Problem 16.3

16.3. a)
Since pressure is a linear quantity in temperature, we can expect a solution of the form
P = A + BT
where A and B are constants.
Plugging in the two values that we have to calibrate with, we find
.9 atm = A + (-80) B
and 1.6 atm = A + (78) B
Solving these equations simultaneously, we find that
A = 1.272 atm
and $B = 4.652 \times 10^{-3} atm/C{}^{\circ}$
Therefore, the equation that we find for pressure in terms of temperature is
$P = 1.272 atm + (4.652 \times 10^{-3} atm/C{}^{\circ}) T$
So, at absolute zero, the pressure is zero. Solving for the temperature of our scale we find
$0 = 1.272 atm + (4.652 \times 10^{-3} atm/C{}^{\circ}) T$
or $T = -273.5{}^{\circ}C$

b)
Af the freezing point of water,
$P = 1.272 atm + (4.652 \times 10^{-3} atm/C{}^{\circ}) (0{}^{\circ}C)$
P = 1.272 atm
and at the boiling point of water,
$P = 1.272 atm + (4.652 \times 10^{-3} atm/C{}^{\circ}) (100{}^{\circ}C)$
P = 1.737 atm