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Nt1310 Unit 1 Lab Report

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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

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