# Adiabatic expansion work far exceeds isobaric of same volume, why?

• I
• MysticDream
But the force the gas exerts at a given volume change is different for a large piston mass than for a small mass. So the acceleration is not simply inversely proportional to the mass.

We are not focused on the initial adiabatic expansion of the gas. We are focused on the final equilibrium state when the piston is no longer moving.

This is not a correct description of the wave mechanics in a gas enclosed in a cylinder and it also neglects the viscous dissipation in the gas enclosed in the cylinder. In any event, it will not tell you the work done by the gas on the piston.
I understand the irreversible work done on the atmosphere will not be affected by the pistons mass, but the total reversible work should be the same regardless of the pistons mass, should it not?

Why would the force be different? You would start with the same exact pressure on the piston face. Do you mean the Pressure-Volume curve would be different?
Do you think that an ideal gas satisfies the ideal gas law in a rapid irreversible expansion (not at thermodynamic equilibrium)?

• Do you think that an ideal gas satisfies the ideal gas law in a rapid irreversible expansion (not at thermodynamic equilibrium)?

Yes.

Why is more work done on a heavier piston?

How do you calculate the work done on a piston with mass and the work done on the atmosphere that is wasted?

Yes.
This is incorrect. The ideal gas law is valid only for thermodynamic equilibrium conditions or for a reversible process (which consists of a continuous sequence of thermodynamic equilibrium states).

The compressive stresses for a gas experiencing a rapid irreversible deformation are given by Newton's law of viscosity (in correct 3D tensorial form). For such a deformation, we learn in fluid mechanics that the principal components of the compressive stress tensor are given locally by: $$\sigma_1=\frac{RT}{V_m}-\left(\frac{4}{3}\mu\right)\frac{\partial v_1}{\partial x_1}-\left(\frac{2}{3}\mu\right)\frac{\partial v_2}{\partial x_2}-\left(\frac{2}{3}\mu\right)\frac{\partial v_3}{\partial x_3}$$$$\sigma_2=\frac{RT}{V_m}-\left(\frac{4}{3}\mu\right)\frac{\partial v_2}{\partial x_2}-\left(\frac{2}{3}\mu\right)\frac{\partial v_1}{\partial x_1}-\left(\frac{2}{3}\mu\right)\frac{\partial v_3}{\partial x_3}$$$$\sigma_3=\frac{RT}{V_m}-\left(\frac{4}{3}\mu\right)\frac{\partial v_3}{\partial x_3}-\left(\frac{2}{3}\mu\right)\frac{\partial v_2}{\partial x_2}-\left(\frac{2}{3}\mu\right)\frac{\partial v_1}{\partial x_1}$$where ##V_M## is the local molar volume of the gas, ##\mu## is the gas viscosity, and the subscripts 1, 2, and 3 represent the 3 principal directions of stress and rate of strain. Note that the first term in each of these equations is the local pressure that one would calculate locally from the ideal gas law at the local molar volume and local temperature. The additional terms are viscous contributions that approach zero in a reversible deformation.

Why is more work done on a heavier piston?

How do you calculate the work done on a piston with mass and the work done on the atmosphere that is wasted?
Are you talking about at each point in time during a rapid irreversible expansion or only at final steady state when the piston has come to rest and the gas is again at equilibrium?

Are you talking about at each point in time during a rapid irreversible expansion or only at final steady state when the piston has come to rest and the gas is again at equilibrium?

Each point in time, but more specifically at max volume of the gas.

Each point in time, but more specifically at max volume of the gas.
Thermodynamics alone can't determine the work the gas does on the piston at each point in time during an irreversible expansion because of the existence of the viscous contributions to the force exerted by the gas. It can however determine the work done by the gas on the piston through final equilibrium when the piston kinetic energy has been damped to zero (provided we know the pressure imposed by the external atmosphere). To get the work the gas does on the piston at each point int time during an irreversible expansion, one must solve the set of partial differential equations represented by the Navier Stokes equations for viscous flow in conjunction with the partial differential equation for overall energy conservation. The independent variables in this fluid mechanics endeavor are spatial position within the cylinder and time, and the dependent variables are temperature, velocity components, and pressure (all of which are functions of the independent variables).

Thermodynamics alone can't determine the work the gas does on the piston at each point in time during an irreversible expansion because of the existence of the viscous contributions to the force exerted by the gas. It can however determine the work done by the gas on the piston through final equilibrium when the piston kinetic energy has been damped to zero (provided we know the pressure imposed by the external atmosphere). To get the work the gas does on the piston at each point int time during an irreversible expansion, one must solve the set of partial differential equations represented by the Navier Stokes equations for viscous flow in conjunction with the partial differential equation for overall energy conservation. The independent variables in this fluid mechanics endeavor are spatial position within the cylinder and time, and the dependent variables are temperature, velocity components, and pressure (all of which are functions of the independent variables).

Can it be approximated with a simple formula? Example: Total work done by gas = work on atmosphere + work on piston.

I still don’t understand why more work would be done on a heavier piston.

Can it be approximated with a simple formula? Example: Total work done by gas = work on atmosphere + work on piston.
No, because you are not going. to know the work on the piston (.e., its kinetic energy) without directly determining the work done by the gas, which involves solving that complicated set of equations interior to the cylinder, and matching the piston velocity to the gas velocity at the inside face of the piston.
I still don’t understand why more work would be done on a heavier piston.
That's only initially during the expansion. At final thermodynamic equilibrium, the work will be the same.

No, because you are not going. to know the work on the piston (.e., its kinetic energy) without directly determining the work done by the gas, which involves solving that complicated set of equations interior to the cylinder, and matching the piston velocity to the gas velocity at the inside face of the piston.

That's only initially during the expansion. At final thermodynamic equilibrium, the work will be the same.

I appreciate your insight. You’ve given me a lot to think about.

• berkeman
I appreciate your insight. You’ve given me a lot to think about.
Check out Transport Phenomena by Bird, et al, Chapter 1, Section 1.2 Generalization of Newton's Law of Viscosity.

• MysticDream and Lord Jestocost