Derive a relation beween the two specific heat capacities of gas on the basis of first law of thermodynamics .

Explanation

Relationship between the two specific heat capacities of gas :
To derive `C_(p)-C_(v)=R` : Let one gram mole of given mass of gas is enclosed within a cylinder with a frictionless air tight - piston . Let P,V be the pressure and volume of the gas at temperature T .
(i) In specific heat at constant volume :
The volume of the given mass of gas must remain constant . Hence , the piston is fixed in position AB .
Let `C_(v)` be the amount of heat energy supplied . It is utilised only to raise the temperature of the gas by `1^(@)C`
`thereforedU=C_(v)dT`

(ii) In specific heat at constant pressure , the pressure must remain constant . Hence the piston is allowed to move freely . The amount of heat energy supplied is used not only to do external work but also to increase the temperature of the gas by `1^(@)C` .
In this case the piston moves forward and work is done againt external pressure .
Suppose by the time the piston moves from AB to CD , the temperature increases by `1^(@)C` .
Let the work done in moving the piston through a distance 'dl' be dW= P dV.
The energy supplied has to increase the internal energy and to do exteranl work .
`thereforeC_(p)dT=dQ=dU+dW`
From first law of thermodynamics
dQ=dU+dW
`thereforeC_(p)dT=C_(v)dT+dW`
`(C_(p)-C_(v))dT=dW=PdV`
But work done `dW=FxxS`
`=PxxAxxdl` (where A is area of cross section of the piston ) = PdV
But PV=RT (for one mole of gas ) .
`thereforedW=PdV=RdT " OR "(C_(p)-C_(v))dT=RdT`
But change of temperature `=dT=1^(@)C` (from definition of specific heat )
`thereforeC_(p)-C_(v)=R`
So difference of molar specific heats of the gas is equals to universal gas constant R .