# Work done by ideal gas under a reversible process which is not isobaric, isochoric, isothermal, adiabatic

First post, sorry for the poor formatting.

Question

Consider a non adiabatic system closed to mass exchange where an ideal gas traverses along a reversible path from state: $P_1,&space;V_1,&space;T_1$ to $P_2,&space;V_2,&space;T_2$. What is the work done by/to the system? (Without resorting to statistical thermo)

Attempt 1

First law is $\Delta&space;U&space;=&space;\delta&space;Q&space;+&space;\delta&space;W$ We cannot simplify this. The internal energy of an ideal gas is a function of temperature hence $\Delta&space;U$ remains. The system is not adiabatic hence $\delta&space;Q$ remains. Finally, the system is not isochoric, hence $\delta&space;W$ remains.

The ideal gas equation for a system closed to mass exchange is $PV&space;=&space;RT$

Because the path is reversible we may use:

$W_{rev}&space;=&space;\int&space;P&space;dV$

However, we must deal with an implicit function and without an additional equation to parameterize it, prevents us from analytically integrating. (I believe?)

We cannot dissect the path into isobaric, isotherm steps because work is not a state function.

My best idea is that if we knew the exact path, we could solve for the work numerically.

Attempt 2

Internal energy for an ideal gas is a function of temperature; because it is a state function we can choose to take a constant volume path between our initial and final temperature.

$\Delta&space;U&space;=&space;\int&space;C_V&space;dT$

What remains is to solve for the heat and by the first law we may obtain the work.

$\delta&space;W_{rev}&space;=&space;\Delta&space;U&space;-\delta&space;Q$

For a reversible path we have the relation:

$dS&space;=&space;\frac{dQ_{rev}}{T}$

Which may be rearranged to give:

$\Delta&space;Q_{rev}&space;=&space;\int&space;T&space;dS$

Again, if we know the exact path and resort to numerical methods, we should be able to solve this by using:

$\Delta&space;S&space;=&space;C_V&space;log(\frac{T_2}{T1})+R&space;log(\frac{V_2}{V_1})$

Thanks!

• Do you need the actual path between the two states or just the end points? – user207455 Jun 18 '19 at 15:10
• Assume you are given only the end points. – HSPrzepa Jun 18 '19 at 15:12
• Isn't the change in internal energy of an ideal gas with constant specific heat $C_V$ going from $p_i, V_i, T_i$ to $p_f, V_f, T_f$ well defined? So, is the opening statement in your Attempt 1 really valid? Also, isn't entropy change a path independent function? – Jeffrey J Weimer Jun 18 '19 at 15:17
• When I wrote(We cannot simplify this) I meant "cancel out due to a quantity becoming zero". Yes, entropy change is path independent. – HSPrzepa Jun 18 '19 at 15:23
• @ Jeffrey J Weimer The entropy change for an ideal gas between these two thermodynamic equilibrium states is unique, and is not path dependent. Any an all reversible paths will give exactly the same entropy change. – Chet Miller Jun 18 '19 at 15:25

Consider a non adiabatic system closed to mass exchange where an ideal gas traverses along a reversible path from state: P_1, V_1, T_1 to P_2, V_2, T_2. What is the work done by/to the system?

You don't have sufficient information. It's not adiabatic but it could be isothermal, isobaric, isochoric, polytropic, or any combination of theses and other possibilities. But I think you have already answered your own question in your two attempts. Without knowing the path from 1 to 2 you cannot determine the work done in going from 1 to 2 because work depends on the path and there are an infinite number of paths between states 1 and 2.

In your Attempt 1 you concluded:

"My best idea is that if we knew the exact path, we could solve for the work numerically"

And that is absolutely correct.

In Attempt 2 you concluded:

"Again, if we know the exact path and resort to numerical methods, we should be able to solve this by using:"

Right again. But as @Chet Miller pointed out the entropy change doesn’t depend on the path just the end points. But knowing the entropy change does not tell you the work done because that does depend on the path.

Hope this helps.

• Perfect answer. – Chet Miller Jun 18 '19 at 15:30
• @BobD "Right again. But as @Chet Miller pointed out the entropy change doesn’t depend on the path just the end points. But knowing the entropy change does not tell you the work done because that does depend on the path." So, numerically speaking, if we had a spreadsheet of the exact P,V,T path that the system followed, why couldn't we calculate the entropy change between each point and approximate the Q_rev integral? – HSPrzepa Jun 18 '19 at 17:05
• @HSPrzepa Sure you can, but the point is you don't NEED to know the path to determine the entropy change, because unlike heat and work, it depends only on the end points like all properties of the system. The same goes for internal energy, temperature, pressure, enthalpy, and every other property of the system. Heat and work are not properties of the system. I just didn't want you to leave you thinking that you needed the path to determine entropy change. Hope this clarifies. – Bob D Jun 18 '19 at 17:16
• @HS Przepa You can do that, but you’ll just get get the exact same result for the entropy change no matter which reversible path you follow. – Chet Miller Jun 18 '19 at 18:21