Obtain the relation between specific heat capacity at constant pressure and specific heat capacity at constant volume for an ideal gas.

Explanation

From first law of thermodynamics for small changes `DeltaQ=DeltaU+DeltaW`….(1)
At constant pressure if volume change `DeltaU`, then `DeltaQ=DeltaU+PDeltaV`…(2)
At constant volume heat absorbed by `DeltaQ`, then `DeltaP=0`
`therefore DeltaQ=DeltaU`...(3)
Internal energy of 1 mole of gas having f degree of freedom `U=(f R DeltaT)/2 therefore DeltaQ=(f R DeltaT)/2`
`therefore ((DeltaQ)/(DeltaT))=(fR)/2`...(4)
`therefore C_V=((DeltaQ)/(DeltaT))_V=((DeltaU)/(DeltaT))_V`
For an ideal gas , the internal enregy U depend on only temperature hence no need to write in subscript V or P.
`therefore` From equation (4) and (5).
`C_V=(fR)/2` ...(6)
Heat is supplied to gas at constant pressure,
`(DeltaQ)_P = DeltaU+PDeltaV`
but for an ideal gas `PV=RT [ because mu=1]`
`therefore PDeltaV=RDeltaT rArr P=(DeltaV)/(DeltaT)=R`...(7)
`therefore (DeltaQ)_P=DeltaU+RDeltaT`
`therefore C_P=((DeltaQ)/(DeltaT))_P=((DeltaU)/(DeltaT))_P+P((DeltaV)/(DeltaT))_P`
`therefore C_P=((fR)/2) +R` ...(8)
`[because ((DeltaQ)/(DeltaT))_P =((DeltaU)/(DeltaT))_P = (DeltaU)/(DeltaT)]`
`therefore` Substrating equation (7) from equation (8),
`C_P-C_V=(fR)/2+R-(fR)/2`
`therefore C_P-C_V=R`