Thermodynamics Cv = Cp + R Question

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Discussion Overview

The discussion revolves around the relationship between the specific heat capacities at constant volume (Cv) and constant pressure (Cp) for ideal gases, specifically examining the equation Cv = Cp + R. Participants explore definitions, derivations, and implications of this relationship within the context of thermodynamics.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that the equation can be expressed as Cp - Cv = nR, emphasizing that it is valid only for ideal gases.
  • Another participant provides formal definitions of Cv and Cp, explaining their derivation from internal energy (U) and enthalpy (H), and how the ideal gas law (pV = nRT) is incorporated into these definitions.
  • A participant questions the relevance of the path taken in deriving the equation, asserting that it follows directly from the definitions and assumptions regarding ideal gases.
  • Another participant suggests proving the equation through consideration of an isobaric process, prompting a discussion on the applicability of different thermodynamic processes.
  • One participant argues that the change in internal energy is independent of the path taken, using the first law of thermodynamics to relate changes in internal energy and heat transfer for both isochoric and isobaric processes.

Areas of Agreement / Disagreement

Participants express differing views on the relevance of the process path in deriving the relationship between Cv and Cp. While some assert that the derivation is straightforward from definitions, others propose that specific processes should be considered. The discussion remains unresolved regarding the necessity of process considerations in proving the equation.

Contextual Notes

Participants reference the ideal gas assumption and the definitions of specific heat capacities, but there is no consensus on the implications of different thermodynamic processes for the validity of the equation.

drcrabs
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Can someone tell me why Cv = Cp + R
 
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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

[tex]C_v = \frac{\partial U}{\partial T}[/tex] and

[tex]C_p = \frac{\partial H}{\partial T}[/tex]

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

[tex]C_p = \frac{\partial (U + nRT)}{\partial T}[/tex]

[tex]C_p = \frac{\partial U }{\partial T} + \frac{\partial (nRT)}{\partial T}[/tex]

[tex]C_p = \frac{\partial U}{\partial T} + nR[/tex]

[tex]C_p = C_v + nR[/tex]
 
Last edited:
What are on?
 
Are you talking about [tex]\partial[/tex] ? 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.
 
Can u prove the equation by the consideration of an isobaric process?
 
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.
 
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
 

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