Undergrad How does ΔU relate to Cv in an adiabatic process?

[Tian En]

For an adiabatic process, Q = 0.
From the first law of thermodynamic,
ΔU = Q + W on the system
when Q = 0,
W = -PΔV,
then why is it that ΔU = Cv ΔT when Cv is meant for the constant volume? We know that when there is work done, the volume is changing, and making use of Cv sounds like an contradiction. Please enlighten me. Thank you.

 

[mjc123]

Let's split it into two processes: Change temperature from T₁ to T₂ at constant volume ΔU₁ = Cv ΔT
Change volume from V₁ to V₂ at constant temperature ΔU₂ = 0
ΔU = ΔU₁ + ΔU₂ = Cv ΔT
As U is a state function, ΔU is independent of pathway.

 

[Chestermiller]

Let's split it into two processes: Change temperature from T₁ to T₂ at constant volume ΔU₁ = Cv ΔT
Change volume from V₁ to V₂ at constant temperature ΔU₂ = 0
ΔU = ΔU₁ + ΔU₂ = Cv ΔT
As U is a state function, ΔU is independent of pathway.

This is true only for an ideal gas (since, for an ideal gas, the internal energy is a function only of temperature).

 

[Tian En]

Since U is a state function, ΔU independent of pathway, it means I can also use CpΔT = -PΔV, right?

 

[Chestermiller]

Since U is a state function, ΔU independent of pathway, it means I can also use CpΔT = -PΔV, right?

No. $$\Delta U=nC_v\Delta T$$. On the other hand, $$\Delta H=nC_p\Delta T$$

 

[Tian En]

I see. I have yet to learn enthalpy.
let's say for an adiabatic process, Q = 0, ΔU = W on the system = -PΔV, and ΔU = nCvΔT = -PΔV.
If we keep the pressure P, number of moles n and molar specific heat capacity Cv constant and we compress the system by ΔV (-ve) and hence we have -PΔV (+ve) and therefore the ΔT (+ve). However, the ideal gas law PV = nRT => PΔV = nRΔT implies ΔV (+ve) and ΔT (+ve) for constant n, R and P.

Previously, we have seen that ΔV (-ve) => ΔT (+ve), but now ΔV (+ve) => ΔT (+ve) from ideal gas law. Is there a contradiction?

 

[Chestermiller]

I see. I have yet to learn enthalpy.
let's say for an adiabatic process, Q = 0, ΔU = W on the system = -PΔV, and ΔU = nCvΔT = -PΔV.
If we keep the pressure P, number of moles n and molar specific heat capacity Cv constant and we compress the system by ΔV (-ve) and hence we have -PΔV (+ve) and therefore the ΔT (+ve). However, the ideal gas law PV = nRT => PΔV = nRΔT implies ΔV (+ve) and ΔT (+ve) for constant n, R and P.

Previously, we have seen that ΔV (-ve) => ΔT (+ve), but now ΔV (+ve) => ΔT (+ve) from ideal gas law. Is there a contradiction?

Please tell me how you think you are going to compress the gas is you keep the pressure constant and don't remove heat.

 

[Tian En]

So, this adiabatic equation no longer hold because by keeping the pressure constant, it violates the ideal gas law and by adding the term Q (-ve) to the equation, it becomes: ΔU = nCvΔT = -PΔV + Q (Isobaric equation)

 

[Chestermiller]

So, this adiabatic equation no longer hold because by keeping the pressure constant, it violates the ideal gas law and by adding the term Q (-ve) to the equation, it becomes: ΔU = nCvΔT = -PΔV + Q (Isobaric equation)

I have no idea what you are talking about. All I can say is that, if you don't increase the pressure, you can't compress the gas (if the system is adiabatic).

 

[HimanshuM2376]

For an adiabatic process, Q = 0.
From the first law of thermodynamic,
ΔU = Q + W on the system
when Q = 0,
W = -PΔV,
then why is it that ΔU = Cv ΔT when Cv is meant for the constant volume? We know that when there is work done, the volume is changing, and making use of Cv sounds like an contradiction. Please enlighten me. Thank you.

∆U=Cv∆T comes from the fact that whatever heat is added to an ideal gas keeping it's volume constant will only increase its internal energy and for an ideal gas internal energy is a function of temperature only.
Now we imagine that the system is undergoing an adiabatic process in which the heat added to the system is zero, and as per our knowledge of first law we know that the change in internal energy is a point function. Therefore if the states 1 and 2 are defined then ∆U will remain the same irrespective of the path undergone by a system . So in an adiabatic process ∆U =Cv∆T = W.
Hope this helps you.

 

[Tian En]

I have no idea what you are talking about. All I can say is that, if you don't increase the pressure, you can't compress the gas (if the system is adiabatic).

I was just trying to relate the formula to physical impossibility. When there are inconsistencies in the formula, it is impossible to happen physically. i.e. By keeping the pressure constant, it can never be adiabatic, because Q is no longer 0.

Hope this helps you.

It helps. Thank you. Since the change internal energy just as a function of temperature and previously I learned that ΔU ≠ nCpΔT but ΔU = nCvΔT, is it possible to relate Cp and ΔT to the ΔU?

 

[Chestermiller]

I was just trying to relate the formula to physical impossibility. When there are inconsistencies in the formula, it is impossible to happen physically. i.e. By keeping the pressure constant, it can never be adiabatic, because Q is no longer 0.

It helps. Thank you. Since the change internal energy just as a function of temperature, I am looking at how to relate Cp and ΔT to the ΔU since previously I learned that ΔU ≠ nCpΔT but ΔU = nCvΔT, is it possible?

$\Delta U=nC_p\Delta T$ is just plain incorrect. If you learned it that way, you learned it wrong.

 

[I like Serena]

It helps. Thank you. Since the change internal energy just as a function of temperature and previously I learned that ΔU ≠ nCpΔT but ΔU = nCvΔT, is it possible to relate Cp and ΔT to the ΔU?

For an ideal gas we have Cp=Cv+R (see e.g. heat capacity on wiki), so ΔU = n(Cp-R)ΔT.

 

[Tian En]

Great, thank you.