Comparison of Change in Entropy

[nazmulhasanshipon]

Homework Statement

If in both an [*isobaric*](https://en.wikipedia.org/wiki/Isobaric_process) and [isothermal](https://en.wikipedia.org/wiki/Isothermal_process) process, the volume of one gram hydrogen gas (which is evidently by the formula $n=\frac{m}{M}$ is $0.5$ mole) is expanded $4$ times the original volume then calculate the change in entropy in the two thermodynamic process and also compare the two change in entropy mathematically.

Relevant Equations

Second law of Thermodynamics, Entropy, Isobaric Process, Isothermal Process

Upon seeing the question in my assignment I knew in a isobaric process, work done by the gas is $W=P\Delta V$ so if volume is increased $4$ times the original considering the original volume as $V$ we can say after expansion the volume is $4V$. Then $W=P(4V-V)=3PV$ and the $Q$ would be $Q=\Delta U +3PV$ but I don't know how to go from here all the way to comparison. Also upon searching in the web I found [this](https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Physical_Chemistry_(Fleming)/05%3A_The_Second_Law/5.04%3A_Calculating_Entropy_Changes) which is helpful but for the isobaric process I don't know anything of the temperature $T$. I am very much confused here. Any help will be appreciated.

 

[Chestermiller]

If the initial and final temperatures are T1 and T2, and the initial and final volumes are V1 and V2, for n moles of an ideal gas, what is the equation for the change in entropy between the two states?

If the pressure is constant and the volume of an ideal gas increases by 4X, from the ideal gas law, by what factor does the temperature increase?

 

[nazmulhasanshipon]

If the initial and final temperatures are T1 and T2, and the initial and final volumes are V1 and V2, for n moles of an ideal gas, what is the equation for the change in entropy between the two states?

If the pressure is constant and the volume of an ideal gas increases by 4X, from the ideal gas law, by what factor does the temperature increase?

We can use Charles law to find out the final temperature. And "what is the equation for the change in entropy between the two states?" I don't know about that. Please help me understand it.

 

[Chestermiller]

We can use Charles law to find out the final temperature.

Yes.

And "what is the equation for the change in entropy between the two states?" I don't know about that. Please help me understand it.

Suppose you increase the temperature of n moles of an ideal gas reversibly from T1 to T2 at constant volume. Do you know how to determine the entropy change for that constant volume process?

 

[nazmulhasanshipon]

Yes.

Suppose you increase the temperature of n moles of an ideal gas reversibly from T1 to T2 at constant volume. Do you know how to determine the entropy change for that constant volume process?

I know the change in entropy is $\cfrac{\Delta Q}{T}$. And I don't need it for the constant volume. I need it for the constant pressure.

 

[Chestermiller]

I know the change in entropy is $\cfrac{\Delta Q}{T}$. And I don't need it for the constant volume. I need it for the constant pressure.

The general equation for the entropy change of a closed system is $$\Delta S=\int{\frac{dQ_{rev}}{T}}$$

I'm trying to help you get it for the constant pressure case. You need to trust me. What is your equation for the constant volume case?

 

[nazmulhasanshipon]

The general equation for the entropy change of a closed system is $$\Delta S=\int{\frac{dQ_{rev}}{T}}$$

I'm trying to help you get it for the constant pressure case. You need to trust me. What is your equation for the constant volume case?

Actually I don't know about the equation of the entropy change case in constant volume. Would you tell me?

 

[nazmulhasanshipon]

The general equation for the entropy change of a closed system is $$\Delta S=\int{\frac{dQ_{rev}}{T}}$$

I'm trying to help you get it for the constant pressure case. You need to trust me. What is your equation for the constant volume case?

and also sir, can you please tell me what is 2nd law of thermodynamics using the definition of entropy?

 

[Chestermiller]

Actually I don't know about the equation of the entropy change case in constant volume. Would you tell me?

To get the change in entropy of a closed system, you need to devise (i.e., dream up) a reversible path between the initial and final states of the system. For a system experiencing a temperature change at constant volume, the reversible change can be brought about by contacting the system with a sequence of constant temperature reservoirs, each at a slightly different temperature running from T1 to T2. For such a constant volume process, the entropy change is described by $$dS=\frac{dQ_{rev}}{T}=\frac{nC_vdT}{T}$$ or $$\Delta S=nC_V\int_{T_1}^{T_2}{\frac{dT}{T}}$$

Does this make sense so far?

 

[nazmulhasanshipon]

To get the change in entropy of a closed system, you need to devise (i.e., dream up) a reversible path between the initial and final states of the system. For a system experiencing a temperature change at constant volume, the reversible change can be brought about by contacting the system with a sequence of constant temperature reservoirs, each at a slightly different temperature running from T1 to T2. For such a constant volume process, the entropy change is described by $$dS=\frac{dQ_{rev}}{T}=\frac{nC_vdT}{T}$$ or $$\Delta S=nC_V\int_{T_1}^{T_2}{\frac{dT}{T}}$$

Does this make sense so far?

So, sir $dT$ means little change in between temperature $T_1$ and $T_2$? I thought that was a differential. (and sorry I didn't call you sir earlier, I didn't know you were a teacher)

 

[Chestermiller]

So, sir $dT$ means little change in between temperature $T_1$ and $T_2$? I thought that was a differential. (and sorry I didn't call you sir earlier, I didn't know you were a teacher)

Yes, that is correct. The system temperature in this reversible path changes gradually from T1 to T2, as the system is contacted with ever increasing temperatures of ideal reservoirs running from T1 to T2.

 

[nazmulhasanshipon]

Yes, that is correct. The system temperature in this reversible path changes gradually from T1 to T2, as the system is contacted with ever increasing temperatures of ideal reservoirs running from T1 to T2.

Yes, sir. It's a wonderful explanation. So far, it makes sense. But the symbol $dT$ makes me to think it is a differential.

 

[Chestermiller]

Yes, sir. So far, it makes sense. But the symbol $dT$ makes me to think it is a differential.

Yes, that is correct. Is that a problem? Do you know how to integrate dT/T from T1 to T2?

 

[nazmulhasanshipon]

Yes, that is correct. Is that a problem? Do you know how to integrate dT/T from T1 to T2?

Yes, $\ln \frac{T_2}{T_1}$ is the integral. But the problem is that in Calculus teacher told that $dx, dy$ has no meaning and $\frac{dy}{dx}$ is not a fraction so we can't separate them.

 

[Chestermiller]

Yes, $\ln \frac{T_2}{T_1}$ is the integral. But the problem is that in Calculus teacher told that $dx, dy$ has no meaning and $\frac{dy}{dx}$ is not a fraction so we can't separate them.

That integration is correct. I have no ideal what your calculus teacher is talking about.

 

[nazmulhasanshipon]

That integration is correct. I have no ideal what your calculus teacher is talking about.

Okay, sir. Then the next step?

 

[Chestermiller]

I have to run an errand in a little while.

The constant pressure process goes from T1 and V1 to T2 and V2. We are going to split the reversible change into two steps. First, reversibly increase the temperature at constant volume V1 from T1 to T2 (we've already done that). Next, determine the entropy change in going from V1 to V2 at the final temperature from step 1, T2. The entropy change for the overall constant pressure reversible process will be equal to the sum of the entropy changes for steps 1 and 2.

 

[nazmulhasanshipon]

I have to run an errand in a little while.

The constant pressure process goes from T1 and V1 to T2 and V2. We are going to split the reversible change into two steps. First, reversibly increase the temperature at constant volume V1 from T1 to T2 (we've already done that). Next, determine the entropy change in going from V1 to V2 at the final temperature from step 1, T2. The entropy change for the overall constant pressure reversible process will be equal to the sum of the entropy changes for steps 1 and 2.

I will be doing it the ways you described. I'm sure I'll face some problem. Can I know when would you be free?

 

[nazmulhasanshipon]

I have to run an errand in a little while.

The constant pressure process goes from T1 and V1 to T2 and V2. We are going to split the reversible change into two steps. First, reversibly increase the temperature at constant volume V1 from T1 to T2 (we've already done that). Next, determine the entropy change in going from V1 to V2 at the final temperature from step 1, T2. The entropy change for the overall constant pressure reversible process will be equal to the sum of the entropy changes for steps 1 and 2.

Sir, if I haven't made a mistake then the change in entropy for the isobaric process is- $$n C_v \int_{T_1}^{T_2} \frac{dT}{T} + nR \ln \left (\frac{V_2}{V_1} \right )$$ and the change in entropy in the isothermal process is $$nR \ln \left (\frac{V_2}{V_1} \right )$$
Am I correct, sir?