How Can You Linearize the Equation v=vw + (v0 - vw)e^-kt for Graphical Analysis?

  • Thread starter Thread starter AlexLM
  • Start date Start date
  • Tags Tags
    Linearizing
Click For Summary
SUMMARY

The discussion focuses on linearizing the equation v = vw + (v0 - vw)e^-kt for graphical analysis in a physical chemistry context. The correct rearrangement is v - vw = (v0 - vw)e^-kt, allowing for the use of ln(v - vw) as the ordinate and time (t) as the abscissa. The y-intercept (b) is ln(v0 - vw) and the slope (m) is -k. Utilizing a semi-log plot can simplify the process, as it eliminates the need for manual calculation of the natural logarithm.

PREREQUISITES
  • Understanding of exponential decay functions
  • Familiarity with natural logarithms
  • Knowledge of graphing techniques, specifically semi-log plots
  • Basic concepts of velocity and time in physical chemistry
NEXT STEPS
  • Learn how to rearrange exponential equations for linearization
  • Study the properties of semi-logarithmic graphs
  • Explore the concept of slope and intercept in linear equations
  • Investigate the application of natural logarithms in physical chemistry
USEFUL FOR

Chemistry students, physics learners, and anyone involved in graphical analysis of exponential decay in physical chemistry experiments.

AlexLM
Messages
4
Reaction score
0
Hi everyone!

I have an issue with my physical chemistry lab. They ask us to rearrange the fallowing equation

v=vw + (v0 - vw)e^-kt
where v is velocity and t is time, vw is a constant

They also ask which quantity would be plotted as ordinate and as absissa. We have to indicated the parameter of the slope and the intercept. Now once I have a equation I'll be able to find all those things but I can't seem to get to that equation.

I was thinking of plotting it for velocity and time. We also need to be able to figure out k from the graph

I tried for a while using the natural log of the function to lower the exponent and remove the e but I still can't seem to get a linear equation in which I'll be able to extract m and b.

As you can see I'm a chemistry student, not very good with anything physics related, and sorry for the broken English, it's my second language :)

Also sorry if this is in the wrong section, I'm new to this
 
Physics news on Phys.org
Welcome to PF!

Hi AlexLM! Welcome to PF! :smile:

Try writing it as v - vw = (v0 - vw)e-kt :wink:

(and btw, your English is fine :smile:)
 
Hi tiny-tim! Thanks a lot for the answer!

So basically if I graph for ln(v-vw)=ln(v0-vw) -kt I could use ln(v-vw) as my ordinate and t as my absissa. Then b would be ln(v0-vw) and the slope would be -k?

Does that make sense?
 
yup! :smile:

(oh, and it's "abscissa", from the Latin "abscindere" :wink:)
 
AlexLM said:
Hi tiny-tim! Thanks a lot for the answer!

So basically if I graph for ln(v-vw)=ln(v0-vw) -kt I could use ln(v-vw) as my ordinate and t as my absissa. Then b would be ln(v0-vw) and the slope would be -k?

Does that make sense?

If you do it on a semi-log plot, you don't actually have to calculate ln (v - v0). Your graphics package should have semi-log plot capabilities.
 

Similar threads

  • · Replies 18 ·
Replies
18
Views
2K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
6
Views
4K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
5K
Replies
41
Views
5K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
7
Views
5K