Question

In a student experiment, a constant-volume gas thermometer is calibrated in liquid nitrogen (−196°c ) and in boiling ethyl alcohol (77°c). the separate pressures are 0.349 atm and 1.634 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?

Answer 1

The given linear equation is:

`P=a+bt`

Determine the value of a and b (constants) by plug in the values of pressure and temperature.

`P_(1)` (Pressure of liquid nitrogen) =
`0.349 atm`

`T_(1)` (Temperature of liquid nitrogen) =
`-196^(0)C`

`P_(2)` (Pressure of ethyl alcohol) =
`1.634 atm`

`T_(2)` (Temperature of ethyl alcohol) =
`77^(0)C`

Put above values in given equation:

`P_(1)=a+bT_(1)`

`0.349 atm=a+b(-196^(0)C)` (1)

`P_(2)=a+bT_(2)`

`1.634 atm=a+b(77^(0)C)` (2)

Subtract equation (1) from equation (2), we get the value of b

`1.285 atm = b(273^(0)C)`

`b = 4.7* 10^(-3) atm/^(0)C`

Now, put the value of b in equation (1)

`0.349 atm=a+(4.7* 10^(-3)atm/^(0)C)(-196^(0)C)`

`a = (0.349) + (4.7* 10^(-3)atm/^(0)C)(196^(0)C)`

`a =921.549* 10^(-3) atm`

Now, at absolute zero, Pressure is equal to zero.

Absolute temperature =
`T_(o) =-(a)/(b)`

`T_(o) =  -(921.549* 10^(-3) atm)/(4.7* 10^(-3) atm/^(0)C)`

=
`-196.074 ^(0)C`

Thus, absolute zero =
`-196.074 ^(0)C`