Change in entropy of the surroundings in an irreversible process

[palaphys]

Homework Statement

A gas is expanding isothermally in an irreversible process. Assume that the gas is ideal. Compute the following
1) change in entropy of the system, if the gas expands from volume V_i to V_f.
2) change in entropy of the surroundings, if the gas expands under a constant external pressure P_ext.

Relevant Equations

Ds= dq/T for reversible procvesses

For the first part, I got the answer pretty easily, it is a well known fact that the change in the entropy of the system does not matter about the process taken, so its the same for both reversible and irreversible processes, i.e nR ln(Vf/Vi). To derive this, I have assumed that the temperature of the system and surroundings is the same. Not sure if this assumption is right,

For part two, I am not sure how to proceed. I have learnt that the change in entropy for a REVERSIBLE process is Q/ T. However as entropy is a state function, we can assume a reversible process.

As a result, $$ dS= dQ_{rev,surroundings} /T $$ where T is the temperature of the surroundings, assumed to be equal to the temperature of the system (not sure again). How to calculate the value of dQ here??

 

[Chestermiller]

Homework Statement: A gas is expanding isothermally in an irreversible process. Assume that the gas is ideal. Compute the following
1) change in entropy of the system, if the gas expands from volume V_i to V_f.
2) change in entropy of the surroundings, if the gas expands under a constant external pressure P_ext.
Relevant Equations: Ds= dq/T for reversible procvesses

For the first part, I got the answer pretty easily, it is a well known fact that the change in the entropy of the system does not matter about the process taken, so its the same for both reversible and irreversible processes, i.e nR ln(Vf/Vi). To derive this, I have assumed that the temperature of the system and surroundings is the same. Not sure if this assumption is right,

During the irreversible expansion, even though the temperature of the system matches the temperature of the surroundings at the interface with the surroundings, within the system there will be temperature non-uniformities, with the highest temperature regions near the boundary with the surroundings and the lowest temperatures within the interior. Only at the initial and final thermodynamic equilibrium states is the gas temperature uniform and equal to the temperature of the surroundings throughout

For part two, I am not sure how to proceed. I have learnt that the change in entropy for a REVERSIBLE process is Q/ T. However as entropy is a state function, we can assume a reversible process.

As a result, $$ dS= dQ_{rev,surroundings} /T $$ where T is the temperature of the surroundings, assumed to be equal to the temperature of the system (not sure again). How to calculate the value of dQ here??

See my Physics Forums Insights article on calculating the changes in entropy of the system and surroundings:
www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

In part 1, the irreversible change described in part 1 is specified in more detail, by indicating that the change takes place at constant external pressure.

The surroundings are always treated as an ideal isothermal reservoir (or group of reservoirs), such that, in this problem, $$\Delta S_{surr}=\frac{Q_{surr}}{T}$$where $Q_{surr}$ is equal to the heat received by the surroundings in the actual irreversible process. Note that, in determining the entropy change for system and surroundings for an irreversible process, they must first be separated from one-another and then each subjected to a new reversible process (for each) between the same initial and final states as in the irreversible process; and the two new reversible processes for each do not have to bear any resemblance whatsoever to the actual irreversible process.

In the irreversible process of part 2, what is the change in internal energy of the system? In part 2, the constant external pressure $P_{ext}$ must be the same as the final pressure $P_f$ in part 1 (and, for course, the initial and final volumes must be the same as in part 1). What is the amount of work done by the constant external pressure on the surroundings? From the first law, what is the amount of heat Q transferred from the surroundings to the system? What is the change in entropy of the system in part 2? What is the change in entropy of the surroundings in part 2?

 

[palaphys]

In the irreversible process of part 2, what is the change in internal energy of the system? In part 2, the constant external pressure $P_{ext}$ must be the same as the final pressure $P_f$ in part 1 (and, for course, the initial and final volumes must be the same as in part 1). What is the amount of work done by the constant external pressure on the surroundings? From the first law, what is the amount of heat Q transferred from the surroundings to the system? What is the change in entropy of the system in part 2? What is the change in entropy of the surroundings in part 2?

Okay sir, just went throught your article. I will try to execute the steps that have been suggested. So firstly I will try to apply the first law of thermodynamics for the surroundings, assuming a reversible path bearing no similarity with the irreversible one, but the initial and finals states are the same.

$$ \Delta U = q_{surroundings} - P_{ext} (\Delta V) $$
Where dV= 0.
Hence
$$\Delta U = q_{surroundings}$$
Change in entropy of system, I need q_surroundings. How do I find that out?

 

[Chestermiller]

Okay sir, just went throught your article. I will try to execute the steps that have been suggested. So firstly I will try to apply the first law of thermodynamics for the surroundings, assuming a reversible path bearing no similarity with the irreversible one, but the initial and finals states are the same.

$$ \Delta U = q_{surroundings} - P_{ext} (\Delta V) $$
Where dV= 0.
Hence
$$\Delta U = q_{surroundings}$$
Change in entropy of system, I need q_surroundings. How do I find that out?

This is not quite correct. For no change in temperature between initial and final states of the system, $$\Delta U_{system}=0$$Therefore, for the path in part 2, from the first law of thermodynamics, we have $$Q_{system}=P_{ext}(V_f-V_i)$$and, $$Q_{surr}=-Q_{system}=-P_{ext}(V_f-V_i)$$So according to this, what is the change in entropy for the surroundings for path 2?

 

[palaphys]

This is not quite correct. For no change in temperature between initial and final states of the system, $$\Delta U_{system}=0$$Therefore, for the path in part 2, from the first law of thermodynamics, we have $$Q_{system}=P_{ext}(V_f-V_i)$$and, $$Q_{surr}=-Q_{system}=-P_{ext}(V_f-V_i)$$So according to this, what is the change in entropy for the surroundings for path 2?

why is it not right?
I used delta U for the surroundings, considering that as a separate system. also, how do we know that
$$ Q_{surr}= -Q_{system}$$ for this reversible path?
but according to this, we get
$$\Delta S_{surr} = Q_{rev}/T= -P_{ext}(V_f- V_i)/T$$

 

[Chestermiller]

why is it not right?

I may have confused you. You are supposed to apply the 1st law to the system first, not the surroundings..

I used delta U for the surroundings, considering that as a separate system. also, how do we know that
$$ Q_{surr}= -Q_{system}$$ for this reversible path?

$Q_{system}$ is the amount of heat transferred from the surroundings to the system. This has to be minus the heat transferred from the system to the surroundings, since the combination of system and surroundings is isolated.

but according to this, we get
$$\Delta S_{surr} = Q_{rev}/T= -P_{ext}(V_f- V_i)/T$$

Since $P_{ext}=P_f$ and $P_iV_i=P_fV_f$, express the entropy change of the surroundings n terms of n, R, Vi, and Vf.

 

[palaphys]

Since $P_{ext}=P_f$ and $P_iV_i=P_fV_f$, express the entropy change of the surroundings n terms of n, R, Vi, and Vf.

Is $P_{ext}= P_f$ the pressure of the system? Then I guess I should just substitute $$P=nRT/V_f.$$
On a side note, is it even possible to apply the first law on the surroundings? By the term surroundings I mean the neighbourhood.

 

[Chestermiller]

Is $P_{ext}= P_f$ the pressure of the system?

$P_{ext}$ is the final pressure of the system at final equilibrium of the irreversible process. At time zero, we drop the external pressure from $P_i$ to $P_f$, and hold it at that value during the irreversible expansion.

Then I guess I should just substitute $$P=nRT/V_f.$$
On a side note, is it even possible to apply the first law on the surroundings?

Yes. $$\Delta U_surr=Q_{surr}$$But the surroundings temperature doesn't change because the surroundings are assumed to have either infinite mass, infinite heat capacity, or both.

$$\Delta S_{surr}=-\frac{P_{ext}}{T}(V_f-V_i)=-\frac{P_f}{T}\left(\frac{nRT}{P_f}-\frac{nRT}{P_i}\right)$$$$=-nR\left(1-\frac{P_f}{P_i}\right)=nR\left(\frac{V_i}{V_f}-1\right)$$
So the entropy change of the surroundings is negative, but not as negative as the positive entropy increase of the system.

 

[palaphys]

At time zero, we drop the external pressure from $P_i$ to $P_f$, and hold it at that value during the irreversible expansion.

I have a few questions. Firstly, what does this statement even mean? I do not understand.
Secondly, I will try to recap the process of solving this question, to see what I have understood.
So we forget about the fact that the process is irreversible, and assume a reversible path from initial state to final state. Hence, between these two final states, using the first law on the system,

$$ \Delta U = q_{system} + W_{system} $$
As work done is $$-P_{ext}(V_f - V_i) $$
And also the internal energy of an ideal gas is a function of temperature assuming no intermolecular forces,

$$ \Delta U = 0, q_{system} = P_{ext} (V_f - V_i) $$

Also, in the final state of the process, heat is absorbed from the surroundings,
Hence $$Q_{surroundings}= - Q_{system} = -P_{ext} (V_f- V_i) $$

To finish, as we have assumed a reversible path,
$$ \Delta S_{surr} = \frac{Q_{surr}}{T} = -\frac{P_{ext}(V_f - V_i)}{T}$$

Here is my final question- how do we know that $P_{ext}= P_f$?
Appreciate the help so far.

 

[Chestermiller]

I have a few questions. Firstly, what does this statement even mean? I do not understand.
Secondly, I will try to recap the process of solving this question, to see what I have understood.
So we forget about the fact that the process is irreversible, and assume a reversible path from initial state to final state. Hence, between these two final states, using the first law on the system,

$$ \Delta U = q_{system} + W_{system} $$
As work done is $$-P_{ext}(V_f - V_i) $$
And also the internal energy of an ideal gas is a function of temperature assuming no intermolecular forces,

$$ \Delta U = 0, q_{system} = P_{ext} (V_f - V_i) $$

Also, in the final state of the process, heat is absorbed from the surroundings,
Hence $$Q_{surroundings}= - Q_{system} = -P_{ext} (V_f- V_i) $$

To finish, as we have assumed a reversible path,
$$ \Delta S_{surr} = \frac{Q_{surr}}{T} = -\frac{P_{ext}(V_f - V_i)}{T}$$

Here is my final question- how do we know that $P_{ext}= P_f$?
Appreciate the help so far.

Because, in the final state, the boundary between the system and surroundings is not moving, so the pressures on both sides of the boundary must be equal .

 

[palaphys]

Because, in the final state, the boundary between the system and surroundings is not moving, so the pressures on both sides of the boundary must be equal .

alright, but isn't that only in the FINAL state? Substituting P_f in that expression makes it seem as if the pressure was P_f throughout the process.

 

[Chestermiller]

alright, but isn't that only in the FINAL state? Substituting P_f in that expression makes it seem as if the pressure was P_f throughout the process.

Do you think that in an irreversible process the gas pressure is described by the ideal gas law? It isn’t. The ideal gas law describes the pressure in an ideal gas only if the gas is at thermodynamic equilibrium. Otherwise, we don’t know what the pressure of the gas is. In the case where we prescribe the pressure externally, we are manually, specifying the pressure of the gas at its interface with the surroundings over the entire process.

 

[palaphys]

In the case where we prescribe the pressure externally, we are manually, specifying the pressure of the gas at its interface with the surroundings over the entire process.

Then it should remain as P_ext right? Why not P_i ? Why specifically P_f?
Don't seem to get the reason behind it. Please explain it in a simpler way

 

[Chestermiller]

Then it should remain as P_ext right? Why not P_i ? Why specifically P_f?
Don't seem to get the reason behind it. Please explain it in a simpler way

Because you used Pf in part 1 ( actually Vf). That way, Parts 1&2 could be considered to represent the same irreversible process.

 

[palaphys]

Because you used Pf in part 1 ( actually Vf). That way, Parts 1&2 could be considered to represent the same irreversible process.

Oh, I get it now. Thanks