Determining entropy as function of pressure

In summary, the entropy of an ideal gas at a certain pressure can be obtained by integrating the quantity ##dS = \frac{dq}{T}## over a reversible path between a reference state and the final state. This is usually done by dividing the process into small infinitesimal steps and summing the ##dq/T##:s. The absolute entropy at an arbitrary pressure can be calculated by assuming the 3rd Law of Thermodynamics and using heat capacity data at low temperatures.
  • #1
Addiw777
2
0
I read this discussion but I am interested in how the entropy is obtained as a function of pressure. Namely, how can you determine a following integral for an ideal gas:

$$S(p) = -\int_{0}^{p} \frac{nR}{p}dp $$

when you need to start from 0 pressure?
 
Chemistry news on Phys.org
  • #2
That integral does not converge if you make the lower integration limit approach 0. To calculate entropy as a function of pressure, you need to know the entropy at some reference state and then imagine a reversible path between that reference state and the wanted final state. Finally you integrate the quantity ##dS = \frac{dq}{T}## over that path.

For example, if you have a reversible isothermal compression of an ideal gas from initial pressure ##p_1## to final pressure ##p_2##, you can divide the process to small infinitesimal steps where the volume decreases by ##dV##, and an infinitesimal heat flow ##dq## out of the system keeps the temperature constant despite the mechanical work done on the system by compressing it. Then you just sum the ##dq/T##:s for the whole process.
 
  • #3
Thank you for that nice example, however I believe that my question still persists but perhaps I put it unclearly in my first post.
So, what I basically do not understand is how I can get the entropy in arbitrary pressure and temperature? When I take my physical
chemistry book I can see quantities like ##S^{o}## at 298 K. Very nice, but how do they know that this is entropy at this specific pressure
of 101325 Pa? Obviously the change in entropy for an ideal gas due to pressure change () in isothermic process and when no reaction or mixing occurs it is ## \Delta S = -nRln(\frac{p_{2}}{p_{1}})##. But how "the absolute" entropy at arbitrary pressure is obtained?
 
  • #4
Absolute entropy is calculated by assuming the 3rd Law of Thermodynamics, which says that the most stable crystalline form of any compound at absolute zero of temperature has exactly zero entropy. Then you need to have heat capacity data of the substance down to low enough temperatures to be able to calculate the absolute entropy at some final state that has a nonzero temperature.
 

1. What is entropy?

Entropy is a measure of the disorder or randomness in a system. It is a thermodynamic property that quantifies the number of possible energy arrangements in a given system.

2. How is entropy related to pressure?

Entropy is directly proportional to pressure. As pressure increases, the number of possible arrangements of energy in a system also increases, resulting in an increase in entropy.

3. How do you determine entropy as a function of pressure?

The most common method for determining entropy as a function of pressure is by using the Clausius-Clapeyron equation, which relates changes in entropy to changes in pressure and temperature.

4. What factors can affect the relationship between entropy and pressure?

The relationship between entropy and pressure can be affected by changes in temperature, volume, and the type of substance being studied. Phase transitions, such as melting or vaporization, can also impact this relationship.

5. Why is it important to determine entropy as a function of pressure?

Determining entropy as a function of pressure is important in understanding the behavior of a system at different pressures and temperatures. This information can be used in various fields, such as thermodynamics, chemistry, and material science, to predict and control the behavior of substances under different conditions.

Similar threads

Replies
14
Views
2K
  • Thermodynamics
Replies
4
Views
383
Replies
56
Views
3K
Replies
9
Views
3K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Thermodynamics
Replies
3
Views
788
  • Introductory Physics Homework Help
Replies
10
Views
1K
Replies
22
Views
2K
  • Materials and Chemical Engineering
Replies
2
Views
282
  • Classical Physics
Replies
6
Views
794
Back
Top