Thermodynamic Process: Is Work Conserved?

[Ian Baughman]

What is wrong with this logic, if any? It does not seem like this should be true but maybe I'm mistaken.

Assuming the process consists of two isobaric processes and two isothermal processes the work from B to C in terms of p₁, p₂, V₁, and V₂ is given by the following.

1) W_BC=p₂(V_C-V_B)
2) Isothermal from A → B so p₁V₁ = p₂V_B so V_B=(p₁V₁)/p₂
3) Similarly V_C=(p₁V₂)/p₂
4) W_BC= p₂((p₁V₂)/p₂-(p₁V₁)/p₂)
5) Factor out p₁/p₂ and simplify the you get;

W_BC=p₁(V₂-V₁)​

 

[Charles Link]

What is wrong with this logic, if any? It does not seem like this should be true but maybe I'm mistaken.

Assuming the process consists of two isobaric processes and two isothermal processes the work from B to C in terms of p₁, p₂, V₁, and V₂ is given by the following.

1) W_BC=p₂(V_C-V_B)
2) Isothermal from A → B so p₁V₁ = p₂V_B so V_B=(p₁V₁)/p₂
3) Similarly V_C=(p₁V₂)/p₂
4) W_BC= p₂((p₁V₂)/p₂-(p₁V₁)/p₂)
5) Factor out p₁/p₂ and simplify the you get;

W_BC=p₁(V₂-V₁)​

The isothermal processes have no change in internal energy. for an ideal gas. Much heat must be added to the system in the isobaric expansion for two reasons: 1)work is being done as it expands 2) It is going to a higher temperature state. In the isobaric compression, much heat must be removed from the system for two reasons 1) The system is going to a lower temperature 2) Work is being done on the system. The energy from this work must also be removed by removing heat. I do think your calculations are correct. Thermodynamic problems can sometimes be tricky, but I believe you analyzed it correctly.

 

[Charles Link]

It should also be noted that you can readily compute $\Delta Q=\Delta W$ for each of the two isotherms, since each has $\Delta U=0$. $\\$ Meanwhile $\Delta U$ for each of the isobars will be $\Delta U=\frac{3}{2}nR \Delta T$ for an ideal monatomic gas, and $\Delta U=\frac{5}{2}nR \Delta T$ for an ideal diatomic gas (where $\Delta T=\Delta (PV)/(nR)$). You can thereby readily compute $\Delta Q$ for each of the isobars, if you know the type of gas. $\\$ I think I have this completely correct, but it's always good to have someone double-check the solution. @Chestermiller Did we get everything correct?

 

[Ian Baughman]

It should also be noted that you can readily compute $\Delta Q=\Delta W$ for each of the two isotherms, since each has $\Delta U=0$. $\\$ Meanwhile $\Delta U$ for each of the isobars will be $\Delta U=\frac{3}{2}nR \Delta T$ for an ideal monatomic gas, and $\Delta U=\frac{5}{2}nR \Delta T$ for an ideal diatomic gas (where $\Delta T=\Delta (PV)/(nR)$). You can thereby readily compute $\Delta Q$ for each of the isobars, if you know the type of gas. $\\$ I think I have this completely correct, but it's always good to have someone double-check the solution. @Chestermiller Did we get everything correct?

Since this is a thermodynamic cycle with the final state being identical to the initial state would the total change in internal energy be equal to zero?

 

[Charles Link]

Since this is a thermodynamic cycle with the final state being identical to the initial state would the total change in internal energy be equal to zero?

Yes. And if you will notice, for the two isotherms $\Delta U=0$ and for the transitions on the two isobars, $\Delta U_1=-\Delta U_2$. $\\$ Editing... perhaps we haven't proved yet that $\Delta (PV)$ is equal and opposite for these two cases...but that is obvious because on one isotherm, $PV=nRT_1$ and on the other isotherm $PV=nRT_2$.Thereby $\Delta (PV)=nR \Delta T$ and $\Delta U=\frac{3}{2} nR \Delta T$ or $\Delta U=\frac{5}{2} nR \Delta T$.

 

[Chestermiller]

What is wrong with this logic, if any? It does not seem like this should be true but maybe I'm mistaken.

Assuming the process consists of two isobaric processes and two isothermal processes the work from B to C in terms of p₁, p₂, V₁, and V₂ is given by the following.

1) W_BC=p₂(V_C-V_B)
2) Isothermal from A → B so p₁V₁ = p₂V_B so V_B=(p₁V₁)/p₂
3) Similarly V_C=(p₁V₂)/p₂
4) W_BC= p₂((p₁V₂)/p₂-(p₁V₁)/p₂)
5) Factor out p₁/p₂ and simplify the you get;

W_BC=p₁(V₂-V₁)​

Yes. So...

And yes, since this a thermodynamic cycle, the change in internal energy for the working fluid has to be zero.