Question about u and h and Its Differential Changes for Ideal Gases

In summary, the specific heats for ideal gases are constant, while for real gases they are functions of temperature and pressure. For ideal gases, the internal energy and enthalpy are functions of temperature and specific volume, while for real gases they also depend on pressure. This is due to the fact that in ideal gases, the molecules are far apart and the potential energy of interaction is negligible, making internal energy and enthalpy solely dependent on temperature. However, for real gases, the potential energy of interaction between molecules cannot be ignored and thus affects the internal energy and enthalpy, making them also dependent on pressure.
  • #1
Red_CCF
532
0
Hello

1. I was wondering why internal energy is usually expressed as a function of temperature and specific volume and enthalpy a function of temperature and pressure (ie why is u(T,v) and h(T,p)) and some other set of two properties?

2. For du = [itex]\frac{∂u}{∂T}[/itex]dT + [itex]\frac{∂u}{∂v}[/itex]dv and dh = [itex]\frac{∂h}{∂T}[/itex]dT + [itex]\frac{∂h}{∂p}[/itex]dp, why are the [itex]\frac{∂u}{∂v}[/itex] and [itex]\frac{∂h}{∂p}[/itex] terms zero for ideal gases only?

3. Why are the specific heats for ideal gases a function of temperature only? For real gases does this become cv(T,v) and cp(T,p)?

Thanks very much
 
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  • #2
Are these homework questions?
 
  • #3
DrClaude said:
Are these homework questions?

No, I am trying to brush up on my thermodynamics knowledge and found myself wondering about these questions that I hadn't considered before. I am currently not in any type of school.
 
  • #4
Red_CCF said:
1. I was wondering why internal energy is usually expressed as a function of temperature and specific volume and enthalpy a function of temperature and pressure (ie why is u(T,v) and h(T,p)) and some other set of two properties?
The first law can be written as
$$
dU = Q + W
$$
where ##Q## is the heat entering the system, and ##W## the work performed on the system. Using
$$
dS = \frac{Q}{T}
$$
and writing compression/expansion work on an ideal gas as ##W = -P dV##, you get
$$
dU = T dS - P dV
$$
so it is "natural" to take ##U## as a function of ##S## and ##V## (not of ##T##, as you wrote). Enthalpy is defined as
$$
H \equiv U + PV
$$
from which you get
$$
dH = dU + P dV + VdP = TdS + V dP
$$
(which is essentially a Legendre transformation), so ##H## is a function of ##S## and ##P## (again, not of ##T## as you wrote).

Red_CCF said:
2. For du = [itex]\frac{∂u}{∂T}[/itex]dT + [itex]\frac{∂u}{∂v}[/itex]dv and dh = [itex]\frac{∂h}{∂T}[/itex]dT + [itex]\frac{∂h}{∂p}[/itex]dp, why are the [itex]\frac{∂u}{∂v}[/itex] and [itex]\frac{∂h}{∂p}[/itex] terms zero for ideal gases only?
For non-ideal gases, the facts that you can't consider them as point particles and that there are interparticle interactions mean that the energy will depend in a more complicated way on such things as volume. Check for instance a van der Waals gas.

Red_CCF said:
3. Why are the specific heats for ideal gases a function of temperature only? For real gases does this become cv(T,v) and cp(T,p)?
The specific heats for an ideal gas are constant. For a real gas, they are mostly a function of temperature, but I guess they must be slowly varying functions of volume and pressure also.
 
  • #5
DrClaude said:
The first law can be written as
$$
dU = Q + W
$$
where ##Q## is the heat entering the system, and ##W## the work performed on the system. Using
$$
dS = \frac{Q}{T}
$$
and writing compression/expansion work on an ideal gas as ##W = -P dV##, you get
$$
dU = T dS - P dV
$$
so it is "natural" to take ##U## as a function of ##S## and ##V## (not of ##T##, as you wrote). Enthalpy is defined as
$$
H \equiv U + PV
$$
from which you get
$$
dH = dU + P dV + VdP = TdS + V dP
$$
(which is essentially a Legendre transformation), so ##H## is a function of ##S## and ##P## (again, not of ##T## as you wrote).

So where/how does temperature come in since the definition of cp and cv come from derivative of u and h as a function of T?

DrClaude said:
For non-ideal gases, the facts that you can't consider them as point particles and that there are interparticle interactions mean that the energy will depend in a more complicated way on such things as volume. Check for instance a van der Waals gas.

I guess my confusion is that, how does increasing pressure/specific volume without changing temperature not change the internal energy of the an ideal gas?

DrClaude said:
The specific heats for an ideal gas are constant. For a real gas, they are mostly a function of temperature, but I guess they must be slowly varying functions of volume and pressure also.

I'm a bit confused by this concept, in my book for, c_p for ideal gases is expressed as a polynomial in terms of temperature c_p/R = A+BT+CT^2+DT^3+FT^4 and the coefficients are tabulated for many ideal gases?
 
  • #6
Red_CCF said:
Hello

2. For du = [itex]\frac{∂u}{∂T}[/itex]dT + [itex]\frac{∂u}{∂v}[/itex]dv and dh = [itex]\frac{∂h}{∂T}[/itex]dT + [itex]\frac{∂h}{∂p}[/itex]dp, why are the [itex]\frac{∂u}{∂v}[/itex] and [itex]\frac{∂h}{∂p}[/itex] terms zero for ideal gases only?

The internal energy of a real gas is determined by the kinetic energy of the molecules plus the potential energy of interaction between the molecules. In the limit of ideal gas behavior (high specific volume, low pressure), the molecules are far enough apart that the potential energy of interaction between the molecules becomes negligible. This leaves only the kinetic energy, which is a function only of temperature. Since h = u + pv, and, for an ideal gas, pv = RT, h is also a function only of temperature for ideal gases.
3. Why are the specific heats for ideal gases a function of temperature only? For real gases does this become cv(T,v) and cp(T,p)?
Most physicists regard what they define as an ideal gas as a material having constant heat capacity. We engineers regard the heat capacities of ideal gases to be functions of temperature (which is the way that real gases behave in the limit of high specific volume and low pressures). At pressures beyond the ideal gas range, enthalpy and internal energy of real gases are functions not only of temperature but also pressure. Since cv is defined as the partial derivative of u with respect to T at constant v, cv is also a function of pressure (or specific volume). Since cp is defined as the partial derivative of h with respect to T at constant p, cp is also a function of pressure (or specific volume).

Chet
 
  • #7
Chestermiller said:
Most physicists regard what they define as an ideal gas as a material having constant heat capacity. We engineers regard the heat capacities of ideal gases to be functions of temperature (which is the way that real gases behave in the limit of high specific volume and low pressures).
My answer was based on an ideal gase defined as
Wikipedia said:
An ideal gas is a theoretical gas composed of a set of randomly moving, non-interacting point particles.
 
  • #8
DrClaude said:
My answer was based on an ideal gase defined as
That's what I thought. We engineers define what we call an ideal gas differently (I invite you to check our texts and literature). There have been other threads on PF in which engineers and physicists have compared notes and finally realized that this difference exists in what they respectively call an ideal gas. Apparently, the OP is learning the engineering version.

Chet
 
  • #9
Chestermiller said:
Most physicists regard what they define as an ideal gas as a material having constant heat capacity. We engineers regard the heat capacities of ideal gases to be functions of temperature (which is the way that real gases behave in the limit of high specific volume and low pressures). At pressures beyond the ideal gas range, enthalpy and internal energy of real gases are functions not only of temperature but also pressure. Since cv is defined as the partial derivative of u with respect to T at constant v, cv is also a function of pressure (or specific volume). Since cp is defined as the partial derivative of h with respect to T at constant p, cp is also a function of pressure (or specific volume).

Chet

So in engineering, cp and cv for ideal gases are actually models of real gases at high temperature and low pressure where p and v have a negligible effect? Is it purely for the sake of convenience/situation that usually we see u(T,v) and h(T,p) and of other state variables? Lastly, what are the differences between the definition of an ideal gases in physics and engineering?

Thanks very much
 
  • #10
Red_CCF said:
So in engineering, cp and cv for ideal gases are actually models of real gases at high temperature and low pressure where p and v have a negligible effect?

Not necessarily at high temperature, but, yes, at low pressure.

Is it purely for the sake of convenience/situation that usually we see u(T,v) and h(T,p) and of other state variables?
In my judgement, yes.

Lastly, what are the differences between the definition of an ideal gases in physics and engineering?
As far as I know, the only difference is related to the temperature dependence of the heat capacity that we discussed (and the related functionalities of u and h on T).

Chet
 
  • #11
The nice thing about using it as a function of T and P is that those quantities are a lot easier to measure than entropy or volume.
 

1. What is the relationship between u and h for ideal gases?

The internal energy (u) and enthalpy (h) of an ideal gas are related by the equation h = u + Pv, where P is the pressure and v is the specific volume. This means that the enthalpy of an ideal gas depends on its internal energy and the pressure it is under.

2. How do u and h change for ideal gases at different temperatures?

For ideal gases, the internal energy (u) and enthalpy (h) are directly proportional to the temperature. This means that as the temperature increases, the internal energy and enthalpy also increase.

3. Is the change in u and h the same for all gases?

No, the change in internal energy (u) and enthalpy (h) varies for different gases. The specific heat capacity, which is a measure of how much energy is required to change the temperature of a substance, is different for different gases. This means that the change in u and h will also be different for each gas.

4. How do u and h change when the volume of an ideal gas is changed?

For ideal gases, the internal energy (u) and enthalpy (h) do not change when the volume changes. This is because ideal gases have no intermolecular forces, so the energy is not affected by the volume of the gas.

5. Can the change in u and h be negative for ideal gases?

Yes, the change in internal energy (u) and enthalpy (h) can be negative for ideal gases. This can occur when the gas is expanding and doing work, causing a decrease in its internal energy and enthalpy. However, the overall change in u and h will always be positive if the temperature is increasing.

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