The given linear equation is:
[tex]P=a+bt[/tex]
Determine the value of a and b (constants) by plug in the values of pressure and temperature.
[tex]P_{1}[/tex] (Pressure of liquid nitrogen) =[tex]0.349 atm[/tex]
[tex]T_{1}[/tex] (Temperature of liquid nitrogen) =[tex]-196^{0}C[/tex]
[tex]P_{2}[/tex] (Pressure of ethyl alcohol) =[tex]1.634 atm[/tex]
[tex]T_{2}[/tex] (Temperature of ethyl alcohol) =[tex]77^{0}C[/tex]
Put above values in given equation:
[tex]P_{1}=a+bT_{1}[/tex]
[tex]0.349 atm=a+b(-196^{0}C)[/tex] (1)
[tex]P_{2}=a+bT_{2}[/tex]
[tex]1.634 atm=a+b(77^{0}C)[/tex] (2)
Subtract equation (1) from equation (2), we get the value of b
[tex]1.285 atm = b(273^{0}C)[/tex]
[tex]b = 4.7\times 10^{-3} atm/^{0}C[/tex]
Now, put the value of b in equation (1)
[tex]0.349 atm=a+(4.7\times 10^{-3}atm/^{0}C)(-196^{0}C)[/tex]
[tex]a = (0.349) + (4.7\times 10^{-3}atm/^{0}C)(196^{0}C)[/tex]
[tex]a =921.549\times 10^{-3} atm[/tex]
Now, at absolute zero, Pressure is equal to zero.
Absolute temperature = [tex]T_{o} =-\frac{a}{b}[/tex]
[tex]T_{o} = -\frac{921.549\times 10^{-3} atm}{4.7\times 10^{-3} atm/^{0}C}[/tex]
= [tex]-196.074 ^{0}C[/tex]
Thus, absolute zero = [tex]-196.074 ^{0}C[/tex]
