MOLAR SPECIFIC HEAT AT CONSTANT PRESSURE


    MOLAR SPECIFIC HEAT AT CONSTANT PRESSURE
    “The a mount of heat energy required to raise the temperature of one mole of a gas
    by one Kelvin at constant pressure is called molar specific heat at constant pressure”.
      It is denoted by Cp.
    MATHEMATICAL EXPRESSION
      If DQp is the amount of heat is supplied at constant pressure to ‘n’ moles of a gas to increase the   temperature bDT K, then
    DQ= n CpDT

    OR
    Cp = D Q/n D T
      heat supplied at constant pressure is consumed in two purposes:
      (1) To raise the temperature of gas.
      (2) To do work against external pressure.
    SHOW THAT CP – CV = R
      Consider ‘n’ moles of an ideal gas contained in a cylinder fitted with a frictionless piston. If the piston is   fixed and the gas is heated, its volume remains constant and all the heat supplied goes to increase the   internal energy of the molecules due to which the temperature of the gas increases. If DQv is the amount   of heat supplied and DT is the rise in temperature then,

    DQ= n CvD T

    The pressure of the gas increases during this process, but no work is done
    because the volume is kept constant.

    Hence D W = 0. 
    applying first law of thermodynamics

    Heat supplied = Increase in internal energy + Work done 
    DQD U + 0
    DQD U
    OR
    nCvD T = D U

      If the piston is free to move, the gas may be allowed to expand at a constant pressure. Let the amount of   heat supplied is now is DQpThe addition of heat causes two changes in the system:

     Increase in internal energy
     Work done against external pressure  According to the first law of thermodynamics:
                                        DQ = DU + D              {But DW = PDV} DQP = DU + PDV  
      Since DQp nCpDT
      and    DU = nCvD, therefore,
    nCpD= nCvDT + PDV………………..(1)
      We know that PV = nRT
      At T1 Kelvin: PV1 = nRT1 …..(a)
      At T2 Kelvin: PV2 = nRT2…..(b)
      Subtracting (a) from (b)
      PV2 – PV1= nRT2 – nRT1  P(V2 – V1)= nR(T2 – T1          {(V2 – V1) = DV and (T2 – T1) = DT }
      PDV = nRDT
      Putting the value of PDV in equation (1)
    nCpD= nCvDT + nRDnCpDnDT(Cv + R)Cp = (Cv + R)Cp – Cv = R