Thermodynamics Cv = Cp + R Question

[drcrabs]

Can someone tell me why Cv = Cp + R

 

[so-crates]

Well first of all its Cp - Cv = nR, where is your moles of gas, and its only true for an ideal gas. You find this about about halfway through the first semester of physical chemistry.

The formal definitions of Cv and Cp are

C_v = \frac{\partial U}{\partial T} and

C_p = \frac{\partial H}{\partial T}

Where U is the internal energy and H is the enthalpy, defined to be H = U + pV. But for an ideal gas, pV = nRT. Substitutiotn this into the defition for Cp we get

C_p = \frac{\partial (U + nRT)}{\partial T}

C_p = \frac{\partial U }{\partial T} + \frac{\partial (nRT)}{\partial T}

C_p = \frac{\partial U}{\partial T} + nR

C_p = C_v + nR

 

[drcrabs]

What are on?

 

[so-crates]

Are you talking about \partial ? Thats the symbol for a partial deriviative. Its like a deriviative but for functions of mroe than one variable. To calculate it, you treat the other variables as constants, except for the one youare differentiating against.

 

[drcrabs]

Can u prove the equation by the consideration of an isobaric process?

 

[so-crates]

No, path has nothing to do with it. It follows directly from the definition of U, H, Cp and Cv, and our assumption of an ideal gas. If our gas it is not ideal, that equation does not hold.

 

[drcrabs]

But what if we consider an ideal gas undergoing a isochoric process and
how the first law of thermodynamics applies to it.
ΔU= Q - W

Since the work is defined by the pressure * the change in volume
W = pΔV
then work is 0

Hence ΔU = Q - 0 = Q

but Q = nCvΔT hence ΔU = nCvΔT

But since the change in internal energy is independent of path taken
for any process ΔU = nCvΔT.

Now let us consider a isobaric process
In this case Q = nCpΔT.

Now taking the definition of work to be W = pΔV
and using the defition of the equation of state pΔV = nRΔT

then W = nRΔT

Now consider the first law of thermodynamics again

ΔU= Q - W

then nCvΔT = nCpΔT + nRΔT

leaving us with Cv = Cp + R