# Gas Specific Heat: When to use Cv versus Cp



## Ijoinedbecausecovid (Apr 25, 2020)

How do you know when to use Cv vs. Cp?

I read the below and it makes complete sense to me, but if I'm given an air flow heater problem how do I know if the system is at fixed volume or fixed pressure?

"Cv for a gas is the change in internal energy (U) of a  system with respect to change in temperature at a fixed volume of the  system i.e. Cv =(∂ U/∂ T)v  whereas Cp  for a gas is the change in the enthalpy (H) of the system with respect  to change in temperature at a fixed pressure of the system i.e Cp = (∂ H/∂ T)p.
We  know that, ΔH = ΔU + PΔV (+ VΔP, ΔP=0 for constant pressure) . So the  enthalpy term is  greater than the internal energy term because of  the PΔV term i.e in case of a constant pressure process more energy is  needed, to be provided to the system as compared to that of a constant  volume process to achieve the same temperature rise, as some energy is  utilized in the expansion work of the system. And the relation that  correlates these two is Cp = Cv + R
But since liquids and solids can practically assumed to be incompressible, Cp and Cv for them have almost same values and hence only a single value of specific heat is used for them."


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## Slay the P.E. (Apr 26, 2020)

Ijoinedbecausecovid said:


> How do you know when to use Cv vs. Cp?
> 
> I read the below and it makes complete sense to me, but if I'm given an air flow heater problem how do I know if the system is at fixed volume or fixed pressure?
> 
> ...


Great question. *For ideal gases* it can be shown that both internal energy, _u_, and enthalpy, _h_ depend on temperature only (you can find this proof in any Thermo textbook).  Since _u_ and _h _depend only on temperature for an ideal gas, the specific heats _Cv_ and _Cp_ also depend on temperature only. Thus, for ideal gases, the partial derivatives in the definitions provided in your post can be replaced by ordinary derivatives:

In other words: For *ideal gases* _Cv_ = d_u_/d_T_ and _Cp_ = d_h_/d_T. _Furthermore, if you assume_ Cv_ and _Cp _are constant then these relationships can be easily integrated to yield:

Δ_u_ = (_Cv_)Δ_T_  and   Δ_h_ = (_Cp_)Δ_T_  

So, where does this leave us? Well... if you need to calculate a change in internal energy, then you do it by using (_Cv_)Δ_T_ -- no matter what the process is. If you need Δ_u_ for a constant pressure process then Δ_u_ is still (_Cv_)Δ_T._

If you need to calculate a change in enthalpy, then you do it by using (_Cp_)Δ_T_ -- no matter what the process is. If you need Δ_h_ for a constant volume process then Δ_h_ is still (_Cp_)Δ_T. _ 

If you are given an air flow heater, then you are dealing with an OPEN system. That is a system which has mass flow through it (unlike, say a tank with closed inlet and outlet valves which would be a CLOSED system). Now, take a look at the energy balance for a generic open system: 

​
You will note that what comes into play for such systems is the enthalpy at the inlet, _h_i and the enthalpy at the exit, _h_e. So, in OPEN systems (such as your steady-flow heater), you are usually interested in how much the enthalpy _changes_ as the gas flows from the inlet to the exit (_h_e - _h_i), regardless of the nature of the process. Since you are working with an ideal gas then (_h_e - _h_i) =  (_Cp_)(_T_e - _T_i) regardless of how the heater is accomplishing its task.

Let me know if this isn't clear. I can come up with some easy examples to walk through if needed.


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## Dean Agnostic (Apr 26, 2020)

This brings back so many memories from Fluid Mechanics, Introduction to Environmental Engineering, Thermodynamics and Physics classes.


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## Ijoinedbecausecovid (Apr 27, 2020)

Wow!! Thank you so much! This is a way better answer than any I got during undergrad and grad school. I think I follow everything you've laid out. It may take a while to wrap my mind around it conceptually, but it seems like the big take away that I will need for the PE (I'm taking the MDM test so a problem like this is unlikely to appear and should be simple enough if it does) is that an open system will require Cp and a closed system will require Cv. Is that a good general assumption?


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## Audi Driver P.E. (May 5, 2020)

Slay the P.E. said:


> Great question. *For ideal gases* it can be shown that both internal energy, _u_, and enthalpy, _h_ depend on temperature only (you can find this proof in any Thermo textbook).  Since _u_ and _h _depend only on temperature for an ideal gas, the specific heats _Cv_ and _Cp_ also depend on temperature only. Thus, for ideal gases, the partial derivatives in the definitions provided in your post can be replaced by ordinary derivatives:
> 
> In other words: For *ideal gases* _Cv_ = d_u_/d_T_ and _Cp_ = d_h_/d_T. _Furthermore, if you assume_ Cv_ and _Cp _are constant then these relationships can be easily integrated to yield:
> 
> ...


I sure wish I had you as an instructor during undergrad. This is excellent!


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