We all know that Cv for a monoatomic ideal gas is 3/2 R and when we

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we all know that Cv for a monoatomic ideal gas is 3/2 R and when we have constant volume Q=deltainternal energy =n Cv delta T.
Ive read that in any case delta internal energy = nCvdelta T why?
 
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It comes from the fist law of thermodynamics. This states

ΔU=ΔQ-ΔW or ΔU=ΔW-ΔQ depending on which convention you use.For an isovolumetric curve ΔW is zero, so ΔQ=ΔU
 


i mean why is it in each case(isobaric, adibaetic ) we can say that delta internal energy is n Cv delta T??
i need a physical explanation please?
 


elabed haidar said:
i mean why is it in each case(isobaric, adibaetic ) we can say that delta internal energy is n Cv delta T??
i need a physical explanation please?

You need to consider the 1st law properly where
ΔU is the change in the internal energy
ΔQ is the heat added or removed from the system
ΔW is the work done on/by the system

ΔU=ΔQ-ΔW

The 1st law of Thermodynamics basically says that in a system energy must be conserved, you need to consider what each component is. For the work done you need to find the area under the P-V graph describing the process. With an isovolumetric curve, then you know that the work is zero, as there is no area under the P-V graph representing an isovolumetric curve.

For isobaric and adiabatic processes you can't use Cv as Cv is the molar heat capacity for a constant volume, in adiabatic and isobaric processes the volume doesn't stay constant, so you can't use Cv. For an Isobaric process you need to consider the molar heat capacity for constant pressure, Cp of the substance.
 


we are on the same page here , so why is it always that delta internal energy = n Cv delta T in all cases?
 


elabed haidar said:
we are on the same page here , so why is it always that delta internal energy = n Cv delta T in all cases?

It's not, that only works for constant volume.

The equation you wrote gives you the heat transferred in a system, hence why Cv is known as the heat capacity.Cv in the above equation means the molar heat capacity at a constant volume, for a constant pressure the heat capacity is different.

The change in internal energy is given by the 1st law of Thermodynamics which is the below:

ΔU=ΔQ-ΔW
 


man my book says ΔU = n Cv ΔT ? in any case as long ΔT is the same in any case
 


elabed haidar said:
man my book says ΔU = n Cv ΔT ? in any case as long ΔT is the same in any case

That only hold when you have a constant volume, which is why the heat capacity in that case is referred to as CV
 


i swear i agree with you but the book says it can be in any case and we used it in class so many times
 
  • #10


Vagn said:
That only hold when you have a constant volume, which is why the heat capacity in that case is referred to as CV

Actually, that equation is always true, regardless of volume. The change in internal energy is always equal to the quantity of gas multiplied by Cv ΔT (so long as the gas is ideal, of course).
 
  • #11


that is what i want to know can you elaborate on that,
i mean why? i would love to hear a physical explanation for that , and thank you very much for your help
 
  • #12


Sure.

The reason is because you are looking at internal energy. It's true that the equation Qin = CvΔT is only true for constant volume. However, if you care about internal energy, it's always true.

To show why this is, think about what internal energy actually means. Internal energy is the energy contained within a system purely because the system's molecules are moving around. Temperature is a measure of the mean kinetic energy per particle in a system. Since the kinetic energy per particle at a given temperature is not dependent on the pressure (or, similarly, the volume), the internal energy should also be independent of pressure or volume.

Now, the reason this is often confused is because of the equation above - Qin = CvΔT only at constant volume. At constant pressure, Qin = CpΔT. However, this is the case because Qin isn't only going to the internal energy. In the case of the constant pressure case, some of the energy that is going into the system is being extracted by the fact that the system is expanding, and therefore doing work on its surroundings. Since the internal energy is equal to ΔQ-ΔW, this work done by the system exactly cancels out the different amount of energy that needed to be added to the system for a given ΔT, and thus, the change in internal energy is the same for the same ΔT, even though the system is not maintaining constant volume.
 
  • #13


cjl said:
this work done by the system exactly cancels out the different amount of energy that needed to be added to the system for a given ΔT, and thus, the change in internal energy is the same for the same ΔT, even though the system is not maintaining constant volume.

?
 
  • #14


It may help if you consider the problem a little different.
The change in internal energy for ideal gas is proportional to the change in temperature (and number of mole), for any transformation
\Delta U = n A \Delta T
where A is a constant which of course does not depend on the process. The energy is a function of state and depends only on initial and final states.
Now, we want to find this constant in terms of specific heats.

For a constant volume process the work is zero so
\Delta U = Q_v

If we write Q_v as a function of a specific heat we have
Q_v=n C_v \Delta T
n A \Delta T= n C_v \Delta T
and it follows that the constant A is equal to C_v.

For other processes the work is not zero but it does not mean that the constant A will be different. It does not need to be proven any further but it may help your understanding to do it for other processes.
For example, for constant pressure
\Delta U =n A\Delta T = Q_p -p\Delta V=nC_p\Delta T-nR\Delta T=n(C_p-R)\Delta T
It follows that
A=Cp-R which for ideal gas we know that is Cv.
 
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  • #15


thank you man very much , i have another question but it has to do with diffraction , do you mind? i just don't understand diffraction in a double slit , ? how is it related to interference?? like if i want to find how many bright fringes within the central bright fringe of the diffraction envelope?
or how many bright fringes are found between the bright fringes of the first and second of diffraction?
 
  • #16


I think that the right procedure would be to start a new thread if you have a new question.
 
  • #17


anyway thank you very much
 
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