Question

In a student experiment, a constant-volume gas thermometer is calibrated in liquid nitrogen (−196°C) and in boiling pentane (36.1°C). The separate pressures are 0.344 atm and 1.423 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

Given, the variation of temperature with pressure is linear and is given by

P = A + BT ...........................(1)

For liquid nitrogen : T = -196 ⁰C and P = 0.344 atm

Putting these values in equation (1), we get

0.344 = A + B(-196)

or, 0.344 = A - 196 B ...............................(2)

For boiling Pentane : T = 36.1 ⁰C and P = 1.423 atm (atmosphere)

Putting the values in equation (1), we get

1.423 = A + 36.1 B .................................(3)

Subtracting equation (2) from equation (3), we get

1.079 = 232.1 B

or, B = 0.00464885

Putting the value of B in equation (3), we get

1.423 = A + 0.16782378

or, A = 1.255176  

Putting the value of A and B in equation (1), we get

P = 1.255176 + (0.00464885) T .....................................(4)

This is the required equation.

(a) At absolute zero, the pressure will become 0 atm.

So, 0 = 1.255176 + (0.00464885) T

or, -1.255176 = (0.00464885)T

or, T = -269.997 ⁰C ( nearly~ 270 ⁰C )

(b) The freezing point of water is 0 ⁰C .

So, at T = 0 ⁰C

P = 1.255176 + (0.00464885) x 0 = 1.255176 atm (nearly 1.23 atm)

(c) The boiling point of water is 100 ⁰C.

So, at T = 100 ⁰C

P = 1.255176 + 0.00464885 x 100 = 1.720061 atm (nearly 1.72 atm)

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