Temperature dependence of Cv at very large volume

In summary, the conversation discusses the equation of state for a gas and shows how it can be used to derive a formula for heat capacity at large volumes. It also raises the question of whether the heat capacity should be considered temperature-dependent at large volumes, which is a topic of debate among engineers and physicists.
  • #1
arpon
235
16

Homework Statement


In the case of a gas obeying the equation of state
$$\begin{align}\frac{Pv}{RT}&=1+\frac{B}{v}\end{align} $$
where ##B## is a function of ##T## only, show that,
$$\begin{align}c_v&=-\frac{RT}{v}\frac{d^2}{dT^2} (BT)+\left(c_v\right)_0\end{align}$$
where ##\left(c_v\right)_0## is the value at large volume.

Homework Equations


$$\begin{align}c_v&=\left(\frac{dU}{dT}\right)_v\end{align}$$
$$\begin{align}TdS &=c_v dT+ T \left(\frac{\partial P}{\partial T}\right)_v dv \end{align}$$

The Attempt at a Solution


$$\begin{align}\left(\frac{\partial c_v}{\partial v}\right)_T & =\left(\frac{\partial}{\partial v}\left(\frac{\partial U}{\partial T}\right)_v\right)_T \\
& =\left(\frac{\partial}{\partial T}\left(\frac{\partial U}{\partial v}\right)_T\right)_v
\end{align}$$
Now,
$$\begin{align}dU
&=TdS-Pdv\\
& = c_v dT+ T \left(\frac{\partial P}{\partial T}\right)_v dv - Pdv
\end{align}$$
So, we have,
$$\begin{align}
\left(\frac{\partial U}{\partial v}\right)_T & = T \left(\frac{\partial P}{\partial T}\right)_v - P
\end{align}$$

Using equation (6) & (9),
$$\begin{align}\left(\frac{\partial c_v}{\partial v}\right)_T & = \left(\frac{\partial}{\partial T}\left(T \left(\frac{\partial P}{\partial T}\right)_v - P\right)\right)_v\\
& = T \left(\frac{\partial^2 P}{\partial T^2}\right)_v + \left(\frac{\partial P}{\partial T}\right)_v - \left(\frac{\partial P}{\partial T}\right)_v \\
&= T \left(\frac{\partial^2 P}{\partial T^2}\right)_v
\end{align}$$

Using the equation of state (1), we obtain,
$$\begin{align}
\left(\frac{\partial^2 P}{\partial T^2}\right)_v &= \frac{R}{v^2}\frac{d^2}{dT^2} (BT)
\end{align}$$

So, using (12) and (13), we have,
$$\begin{align}\left(\frac{\partial c_v}{\partial v}\right)_T & = \frac{RT}{v^2}\frac{d^2}{dT^2} (BT)
\end{align}$$
Now
$$\begin{align} dc_v &= \left(\frac{\partial c_v}{\partial v}\right)_T dv + \left(\frac{\partial c_v}{\partial T}\right)_v dT
\end{align}$$

Integrating from ##T=T,~V=\infty## to ##T=T,~V=V##,
$$\begin{align}c_v&=-\frac{RT}{v}\frac{d^2}{dT^2} (BT)+\left(c_v\right)_0\end{align}$$

Here, ##\left(c_v\right)_0## is the value for ##v=\infty## and ##T=T##.
My question is: Is ##\left(c_v\right)_0## independent of ##T##, and why?
 
Physics news on Phys.org
  • #2
At the limit for infinite volume, the B/v term disappears, therefore the gas law gets independent of T (apart from the usual factor of T).
 
  • Like
Likes arpon
  • #3
For a real gas (which this is supposed to represent), in the limit of very large v, ##c_v## is observed to be a function of T.
 
  • #4
Then you need a different equation of state, not the one from post 1.

Actual gases often have additional complications from frozen-in degrees of freedom, dissociation and so on, but those are all not considered here.
 
  • Like
Likes arpon
  • #5
mfb said:
Then you need a different equation of state, not the one from post 1.
Why not?
 
  • #6
The one from post 1 does not lead to a temperature-dependent c_v in the limit of infinite volumes.
 
  • #7
mfb said:
The one from post 1 does not lead to a temperature-dependent c_v in the limit of infinite volumes.
I don't see anything in the mathematics that precludes Cv from being a function of temperature at large volume.
 
Last edited:
  • #8
Hmm... v->infinity is equivalent to B->0. Setting B=0 leads to c_v = c_v, so we cannot make any statement about c_v just with the given information.

Why exactly would a temperature-dependent c_v be problematic?
 
  • #9
mfb said:
Hmm... v->infinity is equivalent to B->0. Setting B=0 leads to c_v = c_v, so we cannot make any statement about c_v just with the given information.
We could say that at large v, c_v is temperature-dependent. That's certainly consistent with the mathematics. We engineers (I'm an engineer), unlike physicists, regard a temperature-dependent heat capacity as a key aspect of the definition of an ideal gas. In all engineering thermodynamics books, the heat capacities of ideal gases are regarded as being temperature dependent. They are the same functions of temperature that the heat capacities of real gases approach in the limit of low pressures and high volumes. So, in this problem, ##c_{v0}(T)## would be the (temperature-dependent) heat capacity of the real gas in the ideal gas region.
Why exactly would a temperature-dependent c_v be problematic?
In my judgment, it would not be problematic at all. Rather, it would be expected.
 
  • #10
Chestermiller said:
In all engineering thermodynamics books, the heat capacities of ideal gases are regarded as being temperature dependent.
Wait, what? For an ideal monoatomic gas it should be 3/2 R.
 
  • #11
mfb said:
Wait, what? For an ideal monoatomic gas it should be 3/2 R.
As I said, engineers regard an ideal gas an entity which, in general, features a temperature-dependent viscosity. In the special case of an ideal monoatomic gas, the heat capacity would be virtually constant at 3/2R. For the present problem, however, no information was provided concerning the molecular nature of the gas. Therefore, it is prudent to assume that, at large specific volumes, the heat capacity can be temperature-dependent.
 

Related to Temperature dependence of Cv at very large volume

1. How does the specific heat capacity (Cv) change as volume increases?

As volume increases, the specific heat capacity (Cv) tends to decrease. This is because with a larger volume, there is more room for molecules to move around and distribute their energy, resulting in a decrease in the amount of energy required to raise the temperature of the substance by 1 degree Celsius.

2. What is the relationship between temperature and Cv at very large volumes?

At very large volumes, the relationship between temperature and Cv becomes less significant. This is because at high volumes, the molecules have more space to move and the intermolecular forces become weaker, leading to a decrease in the effect of temperature on the Cv of a substance.

3. How does the temperature dependence of Cv at very large volumes differ from that at smaller volumes?

The temperature dependence of Cv at very large volumes is less significant compared to that at smaller volumes. This is because at smaller volumes, the molecules are closer together and the intermolecular forces are stronger, resulting in a greater effect of temperature on the Cv of a substance.

4. Can the temperature dependence of Cv at very large volumes be accurately predicted?

The temperature dependence of Cv at very large volumes can be predicted using theoretical models, but it may not always be accurate due to the complexity of intermolecular interactions and the presence of external factors such as pressure and impurities.

5. How does the temperature dependence of Cv at very large volumes affect the behavior of a substance?

The temperature dependence of Cv at very large volumes can affect the behavior of a substance in various ways. It can determine the amount of energy needed to raise the temperature of a substance, the rate of temperature change, and the overall thermal stability of the substance. It can also affect the phase transitions of a substance at different temperatures and volumes.

Similar threads

  • Advanced Physics Homework Help
Replies
4
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
981
  • Advanced Physics Homework Help
Replies
3
Views
600
  • Advanced Physics Homework Help
Replies
4
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
10
Views
801
  • Advanced Physics Homework Help
Replies
4
Views
2K
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
964
Back
Top