Why the change in internal energy is considered same?

In summary, the heat capacity at constant volume, Cv, and the heat capacity at constant pressure, Cp, are physical properties of a material that are independent of the heat transferred in any specific process. For an ideal gas, these properties are defined as (\frac{\partial U}{\partial T})_V and (\frac{\partial H}{\partial T})_P, respectively. These properties can be applied in practice to calculate changes in enthalpy and Gibbs free energy.
  • #1
Elz
12
0
1.While studying the thermodynamic properties of perfect gases I saw that the change in internal energy for constant volume process is mCvdT. but it is also the same for const pressure, const temp, const entropy and other reversible non-flow processes. Why is it so?? why it is not mCpdT for const pressure process? The value of Cv and Cp are not same. And if we consider the same system working once for const volume and again for constant pressure then will the temp difference be same? I mean will both the process have same initial and final temperature? If the are same then how can be the value of dU be same as the values of Cv and Cp are different?
does it work like if i heat CO2 at constant pressure then the value of change in internal energy and if i heat the same amount of CO2 at const volume or const entropy then the value of change in internal energy will be same? I mean has it been experimentally established?




2. dU=mCvdT

Why not Cp for const pressure process?
 
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  • #2
Welcome to PF;
the change in internal energy for constant volume process is mCvdT. but it is also the same for const pressure, const temp, const entropy and other reversible non-flow processes. Why is it so??
What determines the change in internal energy: where does it come from?
 
  • #3
First, C = dQ/dt, not dU/dt. So you have no basis in assuming C = dU/dt, whether C is constant pressure, constant volume, or anything else.

The proof that dU = CVdT for an ideal gas rests on the following:

1. Definition of an ideal gas. That definition comprises pV = nRT and also ∂U/∂p|T = 0.

From those two equations you can show that U must be a function of temperature only (and not of p or V).

Then, invoking the 1st law,
dU = dQ - pdV so that
dU/dT = dQ/dT - pdV/dT.

However, having shown that U = U(t) only, we know that pdV/dt = 0 and conclude that

dU/dT = CV for an ideal gas, independent of p or V.
 
  • #4
Welcome to PF

thanks :)

Definition of an ideal gas. That definition comprises pV = nRT and also ∂U/∂p|T = 0.

From those two equations you can show that U must be a function of temperature only (and not of p or V).

Then, invoking the 1st law,
dU = dQ - pdV so that
dU/dT = dQ/dT - pdV/dT.

However, having shown that U = U(t) only, we know that pdV/dt = 0 and conclude that

dU/dT = CV for an ideal gas, independent of p or V.


Thanks a lot for replying. can u please explain how ∂U/∂p|T = 0 and pdV/dt = 0 is derived?
 
  • #5
rude man said:
First, C = dQ/dt, not dU/dt. So you have no basis in assuming C = dU/dt, whether C is constant pressure, constant volume, or anything else.

The proof that dU = CVdT for an ideal gas rests on the following:

1. Definition of an ideal gas. That definition comprises pV = nRT and also ∂U/∂p|T = 0.

From those two equations you can show that U must be a function of temperature only (and not of p or V).

Then, invoking the 1st law,
dU = dQ - pdV so that
dU/dT = dQ/dT - pdV/dT.

However, having shown that U = U(t) only, we know that pdV/dt = 0 and conclude that

dU/dT = CV for an ideal gas, independent of p or V.

Thanks a lot for replying. can u please explain how ∂U/∂p|T = 0 and pdV/dt = 0 is derived?
 
  • #6
Elz said:
Thanks a lot for replying. can u please explain how ∂U/∂p|T = 0 and pdV/dt = 0 is derived?

∂U/∂p|T = 0 is as I said part of the DEFINITION of an ideal gas.

Then we show that U ≠ U(V) as follows:
dU/dV|T = dU/dp|T * dp/dV|T
and dp/dV = -nRT/V2 = -p/V ≠ 0 but dU/dp|T = 0,
dU/dV|T = 0 and U ≠ U(V).

So both derivatives of U wrt to p and V are zero, so U ≠ U(p) ≠ U(V).
(Replace all "d" above with "∂").

dQ = dU + pdV = dU at constant volume. So
CV = ∂Q/∂T|V = ∂U/∂T|V

But for the special case of an ideal gas we know U = U(T) only,
so ∂U/∂T|V = dU/dT. QED
 
  • #7
rude man said:
∂U/∂p|T = 0 is as I said part of the DEFINITION of an ideal gas.

Then we show that U ≠ U(V) as follows:
dU/dV|T = dU/dp|T * dp/dV|T
and dp/dV = -nRT/V2 = -p/V ≠ 0 but dU/dp|T = 0,
dU/dV|T = 0 and U ≠ U(V).

So both derivatives of U wrt to p and V are zero, so U ≠ U(p) ≠ U(V).
(Replace all "d" above with "∂").

dQ = dU + pdV = dU at constant volume. So
CV = ∂Q/∂T|V = ∂U/∂T|V

But for the special case of an ideal gas we know U = U(T) only,
so ∂U/∂T|V = dU/dT. QED

Thanks a lot... It explained everything. All my doubts are clear... :)
 
  • #8
My friend Rude Man urged me to submit the following repsonse, which, for completeness, provides an alternate perspective on the answer to the OP's question.

The heat capacity at constant volume Cv and the heat capacity at constant pressure Cp are physical properties of a material, and are not wedded to the heat transferred in any specific process. These properties are precisely defined mathematically as:
[tex]C_v=(\frac{\partial U}{\partial T})_v[/tex]
[tex]C_p=(\frac{\partial H}{\partial T})_p[/tex]
where H is the enthalpy. For an ideal gas, U and H are functions only of temperature.

Here, for the case of a gas that is not necessarily ideal, is how these heat capacities can be applied in practice:
[tex]dH=TdS+VdP[/tex]
If we regard the molar enthalpy as a function of T and P, then

[tex]dH=(\frac{\partial H}{\partial T})_PdT+(\frac{\partial H}{\partial P})_TdP[/tex]

From these two equations, we have:

[tex](\frac{\partial H}{\partial T})_P=C_p=T(\frac{\partial S}{\partial T})_P[/tex]

and [tex](\frac{\partial H}{\partial P})_T=T(\frac{\partial S}{\partial P})_T+V[/tex]

Now, we need to evaluate [itex](\frac{\partial S}{\partial P})_T[/itex]. We obtain this from the equation for the differential change in Gibbs free energy:

[tex]dG=-SdT+VdP=(\frac{\partial G}{\partial T})_PdT+(\frac{\partial G}{\partial P})_TdP[/tex]
From this, it follows that:
[tex]S=-(\frac{\partial G}{\partial T})_P[/tex]
[tex]V=(\frac{\partial G}{\partial P})_T[/tex]
If we take the partial of the first of these equations with respect to P and the partial of the second with respect to T, and sum the equations, we obtain:
[tex](\frac{\partial S}{\partial P})_T=-(\frac{\partial V}{\partial T})_P[/tex]
Substituting back into the equation for dH, we have
[tex]dH=C_P+(V-T(\frac{\partial V}{\partial T})_P)dP[/tex]

Note that the coefficient of dP in this equation is equal to zero for an ideal gas, but it is not equal to zero beyond the ideal gas region. This equation provides us with what we need to include the effect of pressure on enthalpy. But, to evaluate the second term, we need to know the equation of state (EOS) of the material, beyond the ideal gas region.
 
  • #9
Chestermiller said:
My friend Rude Man urged me to submit the following repsonse, which, for completeness, provides an alternate perspective on the answer to the OP's question.

Thanks a lot Chestermiller,and give my thanks to Rude Man too...
 
  • #10
Oops. There is a glitch in the final equation in my recent reply. The final equation should read:

[tex]dH=C_PdT+(V-T(\frac{\partial V}{\partial T})_P)dP[/tex]

Sorry for the error.

Chet
 

FAQ: Why the change in internal energy is considered same?

1. Why is the change in internal energy considered the same?

The change in internal energy is considered the same because it is a state function, meaning it only depends on the initial and final states of a system, and not on the path taken to get there. This means that the change in internal energy will be the same regardless of the process or reactions that occur within the system.

2. How is the change in internal energy calculated?

The change in internal energy is calculated by taking the difference between the initial and final internal energy of a system. It can also be calculated by adding the heat transferred to a system and the work done on the system.

3. Why is the change in internal energy important?

The change in internal energy is important because it is a measure of the total energy within a system. It is used to determine the direction of heat and work flow within a system and is crucial in understanding the thermodynamic properties of a system.

4. Can the change in internal energy be negative?

Yes, the change in internal energy can be negative. This indicates that the system has released energy in the form of heat or work, resulting in a decrease in internal energy. An example of this is when a gas expands and does work on its surroundings, decreasing its internal energy.

5. How is the change in internal energy related to the first law of thermodynamics?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted. The change in internal energy is a reflection of this law, as it represents the net energy transferred or converted within a system. This means that the change in internal energy is in accordance with the first law of thermodynamics.

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