Entropy calculations involving integration (physical chemistry)

[anisotropic]

This isn't a homework question per se, but more of a coursework question. Specifically, I'm a bit at a loss as to how to go about learning a particular section of the coursework for a physical chemistry (i.e. thermochemistry) class I am taking. The section in question is that pertaining to calculations involving entropy (S).

For example, take the following question:

The heat capacity of a given ideal gas can be expressed as shown below.

C_P = (10 + 0.006T) J K^-1 mol^-1

Calculate the change in entropy when 4 moles of this gas are isobarically heated from 200 K to 300 K.

Step 1 of the solution (as given by the solutions manual):

ΔS = ∫d_qrev/T

Where are they getting this equation from? But more importantly, what is it actually saying, conceptually? I haven't done calculus in years, so I am more than rusty when it comes to integration, meaning I don't even understand what "d_qrev" means. Thus, I can't just look at the given expression and figure it out for myself. The actual mechanics of working with integrals, I can figure out on my own; the conceptual part, and where equations are being derived from, not so much.

If someone could help me out, it would be appreciated. Note that I am not looking for a solution to the problem given, as I already have that. Rather, I am requesting an explanation as to why the steps that are involved actually work (an explanation of the integration shown above would be a great start).

For the record, the solution ends up converting d_qrev to C_pdT, and integrates from there (200 K to 300 K). But again, I don't know what d_qrev signifies to begin with, so I can't really make sense of any further steps. I do know, however, that q signifies heat transfer in other questions, and "rev" signifies that it is a reversible process (i.e. theoretical), while T is obviously temperature.

 

[Andrew Mason]

This isn't a homework question per se, but more of a coursework question. Specifically, I'm a bit at a loss as to how to go about learning a particular section of the coursework for a physical chemistry (i.e. thermochemistry) class I am taking. The section in question is that pertaining to calculations involving entropy (S).

For example, take the following question:

The heat capacity of a given ideal gas can be expressed as shown below.

C_P = (10 + 0.006T) J K^-1 mol^-1

Calculate the change in entropy when 4 moles of this gas are isobarically heated from 200 K to 300 K.

Step 1 of the solution (as given by the solutions manual):

ΔS = ∫d_qrev/T

Where are they getting this equation from? But more importantly, what is it actually saying, conceptually?

First you find a reversible process between the beginning and end states. That would be a quasi-static expansion from V = nR(200)/P to V = nR(300)/P so it is an expansion by 1.5 times.

Since T is changing over this path, you have to express Q as a function of T and integrate. The expression for Q as a function of T is, by definition, Q = nCpΔT. So dQ = nCpdT

So the change in entropy is :

\Delta S = \int_{T_0}^{T_f} dQ/T = \int_{T_0}^{T_f} nC_pdT/T

That integral works out to:

ΔS = nCp ln(Tf/T0)

I haven't done calculus in years, so I am more than rusty when it comes to integration, meaning I don't even understand what "d_qrev" means. Thus, I can't just look at the given expression and figure it out for myself. The actual mechanics of working with integrals, I can figure out on my own; the conceptual part, and where equations are being derived from, not so much.

Entropy is defined in this way for reasons having to do with thermodynamics. You will need to take a course or read a good text in Heat and Thermodynamics to understand the concepts involved. Good luck!

Also, you may wish to post on the Classical Physics board on this subject.

AM

 

[anisotropic]

Since T is changing over this path, you have to express Q as a function of T and integrate.

As elementary as this question is, can you please explain this part further? What are you actually doing when you integrate?

I assume if you graphed the function, on the y-axis is q, and on the x-axis is T. Graphically speaking, integration is finding the area under the function/curve from point T₁ (300 K) to point T₂ (400 K), correct? If so, what does the actual area under this curve represent? Is it the entropy, or the change in entropy?

The expression for Q as a function of T is, by definition, Q = nCpΔT. So dQ = nCpdT

Could you substitute "by definition" with "due to having the equation memorized"?

I am indeed taking a thermochemistry course, hence these questions. The problem is that the course is extremely condensed, and the notes are largely mathematical, with little time spent conceptually explaining what is going on, if at all. All the explaining is done mathematically, and given that my calculus is, as mentioned, in serious need of re-learning, I find myself understanding literally nothing during class.

The textbook is helpful, but only to a certain extent, as it also heavily emphasizes the use of formulas to explain concepts. These reasons all combine to lead me to post on this forum.

 

[Andrew Mason]

As elementary as this question is, can you please explain this part further? What are you actually doing when you integrate?

You are taking the area of a surface. In this case, you are adding up a whole lot of dQ/T s.

I assume if you graphed the function, on the y-axis is q, and on the x-axis is T. Graphically speaking, integration is finding the area under the function/curve from point T₁ (300 K) to point T₂ (400 K), correct? If so, what does the actual area under this curve represent? Is it the entropy, or the change in entropy?

Change in entropy resulting from heat flow Q at temperature T is defined as Q/T.

So essentially you want to find the area under a graph that plots 1/T on the vertical axis and Q on the horizontal axis. If T was constant, this would be easy - it is just the area of a rectangle 1/T high and Q wide so it is just Q/T. But here 1/T is a curve. So you have to integrate to find the area.

Could you substitute "by definition" with "due to having the equation memorized"?

Well, there is a reason for defining entropy that way. It is useful in analysing thermodynamic processes.

AM