Question regarding thermodynamics

[utkarshakash]

Homework Statement

Why dU=nCdT true for all processes?

 

[nil1996]

This is similar to https://www.physicsforums.com/showthread.php?t=715192

 

[rude man]

Homework Statement

Why dU=nCdT true for all processes?

It is only true for an ideal gas, and only for a quasi-static process.

And C should read c_v, the specific heat capacity at constant volume.
So it's dU = nc_vdT.

 

[utkarshakash]

It is only true for an ideal gas, and only for a quasi-static process.

And C should read c_v, the specific heat capacity at constant volume.
So it's dU = nc_vdT.

Is ΔQ=nCpΔT true for all processes?

 

[Tanya Sharma]

Is ΔQ=nCpΔT true for all processes?

No...only for isobaric process.

 

[utkarshakash]

No...only for isobaric process.

Is it because the equation involves Cp ? If it is, then returning to my original question, change in internal energy also involves Cv but nevertheless, it remains true for all processes whether or not the volume is constant (assuming ideal gas). How can this be justified?

 

[Tanya Sharma]

Is it because the equation involves Cp ?

Yes...There are infinite number of molar specific heat capacities.The two important and also easy to calculate are C_P and C_V

If it is, then returning to my original question, change in internal energy also involves Cv but nevertheless, it remains true for all processes whether or not the volume is constant (assuming ideal gas). How can this be justified?

You can understand this in three steps.

1. ΔU = (f/2)nRΔT .This applies to all kinds of processes involving ideal gases.

2. Next consider an isochoric process(constant volume process) .

Using First Law of Thermodynamics, ΔQ = ΔU + ΔW

Now, ΔQ = nCvΔT ,ΔU = (f/2)nRΔT and ΔW = 0

So,we have nCvΔT = (f/2)nRΔT

or, Cv = (f/2)R .i.e molar heat capacity at constant volume for an idela gas is a constant .

3. Now come back to isobaric process .

As we have noted in point 1 , ΔU = (f/2)nRΔT , in an isobaric process for an ideal gas .But Cv for the gas (even though the gas is undergoing a constant pressure process) is equal to (f/2)R .

So we have ΔU = nCvΔT.

Thus ,we see that this relation ΔU = nCvΔT applies to all kinds of processes involving an ideal gas ,just like ΔU = (f/2)nRΔT holds for any process involving an ideal gas.

 

[rude man]

Is it because the equation involves Cp ? If it is, then returning to my original question, change in internal energy also involves Cv but nevertheless, it remains true for all processes whether or not the volume is constant (assuming ideal gas). How can this be justified?

Definition of an ideal gas:
1. pV = nRT
2. ∂U/∂p at constant T = 0.

From the above, you can show that ∂U/∂V at constant T also = 0.

So U is a function of T only.

First law: dU = δQ - pdV

and C_V = δQ/dT at constant volume by definition;
but at constant volume, dV = 0, so

C_v = ∂U/∂T at constant V.

But since U = U(T) only, and not a function of V, then the partial derivative can be replaced by the total derivative and we wind up with

dU = C_VdT for all quasi-static processes of an ideal gas.

 

[CAF123]

I have a more conceptual question related to this; Since dU is a state variable, a change in dU of a material between two points on a PV diagram is independent of the path taken. So, given this, why can we not define, for example, a constant pressure process between these two end points and define dU = C_PdT?

 

[MegaDeth]

(deleted)

 

[rude man]

I have a more conceptual question related to this; Since dU is a state variable, a change in dU of a material between two points on a PV diagram is independent of the path taken. So, given this, why can we not define, for example, a constant pressure process between these two end points and define dU = C_PdT?

C_p = δQ/dT at constant p, not dU/dT at constant p.

C_p = δQ/dT|_p by definition; but
δQ = dU + p dV
but if p is constant, V is not, otherwise there is no change in state;

so C_p ≠ dU/dT|_p.

 

[CAF123]

Hi rude man. I understand that, but I am trying to understand how it agrees with the fact that internal energy is a state variable of the substance. As far as I understand, the change in internal energy of a substance between two points on a PV diagram does not depend on the path taken. So why not consider a constant pressure process and write dU = c_pdT?

Alternatively, even for a non constant volume process dU = c_vdT still holds. How does your argument take this into consideration?
Thanks,

 

[rude man]

I think I did my best to explain. Maybe someone else can do a better job.

 

[Chestermiller]

dU and dH are given as follows:
$$dU=TdS-PdV$$
$$dH=TdS+VdP$$
For dH, we can also write:
$$dH=T(\frac{\partial S}{\partial T}dT+\frac{\partial S}{\partial P}dP)+VdP=T\frac{\partial S}{\partial T}dT+(T\frac{\partial S}{\partial P}+V)dP$$
The heat capacity at constant pressure is defined as:
$$C_p\equiv\frac{\partial H}{\partial T}$$
Therefore, from the previous equation,
$$dH=C_pdT+(T\frac{\partial S}{\partial P}+V)dP$$
together with,
$$C_p=T(\frac{\partial S}{\partial T})_P$$
For an ideal gas, it can be shown that:
$$(T\frac{\partial S}{\partial P}+V)=0$$
Therefore, for an ideal gas,
$$dH=C_pdT$$
This is independent of pressure.

Now, for dU we can also write:
$$dU=T(\frac{\partial S}{\partial T}dT+\frac{\partial S}{\partial V}dV)-PdV=T\frac{\partial S}{\partial T}dT+(T\frac{\partial S}{\partial V}-P)dV$$
The heat capacity at constant volume is defined as:
$$C_v\equiv(\frac{\partial U}{\partial T})_V$$
Therefore, from the previous equation,
$$dU=C_vdT+(T\frac{\partial S}{\partial V}-P)dV$$
together with,
$$C_v=T(\frac{\partial S}{\partial T})_V$$
For an ideal gas, it can be shown that:
$$(T\frac{\partial S}{\partial V}-P)=0$$
Therefore, for an ideal gas,
$$dU=C_vdT$$

 

[utkarshakash]

U and H can both be regarded as equilibrium functions of the two intensive variables T and P:

U=U(T,P)
H=H(T,P)

Therefore,
$$dU=\frac{\partial U}{\partial T}dT+\frac{\partial U}{\partial P}dP$$
$$dH=\frac{\partial H}{\partial T}dT+\frac{\partial H}{\partial P}dP$$
The heat capacities C_v and C_p are defined as:
$$C_v\equiv\frac{\partial U}{\partial T}$$
$$C_p\equiv\frac{\partial H}{\partial T}$$
Therefore,
$$dU=C_vdT+\frac{\partial U}{\partial P}dP$$
$$dH=C_pdT+\frac{\partial H}{\partial P}dP$$

For an ideal gas, U = U(T) and H=H(T), and $\frac{\partial U}{\partial P}=0$ and $\frac{\partial H}{\partial P}$=0

Nicely explained.

 

[Chestermiller]

I made several corrections to my original posting #14. Please disregard what I wrote previously. The posting is now correct. Sorry for any confusion I may have caused.

Chet

 

[Radhakrishnam]

The first Law, which is applicable to all processes says: Q = ΔU + W. When only mechanical work (PdV work) is involved we write the first law as:Q = ΔU + PdV. For a constant volume process dV = 0. Therefore, Q = ΔU = nC_VdT.

It is true that U is a state function and has a unique value for ΔU connecting two equilibrium states A, B of a system. If A and B are conneted by a const volume process, then A and B cannot be connected by a const pressure process. A constant pressure process takes the system from A to state C (≠B). ΔU_AB≠ ΔU_AC.

Therefore, ΔU for a constant volume process is not equal to ΔU for a constant pressure process.

However, we can take the system from A to C by a constant pressure process and from C we can take it to state B by one more process (or step). If the system is an ideal gas, then if B and C have the same temperature, that is if B and C lie on a reversible isotherm then ΔU_AB= ΔU_AC, since U of an ideal gas is a function only of temperature and is not a function of volume or pressure.

Thus for an ideal gas: Q_V = nC_VdT = ΔU
Q_P = nC_VdT + PdV = ΔU + PdV.