Question

dervitation of thermodynamic temperture​

Answer 1

Thermodynamic temperature is the absolute measure of temperature and is one of the principal parameters of thermodynamics ...It uses the Kelvin scale for measurement and select the triple point of water at 273.16 k as the fundamental fixing point .

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Answer 2

Answer:

This absolute scale is known today as the kelvin thermodynamic temperature scale. It's noteworthy that Thomson's value of −273 was actually derived from 0.00366, which was the accepted expansion coefficient of gas per degree Celsius relative to the ice point

The derivation goes like this:

We have two thermal reservoirs, one at a high temperature TH and one at a colder temperature TC. Three engines work between these two reservoirs.

Engine 1 extracts QH at TH, produces W1 of work, and dumps QC into TC.

Engine 2 takes in QH heat at TH, does W2 of work, and dumps QM into engine 3 at a temperature TM, where TC<TM<TH.

Engine 3 receives QM at TM, does W3 of work, and dumps QC into TC.

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We now assume (assumption #1) that the efficiency of each engine can be written η=1−f(T1,T2):

η1=1−QHQC=1−f(TH,TC)

η2=1−QHQM=1−f(TH,TM)

η3=1−QMQC=1−f(TM,TC)

Hence we can write:

QHQC=f(TH,TC)=f(TH,TM)f(TM,TC)=QHQMQMQC

Because the LHS is only a function of TH and TC, so must the RHS, meaning that the dependence on TM must cancel out. Therefore, we can write:

f(TH,TC)=ϕ(TH)ψ(TC)=ϕ(TH)ψ(TM)ϕ(TM)ψ(TC)=f(TH,TM)f(TM,TC)

Cancelling, we find that

ψ(TM)=1ϕ(TM),

hence

f(TH,TC)=QHQC=ϕ(TH)ϕ(TC)

.

Apparently, Kelvin took ϕ(T)=T because it satisfies all requirements, and thereby established a thermodynamic temperature scale.