Question

In a student experiment, a constant-volume gas thermometer is
calibrated in dry ice (?78.5°C) and in boiling pentane (36.1°C).
The separate pressures are 0.902 atm and 1.442 atm. *Hint:*
Use the linear relationship *P* = *A* + *BT*,
where *A* and *B* are constants.

(a) What value of absolute zero does the calibration
yield?

°C

(b) What pressure would be found at the freezing point of
water?

atm

(c) What pressure would be found at the boiling point of
water?

atm

Answer #1

0.902 atm = A + B(-78.5 + 273K) = A + B*194.5K

1.442 atm = A + B(36.1 + 273K) = A + B*309.1K

Solving for simultaneous equations yields

A = -0.0144937 atm

B = 0.00471205 atm/°K

Therefore

P = -0.0144937 atm + (0.00471205 atm/K)*T

(a) At absolute zero, there is zero pressure (P = 0), therefore
T = (0. 0144937 atm)/( 0.00471205 atm/K) = 3.08K or **-270C**

(b) P = -0.0144937 atm + (0.00471205 atm/K)*(0 + 273°K) =
**1.272 atm**

(c) P =-0.0144937 atm + (0.00471205 atm/K)*(100 + 273°K) =
**1.743 atm**

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