Question

Show that Cp-Cv = R, where Cp and Cv are molar specific heats at constant pressure and constant volume respectively of an ideal gas , and R is a gas constant.​

Answer 1

                                   FIRST LAW OF THERMODYNAMICS 

The quantity of heat energy of heat energy absorbed by the system is equal to the sum of increases in the internal energy of the system and the external work done by it.

 dQ = du + dw

But, dw = Pdv

∴ dQ = du + Pdv  

{ SIGN CONVENTION }

(1) Heat is absorbed by the system dQ = +ve

(2) Heat is released by the system dQ = +ve

(3) Work done by the system dw = +ve

(4) Work done on the system dw = -ve

(5) Increase on internal Energy du = +ve

(6) Decrease in Internal Energy du = -ve

Now, Let us consider a gas with a massless and frictionless pistion. Let the gas expands by a very small volume dV.

Let p is pressure and V is volume of the gas. A is the area of cross-section of the piston, where dx is the distance moved by the piston. Let the volume expansion is dV.

So, the work done by the gas

       dW = force × dz

       dW = pA × dx = p( A dx) = pdV            [ ∴ Adx = dV]

Now, dW = pdV

For small displacement dx, the pressure is assumed to be constant.

Thus,from first law ΔQ = ΔU + pdV

where,

           ΔQ = heat transition between system and surrounding 

           ΔU = change in internal energy of the system.

{ RELATION BETWEEN Cp and Cv }

→ We called this relation as Mayer's Formula.

We can establish the relation between specific heat capacity at constant volume (Cv) and specific heat capacity at constant pressure (Cp) of a gas.

For an ideal gas, the relation between Cp and Cv is

                           Cp - Cv = R -------------------> (1)

∴ This relation is known as Mayer's Formula.

Now, To establish the relation, we need to begin from the first law of thermodynamics for 1 mole of gas.

                             ΔQ = ΔU + pΔV

If heat ΔQ  is absorbed at constant volume,

∴ pΔV = 0 and ΔQ = CvΔT for one mole of a gas 

Now, ΔV = 0

Then, Cv = (ΔQ/ΔT)v = (ΔU/ΔT)v = (ΔU/ΔT) -----------------> (2)

where the V is dropped in the last step, since U of an ideal gas depends only on the temperature, not on the volume.

Now, heat ΔQ is absorbed at constant  pressure, then

         ΔQ = CpΔT

         Cp = (ΔQ/ΔT)p = (ΔU/ΔT)p = (ΔU/ΔT)p

Now, p can be the dropped from the first term since U of an ideal gas depends only on T, not on pressure.

Now, by using Eq. (2)

or                   Cp = Cv +p(ΔV/Δp)p --------------------> (3)

Now, for 1 mole of an ideal gas, PV = RT

If the pressure is kept constant

         p(ΔV/ΔT)p = R -----------------------> (4)

From Eq. (2) , (3) and (4)

                     { Cp - Cv = R }

Here, Cp and Cv are molar specific heat capacities of an ideal gas at constant pressure and volume and R is the universal gas constant.

The ratio of Cp and Cv  is notified as γ

                                γ = Cp/Cv

And it is also known as heat capacity ratio,

     Cv = R/r-1 and Cp = γ R/r-1

Hope it helps you !