Answer the following:
(a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale )?
(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number $0^{o} C$ and $100^{o} C$ respectively. On the absolute scale one of the fixed points is the triple-point of water, which on the kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (kelvin) scale?
(c) The absolute temperature (Kelvin scale ) T is relate to the temperature $t_{c}$ on the Celsius scale by $t_{c} = T - 273.15$ . Why do we have 273.15 in this relation and not 273.16?
(d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale ?

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Solution

(a) Triple-point has a unique temperature; fusion point and boiling point temperatures depend on pressure.
(b) The other fixed point is the absolute zero itself.
(c) Triple-point is ${0.01}^{o} C$ , not $0^{o} C$ .
(d) $T = 273.15 \times 9 / 5 = 491.69$
The temperature and pressure at which a substance can exist in equilibrium in the liquid, solid, and gaseous states.It is unique and constant. The problem with Celsius scale is that boiling and melting point change with atmospheric pressure at that place.

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Similar questions

Q. Why is it wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the celsius scale)?

Q. What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?

Q. Two absolute scales

A

and

B

have triple points of water defined to be

200 A

and

300 B

. A temperature is measured on these scales as

T_{A}

and

T_{B}

What is the relation between

T_{A}

and

T_{B}

? Given, that triple point of water as

273.16 K

Given that P, V and T stand for vapour pressure, volume and temperature, an ideal gas obeys the law -PV α T. If the volume is kept constant, P α T. With decreasing temperature the vapour pressure drops, until P = 0 at some temperature T₀. Since P cannot be less than zero, or negative, T₀naturally becomes the lowest temperature that we can reach, and can be used as a "natural” lower fixed point while constructing a temperature scale. This is the "absolute zero”, which in the Celsius scale is - 273.15⁰C.

To construct an absolute temperature scale , where the absolute zero is naturally 0. Let us choose the triple point of mercury (-38.8⁰C, at 0.2 mPa) as our upper fixed point, and give assign it a value 100. What will be the boiling point of water on this scale?

Q. Assertion :The lowest attainable temperature is absolute zero,i.e.,

0 K = - {273.15}^{o} C .

Reason: Size of each degree on Kelvin scale is same as that on Celsius scale.

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