Thermodynamic scale of temperature

1.5 THERMODYNAMIC SCALE OF TEMPERATURE

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  • As we know the scale of an thermometer is depends on the physical properties like “Thermal expansion” of the liquid used in the thermometer.
  • Since,  different liquids have different cofficient value of expansion, and also do not expand uniformly over a given range of temperature, therefore thermometers are constructed by using different liquids will show different numerical values on thermometer
  • THE TEMPERATURE AT WHICH AN IDEAL GAS CEASES TO EXSIST IS TAKEN AS ZERO DEGREE ABSOLUTE (-273.15ᵒC)  AND THE SCALE DEVELOPED BY TAKING ZERO DEGREE ABSOLUTE AS THE ZERO  IS KNOWN AS THE IDEAL GAS TEMPERATURE SCALE AND IS INDEPENDENT OF THE IDEAL GAS TAKEN.

LORD KELVIN, developed a scale of temperature based upon the SECOND LAW OF THERMODYNAMICS

  • Based on his temperature scale and efficiency of the reversible heat engine, He named his temperature scale THERMODYNAMIC SCALE OF TEMPERATURE Or KELVIN SCALE OF TEMPERATURE. it is very similar to the ideal gas temperature scale.
  • All the reversible engines are operating between tempratures T₁ and T₂ having the similar efficiency i.e. , η = f ( T₁ – T₂ )
  • where f is a universal function independent of the working fluid.
  • since,
    • η = q₂–q₁/q₂ = 1 – q₁/q₂ = 1 + q₂/q₁,
  • therefore, q₂/q₁ must also be a function of temperatures of the two reservoirs. A number of different temperature scale that satisfy this relation can be deviesed ; but by taking temperature as directly proportional to the ideal gas temperature scale.
  • If q₂ is the heat transfer from the source (heat resevior at higher temperature) and q₁ is the heat transfer to the sink (reservoir at lower temperature) and θ₁ and θ₂ are their temperatures on kelvin scale, then
    • q₂/q₁ = θ₁/θ₂
  • in other words , we have defineed the ratio of two temperatures in a way is independent of nature of the thermometric substance.
  • Taking the reciprocal of equation 1 , we have
    • q₁/q₂ = θ₁/θ₂ or 1- q₁/q₂ = 1- θ₁/θ₂
    • q₂–q₁/q₂ = θ₂-θ₁/θ₂
  • Now, if θ1 = 0 (which corresponds to zero of nwe scale), then
    • q₂-q₁/q₂= θ₂/θ₂= 1
  • In other words, this means that zero on the kelvin scale is the temperature of the sink for a reversible engine whose efficiency is unity but according to equatiion (15) section 1.3,
    • Efficiency(η) = q₂-q₁/q₂ = T₂ – T₁/T₂
  • it is possible only at absolute zero on the ideal gas scale of temperature. from this, it can be concluded that kelvin scale and gas scale are identical; provided the gas is supposed to be ideal, The temperature on the thermodynamic scale is , therefore expressed in kelvin (K).

Learning fact

The only two non-silvery metals are gold and copper. …