[Thermodynamics] Doubt about the relation between internal and external work

[valleyman]

Hello, while studying thermodynamics I was wondering why, in a system where the pression is not univocally determined (for example in irreversible trasformations), to measure the total work of the trasformation, we use to consider only the external forces. I mean, this sounds conceptually right to me, as we study the effects done by the trasformation on the environment, but looking trhough it, shouldn't the effective value be different?
If an incognite force F1 is pushing toward a side of the surface the work of that force should be F1*S, where S is the movement of the surface in the direction of the force. Now, externally, another force F2 is pushing on the other side of the same surface, so we have that the work done by F2 is F2*S which should be *different*, as long as F1 is not equal to F2.
So what is the right value of the work? shouldn't it be W(F1) = -W(F2)?
Thanks for the help,
valleyman

 

[Andrew Mason]

Hello, while studying thermodynamics I was wondering why, in a system where the pression is not univocally determined (for example in irreversible trasformations), to measure the total work of the trasformation, we use to consider only the external forces. I mean, this sounds conceptually right to me, as we study the effects done by the trasformation on the environment, but looking trhough it, shouldn't the effective value be different?
If an incognite force F1 is pushing toward a side of the surface the work of that force should be F1*S, where S is the movement of the surface in the direction of the force. Now, externally, another force F2 is pushing on the other side of the same surface, so we have that the work done by F2 is F2*S which should be *different*, as long as F1 is not equal to F2.
So what is the right value of the work? shouldn't it be W(F1) = -W(F2)?
Thanks for the help,
valleyman

I think your question is: You are considering irreversible processes in which the internal pressure of the gas is greater than the external pressure. Why is the work done measured by the work done against the external pressure and not the work done by the internal pressure?

If the forces/pressures are not balanced, the system is not in thermodynamic equilibrium while the process is going on. If the internal pressure is greater than the external force, you have a dynamic rather than a quasi-static process. This means that some of the energy of the system is kinetic. To apply conservation of energy (first law) you have to take into account the kinetic energy in the dynamic system.

If you apply the first law of thermodynamics now to the dynamic system using the internal pressure to determine work done (W = PdV) and take into account the kinetic energy as part of the work done, the first law will be maintained.

In the case of an explosion of a gas in a cylinder doing work against a piston, the internal pressure produces kinetic energy of the piston and of the molecules of exploding gas. Suppose the piston is lifted by the exploding/expanding gas. At some point the internal and external pressure become equal (assuming the cylinder is long enough). But because the piston is moving with some speed, it goes higher before it stops. At that point, the external pressure is higher than the internal pressure and the piston comes back down, picking up speed. It goes past the point where internal and external pressure are equal and eventually stops when its kinetic energy is used up in compressing the gas in the cylinder. It repeats that oscillation until all the energy is used up in the form of heat.

If you apply the first law of thermodynamics now to the stationary system using the external pressure to determine work done (W = PdV) and the final temperature (ie with the extra heat added by the conversion of kinetic energy into heat), the first law will be maintained.

AM

 

[valleyman]

Hey Andrew, thanks for the explanation, you understood the question perfectly (even if I didn't explain so well, I admit) and u've been much clearer than many texts I've read around, but... I've still got a doubt
You said I've to consider also the kinetic energy in the total amount of the work done by internal pressure, but, if that pressure, P_int, is greater, the value P_int*dV should be greater than P_ext*dV. So, when u add another positive quantity, the kinetic energy, to that product (P_int*dV) shouldn't you still get a greater value? I think I didn't get well how you consider this kinetic energy in the count...

 

[Andrew Mason]

Hey Andrew, thanks for the explanation, you understood the question perfectly (even if I didn't explain so well, I admit) and u've been much clearer than many texts I've read around, but... I've still got a doubt
You said I've to consider also the kinetic energy in the total amount of the work done by internal pressure, but, if that pressure, P_int, is greater, the value P_int*dV should be greater than P_ext*dV. So, when u add another positive quantity, the kinetic energy, to that product (P_int*dV) shouldn't you still get a greater value? I think I didn't get well how you consider this kinetic energy in the count...

No. P_{int}\Delta V = P_{ext}\Delta V + KE. In other words, the internal pressure does work against the external pressure AND does additional work in adding kinetic energy to the system.

AM

 

[valleyman]

Oh well, so that's the relation, but so, shouldn't I consider also the KE as part of the total work of the transformation? this would mean I would be constricted to calculate p_int * V anyways, where I'm wrong?

 

[Andrew Mason]

Oh well, so that's the relation, but so, shouldn't I consider also the KE as part of the total work of the transformation? this would mean I would be constricted to calculate p_int * V anyways, where I'm wrong?

It depends on when you want to analyse the system. If it is after everything has returned to equilibrium and the kinetic energy has dissipated into heat in the system, the result is the same from an energy perspective (first law analysis) as if it had occurred under a quasi-static process (i.e. where Pext = Pint). If it is before things have returned to equilibrium, you have to use dW = P_{int}dV for the work done. This results in work done against the external pressure plus kinetic energy in the dynamic system.

AM

 

[valleyman]

ok, now is it clear, thanks a lot for the help!

 

[inquisitive_i]

christmas folks..!
i was sippin ma cupa coffee when i tangled this one..
a cup made of china clay allows the coffee to cool out far more rapidly than a flask of steel does.. my conscience tell me that metal being a good conductor of heat(a better conductor of heat in dis case) must radiate the heat to d vicinity far more efficiently than a clay pot..
than is d above mentioned situation violating a law of THERMODYNAMICS??