/197,1/200,1/203,1/209,1/215,1/221,1/224,1/228,1/232,1/236}
{3.86703,4.10264,4.55388,5.05688,5.07142,5.46975,6.00857,6.07787,6.08382,6.31246}
{{1/197,3.86703},{1/200,4.10264},{1/203,4.55388},{1/209,5.05688},{1/215,5.07142},{1/221,5.46975},{1/224,6.00857},{1/228,6.07787},{1/232,6.08382},{1/236,6.31246}}
…show more content…
Inverse Temperature"]
fitammonia=LinearModelFit[graph1,{x},x]
FittedModel[18.65 -2888.13 x]
Show[ListPlot[graph1,AxesLabel->{"1/T(K^-1)","lnP(torr)"}, PlotStyle->Green],Plot[fitammonia["BestFit"],{x,0.00420,0.0051}]]
b.)
Y=18.65-2888.13x
c.)
Y = 18.65 - (2888.13)x, corresponds to lnP = constant - ( (Subscript[\[CapitalDelta], vap]H)/R) 1/T,
Therefore Subscript[\[CapitalDelta], vap]H = 2888.13 * R. R = 8.3144621 J mol^-1 K^-1.
So, 2888.13 * 8.3144621 =
2888.13*8.3144621
24013.2
Estimate of molar enthalpy of vaporization of ammonia is 24.0132 kJ/mol.
d.)
Another form of the Clausius-Clapeyron equation is ln Subscript[P, 2]/Subscript[P, 1]=\[CapitalDelta]H/R ((1/Subscript[T, 1])- (1/Subscript[T, 2])).^2
To solve this for the normal boiling point, the pressure at the normal boiling point needs to be known. By definition, the pressure at the normal boiling point is 1 atm, or 760 torr.
So, plugging in a value from the temperature/pressure data, and the value obtained for Subscript[\[CapitalDelta], vap]H, we
…show more content…
0th Order 1st Order 2nd Order
b.) fitzerothorder["BestFit"] 0.566307 -0.0012528 x fitfirstorder["BestFit"] 0.899558 E^(-0.00903606 x) fitsecondorder["BestFit"] 0.98437/(1+0.0230931 x) fitzerothorder["ParameterConfidenceIntervalTable"] Estimate Standard Error Confidence Interval
1 0.566307 0.100062 {0.33995,0.792664} x -0.0012528 0.000338273 {-0.00201802,-0.000487571}
fitfirstorder["ParameterConfidenceIntervalTable"] Estimate Standard Error Confidence Interval a0 0.899558 0.0742779 {0.731529,1.06759} k 0.00903606 0.00130481 {0.00608437,0.0119878}
fitsecondorder["ParameterConfidenceIntervalTable"] Estimate Standard Error Confidence Interval a0 0.98437 0.0176884 {0.944356,1.02438} k 0.0234598 0.000935528 {0.0213435,0.0255761}
fitzerothorder["ANOVATable"] DF SS MS F-Statistic P-Value x 1 0.431612 0.431612 13.716 0.00489388
Error 9 0.28321 0.0314678
Total 10 0.714822
fitfirstorder["ANOVATable"] DF SS MS
Model 2 1.36018 0.680092
Error 9 0.05934 0.00659333
Uncorrected Total 11 1.41952
Corrected Total 10 0.714822
fitsecondorder["ANOVATable"] DF SS MS
Model 2 1.41666 0.708328
Error 9 0.0028678 0.000318644
Uncorrected Total 11 1.41952
Corrected Total 10 0.714822
Error magnitude relative to
6. Select the lab book and click on the data link for Ideal Gas 1. In the Data Viewer window, select all the data by clicking on the Select All button and copy the data using CTRL-C for Windows or CMD-C for Macintosh. Paste the data into a spreadsheet program and create a graph with volume on the x-axis and pressure on the
The Lithium Nickel Manganese oxide battery is still in its experimental stages. It consists of a 25% nickel substituted in a LiMn2O4 spinel. This is because Manganese will have 4 electrons in its valence shell which will avoid the Jahn-Teller distortion caused due to the Mn3+. Due to the oxidation or reduction of Nickel ions which leads to the transfer of electrons which corresponds to electric current. LiNi0.5Mn1.5O4 takes shape in two conceivable crystallographic structures concurring the cationic sub lattice: the face-focused spinel (S.G. Fd3m) named as "cluttered spinel" furthermore, the straightforward cubic stage (S.G. P4332) named as "requested spinel". This addition allows
There is a problem with using this equation due to the fact that we are not using an ideal gas. We are using Helium, so adjustments must be made to the ideal gas law in order for proper calculations to be computed. A route to
Finding and balancing equations to find the ideal gas constant using PV=nRT. Using Excel to format and graph various types of data, both given and calculated. The data given for the decomposition of NaHCO3 and Na2CO3 provided the information needed to determine the experimental constant “R” that can then be compared to the textbook definition of “R”.
To determine the pressure of H2 in the cylinder, the vapour pressure of the water was subtracted from the atmospheric pressure. This formula was derived from Dalton’s Law of partial pressure,
where the logarithm on base 10 has been converted to the "natural" logarithm. Multiplying by f , Equation (2) yields: ε 2.51 f 0.869 ln + 3.7 D ℜ f =0
Q = c x m x t q = (4.18)(1.02 g/ml x 50ml )(3.9 oC) = -831 J
Mass on (gm) 50 100 150 200 250 Volume of (l) 5 5 5 5 5 Time (s) 36.11 25.55 19.53 17.97 15.92 Flowrate, Q (m³/s) 1.385E‐04 1.957E‐04 2.560E‐04 2.782E‐04 3.141E‐04 Q² Fy =Mg Fy Ratio (m6/s2) (N) (N) exp to Theoretical Exp theoretical 1.917E‐08 0.491 0.381 0.8 3.830E‐08 0.981 0.762 0.8 6.554E‐08 1.472 1.304 0.9 7.742E‐08 1.962 1.540 0.8 9.864E‐08 2.453 1.962 0.8
where rk is the residual at the particular temperature value, qk is the measured temperature and pk is the
The model is therefore M≈138.75sin2π9x-1989+495.45 and we can superimpose it on the original data as follows.
From equation 6 I obtained the linear equation 7, model to be used in the study.
Model 1: GSIS t= ά0 +ά1 VAt + ά2EXt + ά3EMPt + ά4FDIt + β5ecm (-1) + ἑt-------equation (3)
7 y_trn = np_ut i ls . t o _ cat ego r i cal ( y_trn )