# Absolute zero calibration

## Homework Statement

A constant-volume gas thermometer is cal-
ibrated in dry ice (which is carbon dioxide
in the solid state and has a temperature of
−80◦C) and in boiling ethyl alcohol (78◦C).
The two pressures are 1 atm and 1.542 atm.
What value of absolute zero does the cali-
bration yield?
003 (part 2 of 2) 10.0 points
What is the pressure at the boiling point of
water?

y-y1=m(x-x1)
m = dy/dx

## The Attempt at a Solution

pt 1.
(-80C,1.0atm)
(78C,1.542atm)

m = (1.542-1.0)/(78-(-80)) = .0034

y-1.542 = .0034(x-78)

y-1.542 = .0034x -.2652

y = .0034x + 1.2768

if we assume temp v. pressure is linear.. and according to the text this is a good assumption

so i put the pressure equal to zero at absolute zero

0 = .0034x +1.2768
-1.2768 = .0034x
x = -1.2768/.0034 = -375.5294 C
this is wrong according to the homework program

but!... I got the second half right using that equation
boiling point of water in C is 100C so..

y=.0034(100) +1.2768 = 1.6168 atm

and this leads me to believe that my equation is correct.. what am i doing wrong?

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Redbelly98
Staff Emeritus
Homework Helper

## The Attempt at a Solution

pt 1.
(-80C,1.0atm)
(78C,1.542atm)

m = (1.542-1.0)/(78-(-80)) = .0034
If that were actually 1.000 atm (is it?), you would be dividing 0.542/158. How many significant figures are appropriate when you do that calculation? This will change your values slightly.

Your answer of -375 C is only slightly off from what it should be given this data.