T26 The Transformation of an Amount of Heat Added in an Isochoric Process
Let us add in an isochoric process some heat energy ΔQ to an ideal gas. Using the results of 22 to 25 and our assumption of constant volume the following relations hold:
- ΔV = 0 = ΔV'
- ΔE = ΔQ + ΔW = ΔQ – P · ΔV = ΔQ = ΔH
- ΔE' = ΔE / √ = ΔQ / √ = ΔH / √
- ΔE' = ΔQ' + ΔW' = ΔQ' + v · Δp – P' · ΔV' = ΔQ' + v · Δp
- v is constant
- Δp = ΔE' · v / c2
Even in an isochoric process ΔW' does not vanish. In section 24 we pointed out that any increase in energy necessarily goes along with an increase of momentum and kinetic energy. From these relations the transformation of ΔQ is easily deduced:
ΔQ' = ΔE' – v · Δp = ΔE' – ΔE' · v2 / c2 = ΔE' · ( 1 – v2 / c2 ) = ( ΔE / √ ) · ( 1 – v2 / c2 ) =
= ΔE · √ = ΔQ · √
Heat energy is transformed by multiplication by the root term !
Q' = Q · √
The difference
ΔE' – ΔQ' = ΔQ / √ – ΔQ · √ = ( ΔQ / √ ) · v2 / c2 = ΔQ' · v2 / ( c2 - v2 )
goes to the kinetic energy. All those who write Q' = Q / √ pretend all the energy added being thermic, i.e. representing kinetic energy of the particles relative to the center of mass of the gas. Nothing is left for the kinetic energy of the whole ensemble, total momentum of the gas does not increase. This contradicts our results of section 24.
In section 30 we will follow a completely different line of arguments and show again that heat contents appear smaller to a fast moving observer.
