Could someone help me study this cyclic process of an ideal gas?

[Thermofox]

Homework Statement

$0.5$ moles of an ideal gas, at $P_1= 1atm$ and $T_1= 20 °C$, go through the following cycle:
From 1 to 2: irreversible adiabatic;
From 2 to 3: reversible isovolumetric;
From 3 to 1: reversible isothermic.
In this cycle the system absorbs $Q_h=890J$ and $V_2 = 3V_1$

Relevant Equations

$\Delta U = W + Q$

I need to determine:
1)$\Delta S$ of the system
2)Total work done by the system
3)$\Delta S$ of the surroundings

For point 1 $\Delta S_{system}=0$ because it is a cyclic process.

For point 2 $W_{net}= W_{1-2} + W_{2-3} + W_{3-1}$.
Since 2-3 is an isovolumetric process, $W_{2-3}=0$
=> $W_{1-2}= \Delta U= nc_v\Delta T_{1-2}$ and $W_{3-1}= -ln(\frac {V_1} {V_3})$.

I'm not sure about point 3 though. Because surely $\Delta S_{1-2}^{surr.}= 0$ since in the process there is no heat exchange between system and surroundings. But I don't know how to determine the change for the other 2 transformations. I assume I should be using $Q_h$, that in this case must be all absorbed during 2-3, but still I don't know how to proceed forward.

 

[Philip Koeck]

Homework Statement: $0.5$ moles of an ideal gas, at $P_1= 1atm$ and $T_1= 20 °C$, go through the following cycle:
From 1 to 2: irreversible adiabatic;
From 2 to 3: reversible isovolumetric;
From 3 to 1: reversible isothermic.
In this cycle the system absorbs $Q_h=890J$ and $V_2 = 3V_1$
Relevant Equations: $\Delta U = W + Q$

I need to determine:
1)$\Delta S$ of the system
2)Total work done by the system
3)$\Delta S$ of the surroundings

For point 1 $\Delta S_{system}=0$ because it is a cyclic process.

For point 2 $W_{net}= W_{1-2} + W_{2-3} + W_{3-1}$.
Since 2-3 is an isovolumetric process, $W_{2-3}=0$
=> $W_{1-2}= \Delta U= nc_v\Delta T_{1-2}$ and $W_{3-1}= -ln(\frac {V_1} {V_3})$.

I'm not sure about point 3 though. Because surely $\Delta S_{1-2}^{surr.}= 0$ since in the process there is no heat exchange between system and surroundings. But I don't know how to determine the change for the other 2 transformations. I assume I should be using $Q_h$, that in this case must be all absorbed during 2-3, but still I don't know how to proceed forward.

The work from 3 to 1 has to be negative. Check what you write.

If Q = 0 for an irreversible process you can't conclude that ΔS_sys = 0 for that process.
That would only be true for reversible processes.

To me it's not clear what is meant by irreversible, adiabatic. Does it just mean Q = 0, or does the process follow the adiabatic curve in the pV-diagram? If the latter is true you can calculate W immediately (with one extra assumption).

What is Q_h? Is it the total heat transfer during one cycle or it just the heat going from the surroundings to the system?

 

[Thermofox]

The work from 3 to 1 has to be negative. Check what you write.

If I'm using $\Delta U= W + Q$ then mustn't a compression mean positive work? The work is
$W_{3-1}= -ln(\frac {V_1} {V_3})nRT= -ln(\frac {1} {3})nRT$, because $V_3=V_2=3V_1$

If Q = 0 for an irreversible process you can't conclude that ΔSsys = 0 for that process.
That would only be true for reversible processes.

I said that $\Delta S{sys}=0$ because entropy is a state variable, therefore if I start at 1 and end at 1 there must be no change in entropy. From 1-2 I have a change in entropy in the system, because, since the process is irreversible, entropy has to increase. Thus $\Delta S{sys}^{1-2} \gt 0$.

To me it's not clear what is meant by irreversible, adiabatic. Does it just mean Q = 0, or does the process follow the adiabatic curve in the pV-diagram? If the latter is true you can calculate W immediately (with one extra assumption).

It means that the process occurs rapidly, but with no heat exchange. So $Q=0$.

What is Qh? Is it the total heat transfer during one cycle or it just the heat going from the surroundings to the system?

It is the heat that has been transferred from the surroundings to the system in on cycle.

 

[Philip Koeck]

If I'm using $\Delta U= W + Q$ then mustn't a compression mean positive work? The work is
$W_{3-1}= -ln(\frac {V_1} {V_3})nRT= -ln(\frac {1} {3})nRT$, because $V_3=V_2=3V_1$

You're right of course. I was assuming Q = W + ΔU.
Sometimes I get too fixed in my ways.

 

[Philip Koeck]

I said that $\Delta S{sys}=0$ because entropy is a state variable, therefore if I start at 1 and end at 1 there must be no change in entropy. From 1-2 I have a change in entropy in the system, because, since the process is irreversible, entropy has to increase. Thus $\Delta S{sys}^{1-2} \gt 0$.

I must have misread something. For the whole cycle ΔS = 0 for the system, just as you say.

 

[Philip Koeck]

It is the heat that has been transferred from the surroundings to the system in on cycle.

Still not clear. Is Q_h the total net heat transfer from the surroundings to the system in one cycle or is it just the heat transfer during endothermal steps?
I'm guessing the latter, so Q_h would be only Q_2-3, whereas Q_c would be Q_3-1.
Is that right?

 

[Thermofox]

Still not clear. Is Q_h the total net heat transfer from the surroundings to the system in one cycle or is it just the heat transfer during endothermal steps?
I'm guessing the latter, so Q_h would be only Q_2-3, whereas Q_c would be Q_3-1.
Is that right?

Yes

 

[Philip Koeck]

Yes

Good. That means you can calculate T and p for state 2.
Would that help?

 

[Thermofox]

Good. That means you can calculate T and p for state 2.
Would that help?

With that I can complete point 2, but I still don't know how to find $\Delta S_{surr}$

 

[Philip Koeck]

With that I can complete point 2, but I still don't know how to find $\Delta S_{surr}$

I would go for ΔS_{surr 1-2} = 0 since no heat enters or leaves the surroundings and, by definition, the surroundings are unchanged.
The other two processes are reversible so they're you know ΔS for the surroundings.

 

[Thermofox]

I would go for ΔS_{surr 1-2} = 0 since no heat enters or leaves the surroundings and, by definition, the surroundings are unchanged.
The other two processes are reversible so they're you know ΔS for the surroundings.

Can you tell me how to find $\Delta S_{surr,2-3}$ and $\Delta S_{surr,3-1}$, because I can't see it.

 

[Chestermiller]

You know that the heat for the cycle is 890 J and you know the work for the isothermal reversible step. The change in internal energy for this step is zero, so that means that you know the heat for this step. You also know that the heat for the irreversible step is zero. This all then tells you the heat for the constant volume step.

 

[Philip Koeck]

Can you tell me how to find $\Delta S_{surr,2-3}$ and $\Delta S_{surr,3-1}$, because I can't see it.

Both those steps are reversible so for each of them the entropy changes for system and surroundings add up to 0.
Is that what you needed?

 

[Chestermiller]

I think you need to know the degrees of freedom of the gas.

 

[Thermofox]

Both those steps are reversible so for each of them the entropy changes for system and surroundings add up to
Is that what you needed?

Then $\Delta S_{sys}=- \Delta S_{surr.}$?

 

[Thermofox]

I think you need to know the degrees of freedom of the gas.

It is a monoatomic gas, I forgot to say it

 

[Philip Koeck]

Then $\Delta S_{sys}=\Delta S_{surr.}$?

Yes, with a minus on one side. That's true for all reversible processes.

 

[Thermofox]

Yes, with a minus on one side. That's true for all reversible processes.

Yeah I forgot to add it.

 

[Thermofox]

Yes, with a minus on one side. That's true for all reversible processes.

But now I get a negative change in entropy, shouldn't it increase?

 

[Chestermiller]

It is a monoatomic gas, I forgot to say it

Then once you know the the heat for the isochoric step, you know the temperature change for that step. Then you can get the temperature at point 2 and then the entropy change for this step.

 

[Philip Koeck]

But now I get a negative change in entropy, shouldn't it increase?

In total for system and surroundings it should increase.

For the 2 reversible processes it stays the same in total.
For the first process ΔS = 0 for the surroundings, whereas for the system the entropy should increase.

How do you calculate the system's entropy change during the adiabatic expansion?

 

[Chestermiller]

Let’s see some numbers.

 

[Thermofox]

How do you calculate the systems entropy change during the adiabatic expansion?

I thought of breaking up 1-2 into 2 processes, an isovolumetric and an isobaric one.
=> $\Delta S_{1-4} + \Delta S_{4-2} = \Delta S_{1-2}$

 

[Philip Koeck]

I thought of breaking up 1-2 into 2 processes, an isovolumetric and an isobaric one.
=> $\Delta S_{1-4} + \Delta S_{4-2} = \Delta S_{1-2}$

Okay. What do you get?

 

[Chestermiller]

Where does point 4 come in and why?

 

[Thermofox]

Where does point 4 come in and why?

Point 4 as the same pressure of point 1 and the same volume of point 2( and 3). I used it so that I could determine $\Delta S$. Otherwise I don't know how I could determine $\Delta S_{1-2}$

 

[Philip Koeck]

Point 4 as the same pressure of point 1 and the same volume of point 2( and 3). I used it so that I could determine $\Delta S$. Otherwise I don't know how I could determine $\Delta S_{1-2}$

An isotherm and an isochor (isovolumetric) would be another option.

 

[Thermofox]

Okay. What do you get?

$T_4= \frac {P_4 V_4} {nR}= \frac {(1)(36.1)} {(0.5) (0.0821)}= 879,4K$, I found V_1 in the same manner.

$\Delta S_{1-4} + \Delta S_{4-2} = \Delta S_{1-2}$

=> $\Delta S_{1-4}= nc_p ln(\frac {T_4} {T_1})= 0.5 \frac 5 2 ln(\frac {879,4} {293.16})= 11,41 \frac J K$

Since 1-2 is adiabatic, $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$

=> $T_2= T_1 (\frac {V_1} {V_2})^{\gamma - 1}= 293,16(\frac 1 3)^{\frac 5 3 - 1}= 140,9(K)$

$\Delta S_{4-2}=nc_v ln(\frac {T_2} {T_4})= 0.5 \frac 3 2 8,31 ln(\frac {140,9} {879,4})= -11,41 \frac J K$

=> $\Delta S_{1-2}= 0$ but it should be greater than 0, maybe it's because of how I determined $T_2$?

 

[Philip Koeck]

Since 1-2 is adiabatic, $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$

It's not reversible adiabatic so you can't use those expressions.

And you don't need to since you already know states 1 and 2 from other information.

 

[Thermofox]

It's not reversible adiabatic so you can't use those expressions.

Ok.
Q_{2-3}=Q_h
=> $Q_h= nc_v \Delta T= 0.5 \frac 3 2 8.31 (T_3 - T_2)$, hence $$ T_2 = \frac {nc_vT_3 - Q} {nc_v}= 150.36 \frac J K$$

If I use this temperature I obtain that $\Delta S_{4-2}= -11\frac J K$
=>$\Delta S_{1-2}= 0.41\frac J K$