Calculation of thermodynamic variables for irreversible adiabatic process of ideal gas

[zenterix]

Homework Statement

An ideal monoatomic gas at $1\text{bar}$ and $300\text{K}$ is expanded adiabatically against a constant pressure of $0.5\text{bar}$ until the final pressure is $0.5\text{bar}$.

Relevant Equations

What are the values of $q$ per mole, $w$ per mole, $\Delta \overline{U}$ and $\Delta\overline{H}$?

We start at state 1 and end at state 2.

We are given the information that

$$P_1=1\ \text{bar}$$
$$T_1=300\text{K}$$
$$P_2=0.5\ \text{bar}$$

From the ideal gas law, we can obtain $V_1$.

$$V_1=\frac{nRT_1}{P_1}$$

$$=2494.2\cdot 10^{-5}\mathrm{m^3}$$

My first question arises here. The process is irreversible: we don't know intermediate equilibrium states. There are, however, initial and final equilibrium states. Can we assume that these two states are on the same adiabat?

In other words, starting at state 1, assuming the system is adiabatic, is it correct to say that no matter what, every reachable equilibrium state must be on the same adiabat that state 1 is on?

If yes, then this adiabat is represented by the relationship

$$PV^\gamma=k$$

where $\gamma=\frac{5}{3}$ for our ideal gas.

We can determine $k$ using data from the initial equilibrium

$$k=P_1V_1^\gamma=212.63n^{5/3}\ \mathrm{Pa\ (m^3)^{\gamma}}$$

Since we know the final state pressure and we now know $k$ we can determine the final volume $V_2$

$$P_2V_2^\gamma=k$$

$$\implies V_2=3780.5\cdot 10^{-5}n\ \mathrm{m^3}$$

Next, we can determine $T_2$ from the ideal gas law

$$T_2=\frac{P_2V_2}{nR}=227.35\text{K}$$

To calculate the total work, we have

$$w=-\int PdV=-\int_{V_1}^{V_2} P_2 dV=-643.14n\ \text{J}$$

Since we have an adiabatic process, then $q=0$ and so work equals change in internal energy.

$$\Delta U=w=-643.14n\ \text{J}$$

To obtain the variables per mole we divide by $n$

$$\Delta\overline{U}=\overline{w}=-643.14\ \mathrm{\frac{J}{mol}}$$

The end of the book says this should be $-748\mathrm{\frac{J}{mol}}$.

I am less worried about the discrepancy in the values than in the reasoning used above. Is it correct?

Finally, let's calculate the change in enthalpy per mole of gas.

I am not so sure about the calculation that follows.

$H$ is a state function and its differential is

$$dH=dU+PdV+VdP$$

It is not clear to me yet when we can or can't use this expression and integrate it.

I do know we can integrate it when we have a reversible process.

Since we know the initial and final state variables, and since the change in enthalpy is the same whether we move between the states reversibly or irreversibly, we can integrate assuming we have a reversible process.

$$\Delta H=\Delta U+\int_{V_1}^{V_2}PdV+\int_{P_1}^{P_2} VdP$$

The first two terms represent $\Delta U-w=0$. Thus

$$\Delta H=\int_{P_1}^{P_2} VdP=\int_{1\text{bar}}^{0.5\text{bar}}\left ( \frac{k}{P} \right )^{1/\gamma} dP$$

$$=-1508.6n\ \text{J}$$

and so change in enthalpy per mole of gas is

$$\Delta\overline{H}=-1508.6\mathrm{\frac{J}{mol}}$$

which is also different from the end of the book answer of $-1247\mathrm{\frac{J}{mol}}$.

 

[mjc123]

I think the point is that your assumption in bold is incorrect. PV^γ = constant applies only to an equilibrium process, i.e. if the external pressure was infinitesimally less than the gas pressure at all points.

I did it a different way, which gets the book answer. Let the gas expand isothermally to point 3 (0.5 bar, 300K). Calculate the work, which equals q as ΔU = 0. To get to the adiabatic endpoint 2, we must lose heat q, while maintaining constant pressure of 0.5 bar. Calculate the change in temperature, hence T2 and V2. Calculate the work, and add the work for the 2 steps.

 

[zenterix]

This problem shows me how I don't quite understand the first law.

This is what I understood that you did (though I think I understood incorrectly).

First of all, let me just say this.

The work in the irreversible expansion at constant pressure of $0.5\ \text{bar}$ is

$$w_{irr}=-\int_{V_1}^{V_f} \left ( 0.5\cdot 10^5 \mathrm{\frac{N}{m^2}}\right ) dV\ \mathrm{m^3}$$

I am not assuming that I know what $V_f$ is.

The isothermal expansion is from $a$ to $b$.

The work done and heat in this expansion are

$$w_{ab}=-nRT_1\ln{\frac{V_2}{V_1}}=-1728n\ \text{J}$$

$$q_{ab}=-w_{ab}=1728n\ \text{J}$$

You refer to the "adiabatic endpoint 2". It is not clear what this point is.

In the picture above, there is the point $c$ which is the final point of a reversible adiabatic expansion.

At $c$ we have

$$T_3=T_1\left (\frac{P_2}{P_1}\right )^{\frac{\gamma-1}{\gamma}}$$

$$=227.35\text{K}$$

$$V_3=\frac{nRT_3}{P_2}=0.03780\ \mathrm{m^3}$$

The change in internal energy for the reversible adiabatic expansion is

$$\Delta U_{ac}=w_{ac}=-\int_{V_1}^{V_3}\frac{k}{V^\gamma}dV=-905.92n\ \text{J}$$

So, my question is, after the isothermal expansion, are you moving from $b$ to $c$ or is the final point some other point different from $c$?

If the final point is $c$ then we have

$$w_{bc}=-P_2(V_3-V_2)=603.94n\ \text{J}$$

$$\Delta U_{bc}=w_{bc}+q_{bc}=\Delta U_{ac}=w_{ac}$$

which we can solve for $q_{bc}$

$$q_{bc}=-1509.87n\ \text{J}$$

To recap the calculations, to move from $a$ to $c$ the change in internal energy is $-905.92n\ \text{J}$. If we do this reversibly and adiabatically, this is the same as the work done.

If we go from $a$ to $b$ to $c$ then we do $-1124.89n\ \text{J}$ of work and absorb $218.97n\ \text{J}$ of heat. The end result is also $-905.92n\ \text{J}$ of internal energy change.

However, again, it is not clear to me what the final state is when we do the irreversible expansion. Is it point $c$?

If it is then final volume is $V_3$ and so

$$w_{irr,ac}=-\int_{V_1}^{V_3} \left ( 0.5\cdot 10^5 \mathrm{\frac{N}{m^2}}\right ) dV\ \mathrm{m^3}=-643.15n\ \text{J}$$

which is smaller than the change in internal energy. Where does the rest of the change in internal energy come from?

I am not sure what to conclude.

 

[Chestermiller]

Your original post assumes a reversible adiabat, which is not the case. The process. is irreversible. The work, assuming that the external pressure is dropped to P2 for the entire expansion is $$W=P_2(V_2-V_1)=P_2\left(\frac{nRT_2}{P_2}-\frac{nRT_1}{P_1}\right)$$The heat Q is zero because the process is adiabatic. The change in internal energy is $$\Delta U=nC_V(T_2-T_1)$$

 

[mjc123]

Apologies for not making it clearer. I hadn't fully thought it through myself.
By "adiabatic endpoint 2" I meant what you called "state 2" in your original post; i.e. the end point of the irreversible adiabatic expansion at 0.5 bar.
I should have specified that I intended an irreversible isothermal expansion at constant external pressure of 0.5 bar. The work would be -0.5*(V2-V1) = -1247 J/mol.
At this point let me insert your picture, with some extra points labelled for clarity. The difference here is that by point c (also labelled y) I intend "state 2", the endpoint of the irreversible adiabatic expansion. This will be different from your point c because the adiabat PV^γ = constant is only followed in a reversible process.
Going from a to b absorbed heat 1247 J/mol. We now lose this heat isobarically and the temperature changes
ΔT = q/Cp = -1247/20.785 = -60 K; T3 = 240K; V3 = 0.0399 m³/mol
w_bc = -0.5*(V3-V2) = 499 J/mol
w = w_ab + w_bc = -748 J/mol
Diagrammatically, w equals the area of the rectangle WXYZ, which is equal to WXRS minus ZYRS.
ΔU_ac = w_ac as you said.
ΔH_ac = ΔH_ab + ΔH_bc
= 0 + CpΔT = -1247 J/mol

 

[zenterix]

In the irreversible adiabatic expansion we have no heat exchange and only work is done

$$w_{irr,af}=-\int_{V_1}^{V_f} P_2dV=-P_2(V_f-V_1)\tag{1}$$

This is the same as the change in internal energy of the system between the initial state and the unknown final state.

We don't know what $V_f$ is at this point.

Note that from the ideal gas law

$$T_f=\frac{P_2V_f}{nR}\tag{2}$$

If we construct a reversible path to this final point then we know that the change in internal energy must be the same as in the irreversible case.

One reversible path is an isothermal expansion to pressure $P_2$ and volume $V_2$ and then an isobaric process from $V_2$ to $V_f$.

The total change in internal energy is

$$\Delta U_{af}=\Delta U_{ab}+\Delta U_{bf}=\Delta U_{bf}=w_{bf}+q_{bf}\tag{3}$$

$$=-P_2(V_f-V_2)+\frac{5}{2}R(T_f-T_1)\tag{4}$$

where I used

$$q_{bf}=\Delta H_{bf}=\int_{T_1}^{T_f}C_p\cdot n dT=\frac{5}{2}nR(T_f-T_1)$$

We equate $\Delta U_{af}$ to $w_{irr,af}$ and solve for $V_f$.

We get

$$V_f=0.0399072n\ \mathrm{m^3}\tag{5}$$

So the final state is a point that is between points $c$ and $b$ in the picture.

The total work done equals the change in internal energy and

$$w_{irr,af}=\Delta U_{af}=w_{bf}+q_{bf}=-748.26n\ \text{J}$$

The total change in enthalpy is

$$\Delta H_{af}=\Delta H_{ab}+\Delta H_{bf}$$

$$=0+q_{bf}$$

$$=-1247.1n\ \text{J}$$

Here is a picture illustrating what I have done above.

 

[Chestermiller]

I can't figure out what you did in your analysis of the previous post.

For the irreversible path, if you combine the two equations I presented in post #3 using the first law of thermodynamics, you get $$nC_v(T_2-T_1)=-nR\left(T_2-\frac{P_2}{P_1}T_1\right)$$or $$T_2=\frac{C_v+R(P_2/P_1)}{C_p}T_1$$
This can then be used to get V2.

There is no need to consider a reversible path in this analysis unless you also want to determine the entropy change for. the system. In that case, the reversible path is not adiabatic.

 

[zenterix]

Apologies for not making it clearer. I hadn't fully thought it through myself.
By "adiabatic endpoint 2" I meant what you called "state 2" in your original post; i.e. the end point of the irreversible adiabatic expansion at 0.5 bar.
I should have specified that I intended an irreversible isothermal expansion at constant external pressure of 0.5 bar. The work would be -0.5*(V2-V1) = -1247 J/mol.
At this point let me insert your picture, with some extra points labelled for clarity. The difference here is that by point c (also labelled y) I intend "state 2", the endpoint of the irreversible adiabatic expansion. This will be different from your point c because the adiabat PV^γ = constant is only followed in a reversible process.
Going from a to b absorbed heat 1247 J/mol. We now lose this heat isobarically and the temperature changes
ΔT = q/Cp = -1247/20.785 = -60 K; T3 = 240K; V3 = 0.0399 m³/mol
w_bc = -0.5*(V3-V2) = 499 J/mol
w = w_ab + w_bc = -748 J/mol
Diagrammatically, w equals the area of the rectangle WXYZ, which is equal to WXRS minus ZYRS.
ΔU_ac = w_ac as you said.
ΔH_ac = ΔH_ab + ΔH_bc
= 0 + CpΔT = -1247 J/mol
View attachment 348778

This is an interesting way to do the calculation, different from my calculation above.

The ultimate goal is to find the final state.

In my calculation, I computed the total adiabatic work to reach the final state and then constructed a reversible path to this final state, solving for the unknown $V_f$.

Since we know the pressures at states $a$, $b$, and $d$, and we know the temperature at $a$ and $b$, and since the temperature at $d$ depends on $V_f$, then finding $V_f$ gives us $T_f$.

At this point we know the thermodynamic state variables in every state.

Before going on to your calculation, note that

$$\Delta U_{ad}=w_{ad}\tag{1}$$

$$\Delta U_{ad}=w_{ad}+q_{ad}=w_{ab}+w_{bd}+q_{ab}+q_{bd}\tag{2}$$

These equations are true for any process that goes from $a$ to $d$.

In your calculation, you used the following for your two-step irreversible process

$$q_{ad}=0\tag{3}$$

and so

$$q_{bd}=-q_{ab}\ \ \ \ \ \text{and}\ \ \ \ \ w_{ad}=w_{ab}+w_{bd}\tag{4}$$

Now, it is still a bit bewildering to me that you can use this for your two-step irreversible process. These relationships are definitely true for the single-step irreversible process.

In my calculation, I did not use $q_{ad}=0$ for the reversible process that leads to the final state. In fact, in my calculation we have

$$w_{ab}=-1748n\ \text{J}$$

$$w_{bd}=498n\ \text{J}$$

$$q_{ab}=1728n\ \text{J}$$

$$q_{bd}=-1247n\ \text{J}$$

$$w_{ad}=-748.26n\ \text{J}$$

$$q_{ad}=0$$

These values satisfy (1), (2), and (3) but not (4).

In what follows, I show the calculations you apparently did.

Isothermal irreversible expansion from $a$ to $b$.

The work is

$$w_{ab}=-P_2(V_2-V_1)=-q_{ab}=-1247.1n\ \text{J}$$

$$\Delta H_{ab}=U_2+P_2V_2-U_1-P_1V_1=P_2V_2-P_1V_1=nR(T_2-T_1)=0$$

The next step you did was to say that to get from $b$ to $d$, the heat absorbed in the isothermal expansion must be released from the system.

The resulting temperature change is

$$\Delta T=T_d-T_b=\frac{-q_{ab}}{C_pn}$$

$$=\frac{-1247.1n}{\frac{5}{2}Rn}$$

$$=-60\text{K}$$

Thus, $T_d=240\text{K}$.

At this point we can use the ideal gas state equation to find $V_f=0.03990\mathrm{m^3}$.

We have all the state variables at each of the equilibrium points of interest.

The work to get from $b$ to $d$ is

$$w_{bd}=-P_2(V_d-V_2)=498.84n\ \text{J}$$

The total work for the two stages of your irreversible process is

$$w_{ad}=w_{ab}+w_{bd}=-1247.1n\ \text{J}+498.84n\ \text{J}=-748.26n\ \text{J}$$

This is the same value as

$$w_{ad}=-P_2(V_d-V_a)=-748.26n\ \text{J}$$

Here is a picture showing your calculations

 

[zenterix]

There is no need to consider a reversible path in this analysis unless you also want to determine the entropy change for. the system. In that case, the reversible path is not adiabatic.

I think your solution is yet a third way to solve the problem. And it is the simplest and most efficient of the three.

As you mention, and as I also noted in my post #8, the reversible path I considered is not adiabatic. But it does reach the same final state, which seems to be what matters.

All you did was compute the irreversible work for the single irreversible adiabatic process and you equated that to the change in the internal energy which for an ideal gas is change in temperature times $C_V$.

The variable in this equation is the final temperature, so you solve for that first.

Everything else follows from that.

Final volume is obtained from the ideal gas law.

Internal energy change can now be computed with the final temperature.

Change in enthalpy is just $C_P\Delta T$, which we also know because of the final temperature we discovered.

Note that I added a new picture to show what I did

Maybe this clears it up for you.

 

[mjc123]

I think your method is neater than mine because it doesn't require the assumption of an irreversible isothermal process, only using the state functions of a and b, and the work of ac and bc. Chestermiller's is the most succinct I think.