# Why is the work done on the system not equal in frame of gas and surroundings?

Consider an isothermal irreversible process for an ideal gas in a cylindrical container closed with a piston (system). Initial and final states are $$(P_1,V_1,T)$$ and $$(P_2,V_2,T)$$ respectively. Initially the pressure of surroundings was $$P_1$$ and which is increased to $$P_2$$,

Work done by surroundings on the system is $$W_1=P_2(V_1-V_2)$$ which is in the frame of surroundings. If we see in the frame of gas, work done on system is $$W_2=\int -P_{\text{gas}} dv$$

Since, process is isothermal and $$P_{\text{gas}}V=nRT$$ $$W_2=nRT\ln{\frac{V_1}{V_2}}$$

I am unable to figure it out that why is $$W_1≠W_2$$ because according to the first law it should be same (change in internal energy and heat exchange is same in both the frames) . Can please someone help me out?

• You used the ideal gas law to get Pgas for the irreversible non-equilibrium process? Do you think that is allowed? Commented Oct 26, 2023 at 19:22
• @ChetMiller why not? Commented Oct 27, 2023 at 8:16
• Are you familiar with the dissipative property of fluids (liquids and gases) known as viscosity? Have you ever heard the term "viscous Newtonian fluid?" Commented Oct 27, 2023 at 10:18
• @ChetMiller Yes. Commented Oct 27, 2023 at 10:21
• If you are familiar with both of these, then are you familiar with the 3D mathematical form of the equations for the Cartesian components of stress tensor in terms of the thermodynamic pressure, the fluid viscosity, and the velocity gradients in the fluid? Commented Oct 27, 2023 at 10:49

This is for @Chesx

For non-polymeric fluids (liquids and gases), Newton's law of viscosity very accurately describes the state of stress within a fluid as a function of the local fluid deformation and local rate of deformation. It also reduces to the ideal gas law (or other thermodynamics equilibrium equation of state) in the limit of low rates of deformation (i.e., low velocity gradients).

In Cartesian coordinates, the components of the compressive stress tensor are related to the thermodynamic pressure p(v,T) and the fluid velocity gradients by:

$$\sigma_{xx}=p(v,T)-\frac{2}{3}\mu\left[2\frac{\partial u_x}{\partial x}-\frac{\partial u_y}{\partial y}-\frac{\partial u_z}{\partial z}\right]$$ $$\sigma_{yy}=p(v,T)-\frac{2}{3}\mu\left[-\frac{\partial u_x}{\partial x}+2\frac{\partial u_y}{\partial y}-\frac{\partial u_z}{\partial z}\right]$$ $$\sigma_{zz}=p(v,T)-\frac{2}{3}\mu\left[-\frac{\partial u_x}{\partial x}-\frac{\partial u_y}{\partial y}+2\frac{\partial u_z}{\partial z}\right]$$ $$\sigma_{xy}=\mu\left[\frac{\partial u_y}{\partial x}+\frac{\partial u_x}{\partial y}\right]$$ $$\sigma_{xz}=\mu\left[\frac{\partial u_z}{\partial x}+\frac{\partial u_x}{\partial z}\right]$$ $$\sigma_{yz}=\mu\left[\frac{\partial u_z}{\partial y}+\frac{\partial u_y}{\partial z}\right]$$where the u's are the components of gas velocity, v is specific volume, and $$\mu$$ is the gas viscosity.

In the limit of low gas velocities, these equations reduce to the isotropic thermodynamic equilibrium state values of the stress components:$$\sigma_{xx}=\sigma_{yy}=\sigma_{zz}=p(v,T)$$and$$\sigma_{xy}=\sigma_{xz}=\sigma_{yz}=0$$ For a fluid satisfies the ideal gas thermodynamic equation of state $$p=\frac{RT}{v}$$

We also see from all this that the isotropic thermodynamic pressure p does not capture the state of stress in a gas that is deforming at finite rate. In fact, what we call the pressure is not the force per unit area on an internal surface of arbitrary orientation, but rather it is merely the isotropic part of the overall stress tensor.