Developing equation for rare gas thermometer using F

[drewmacq]

Homework Statement

Develop an analogous equation for a rare gas thermometer using the Fahrenheit scale as reference points.

Homework Equations

T = aL + b where L is length of metal rod used to determine scale, a and b are constants

expansion coefficient = 2.7315

The Attempt at a Solution

So we derived the formula for Kelvin scale in class using T = aL + b, and now we're supposed to do the same with Fahrenheit. I've been working for a little over a week now and have come up with little to nothing. I don't need answers, I need somewhere to start

I tried using 32, the freezing point in the F scale, in that first equation to get

32 = aL + b

When we used Celsius to get the Kelvin scale, this worked well because freezing point was zero and that meant b = -aL ... from there I could figure the rest of the derivation out..

I tried to use 212 as the boiling point to get:

212 = aL + b

I then substituted b from the second equation into the first, took up a few pieces of paper with math, and came out with a bunch of hot mess... any suggestions?

 

[Jhero]

Hello, thank you for posting this question. I understand that you are trying to develop an analogous equation for a rare gas thermometer using the Fahrenheit scale as reference points. Here are some steps that may help you in solving this problem:

1. Review the equation for the Kelvin scale that you derived in class: T = aL + b.

2. Recall that the Fahrenheit scale has two reference points: the freezing point of water at 32°F and the boiling point of water at 212°F.

3. Use these reference points to set up two equations:

32 = aL + b
212 = aL + b

4. Rearrange the equations to solve for b in terms of aL:

b = 32 - aL
b = 212 - aL

5. Substitute the second equation into the first equation to eliminate b:

32 - aL = 212 - aL

6. Solve for aL:

aL = 180

7. Substitute this value of aL into the second equation to solve for a:

212 = a(180)

a = 212/180 = 1.1778

8. Now that you have the values for a and b, you can write the analogous equation for the rare gas thermometer using the Fahrenheit scale as reference points:

T = 1.1778L + (32 - 1.1778L)

T = 32 + 0.1778L

9. This equation can be used to convert temperatures from the Fahrenheit scale to the rare gas thermometer scale. For example, if you have a temperature of 100°F, you can plug it into the equation to get the corresponding temperature on the rare gas thermometer:

T = 32 + 0.1778(100) = 50.78

Therefore, a temperature of 100°F is equivalent to 50.78 on the rare gas thermometer scale.

I hope this helps you get started on solving your problem. Good luck!