State first law of thermodynamics and apply this law to obtain a relationship between two specific heats of a gas. 

Explanation

First law of thermodynamics : It states that, the amount of heat absorbed by a system is equal to the sum of increase in internal energy of system and work done by the system. (i.e.) `dQ = dU + dW`.
Relation between `C_(P)`  and `C_(V)` :  Consider one mole of an ideal gas. Let the gas is heated at constant volume so that its temperature increases by dT.
If `Q_(1)` = heat supplied to 1 mole of gas at constant volume, then `Q_(1) = C_(v)dT`. …….(i)
  Now, let the gas be heated at constant pressure, so that its temperature increases by dT.
  If `Q_(2) =`  heat supplied to 1 mole of gas at constant pressure, then `Q_(2) = C_(P)dT` ……….(ii)
When gas is heated at constant volume, it will not perform external work. According to first law of thermodynamics, the heat supplied will just increase the internal energy of the gas.
Therefore, equation (i) becomes, `dU = C_(v)dT` ……..(iii)
  When heat is supplied at constant pressure, it will increase internal energy as well as enable the gas to perform work (dW). If dV is increase in volume, then work done by the gas
dW = PDV ....(iv) According to first law of thermodynamics
`Q_(2) = Q_(1) + dW`
`Q_(2) = dU + dW`
`C_(P)dT = C_(v)dT + PdV`
`(C_(P) - C_(V)) dT = PdV`
or `C_(P) - C_(V) = (PdV)/(dT)` …….(v)
For an ideal gas, PV = RT
  Differentiating above equation, we get PdV = RdT.
Substituting value of PdV in equation (v), we get
`C_(P) - C_(V) = (RdT)/(dT)  = R` or `C_(P) - C_(V) = R` ……..(vi)
If `C_(P)` and `C_(V)` are measured in heat units and R in units of work, then `C_(P) - C_(V) = (R)/(J)`