# Regarding the equivalence of expressions for the change in internal energy of a system when the process is not quasi-static

It is well-established that some authors prefer expressing the change in internal energy of a thermodynamic system in terms of the work done ON the system and others will express it in terms of the work done BY the system. I have also read that these two corresponding expressions are equivalent as long as you flip the sign.

That is, $$\Delta U = q + W_{on} = q - W_{by}$$

My text takes great care to note that most of the formulas it derives are subject to the assumption that the process is quasi-static, and it is immediately clear to me that given that assumption the two expressions above are equivalent. I'm wondering whether this holds when the process is not. In particular, I feel like a system would not generally expand unless there was a pressure differential at its boundary. Wouldn't equal pressures imply that average force per unit area on both sides of the boundary would be the same, giving a net force of zero? It's tough for me to visualize because the expansion/contraction itself would tend to oppose the change in pressure. But, if you had that $$p_{int} > p_{ext}$$ during an expansion its not clear to me that you would have the magnitude of the work, using the formula $$W = \pm\int p \ dV$$, be the same for both.

• Just an observation about the sign of W in the first law. I have found that in chemistry they state the first law as Delta U = Q + W and state that W is positive if done on the system thereby increasing internal energy. In thermodynamics the first law is Delta U = Q - W and state that W is negative if done on the system thereby increasing internal energy. Net result is same as long as W is consistently defined in each area. – Bob D Oct 5 '18 at 22:05
• @bobD Whether W is positive or negative depends on whether W refers to the work done on the system by its surroundings or the work done on the surroundings by the system. It appears based on the answer below that they do in fact turn out to be the same PROVIDED the system eventually returns to equilibrium. – David Reed Oct 5 '18 at 23:30

If the piston is massless, then, by Newton's 3rd law, the force per unit area exerted by the gas on the inside face of the piston must be equal and opposite to the force per unit area exerted by the surroundings on the outside face of the piston, irrespective of whether the process is irreversible. Therefore, the work will be the same, but of opposite sign.

If the piston has mass, the force per unit area exerted by the gas on the inside face of the piston minus the force per unit area exerted by the surroundings on the outside face of the piston will equal the mass times acceleration of the piston. But, if the system eventually equilibrates with its surroundings (as a result of viscous stresses within the gas causing the dissipation of the kinetic energy of the piston), the net work done by the gas will, again be equal in magnitude and opposite in sign to the work done by the surroundings. This is true even if the piston contact with the cylinder is frictionless.

So, what is going on? Well, first of all, even for an ideal gas, the ideal gas law only applies when the gas is at thermodynamic equilibrium. In a reversible process, the gas passes through a continuous sequence of thermodynamic equilibrium states, so the ideal gas law always applies. But, in an irreversible process, the ideal gas is not at thermodynamic equilibrium, so the ideal gas law does not apply.

Then what does apply in an irreversible process? Well, gases (even ideal gases) are viscous fluids in which the state of stress is described by the equations for a Newtonian fluid. The Newtonian fluid equation predicts that the normal compressive stress (normal force per unit area) at the piston face is given, not by the ideal gas law, but by: $$\sigma=\frac{RT}{v}-2\mu\frac{\partial V}{\partial z}\tag{1}$$where v is the specific molar volume of the gas evaluated at the inside piston face, T is the temperature of the gas at the piston face, $$\mu$$ is the gas viscosity, V is the velocity of the gas (which varies with spatial position within the gas), and z is axial position along the cylinder. The first term on the right hand side of this equation is the pressure predicted by the ideal gas law (with all the parameters evaluated locally at the inside piston face), and the second term is the contribution of viscous stresses to the overall compressive stress at the piston face. Note that, for a reversible (quasi static, very slow deformation), the second term is negligible. Furthermore, for a reversible process, the specific volume, temperature, and pressure are uniform within the cylinder.

The net result of all this is that, for an irreversible process, the force per unit area that the gas exerts on the inside piston face is not determined solely by the volume of the gas but also by the rate of change of volume with time. This is the reason that the ideal gas law does not describe the force per unit area of the gas on the piston in an irreversible process. However, as indicated above, in the end, if the final state of the gas is again one of thermodynamic equilibrium, the work done by the gas on the piston will be equal in magnitude and opposite in sign to the work done by the surroundings on the piston (assuming the piston is oriented horizontally, so that its potential energy does not change).

For a horizontal cylinder, the instantaneous force balance on the piston is $$(P_g-P_o)A=m\frac{dv}{dt}$$where $$P_g$$ is the force per unit area that the gas exerts on the inside face of the piston (a function of time, referred to as $$\sigma$$ in Eqn. 1 above), $$P_0$$ is the force per unit area exerted externally on the outside face of the piston, A is the cross sectional area of the piston, m is the mass of the piston, and v is the velocity of the piston. If we multiply both sides of this equation by the velocity $$v=\frac{L}{dt}$$ and integrate with respect to time from t = 0 to t = t, we obtain: $$\int_{V(0)}^{V(t)}{P_gdV}=\int_{V(0)}^{V(t)}{P_odV}+m\frac{v^2(t)}{2}\tag{2}$$where V(t)=AL(t) is the gas volume at time t. The left hand side of this equation is $$W_g(t)$$, and represents the work done by the gas on the piston up to time t, and the integral on the right hand side represents $$W_o(t)$$, the work done by the outside face of the piston on the surroundings up to time t: $$W_g(t)=W_o(t)+m\frac{v^2(t)}{2}\tag{3}$$The piston may oscillate about its final equilibrium position as time progresses, but, eventually, as a result of the viscous damping stresses in the gas, the piston will come to rest (even if the piston is frictionless), and its kinetic energy will have dissipated. At this point, the gas will again be at equilibrium, and we will have $$W_g(\infty)=W_o(\infty)$$Thus, at final equilibrium, the work done by the gas on the inside face of the piston will exactly match the work done by the outside face of the piston on the surroundings.
For the case of a vertical piston, the analogous result is: $$W_g(\infty)=W_o(\infty)+\frac{mg}{A}(V(\infty)-V(0))$$In this case, if the piston is regarded as part of the surroundings, then the left hand side of this equation again represents the work done by the gas on its surroundings.
For the answer to your second question, if L(t) represents the distance of the piston from the dead end of the cylinder, the velocity of the piston is $$v(L)=\frac{dL}{dt}$$If the deformation of the gas is assumed to be homogeneous (a uniform function of x), we have: $$\frac{dv}{dx}=\frac{1}{L}\frac{dL}{dt}$$But, since the volume of gas is V=AL, we have that:$$\frac{dv}{dx}=\frac{1}{V}\frac{dV}{dt}\tag{4}$$ This establishes the connection between the viscous term in the expression for the compressive stress and the rate of change of volume.