Recent content by Chestermiller

Yes, neglecting dissipation factors such as air drag.

A quasi-static path is considered to represent a continuous sequence of thermodynamic equilibrium states.

Who says that in thermodynamics, we study gases that are in thermodynamics equilibrium? The states of the gas in irreversible processes do not have to be equilibrium states; only the initial and final status have to the thermodynamic equilibrium states. There are 2 thermodynamic equilibrium...

Was it also called the Missouri School of Mines, or is that something different?

For large deformations, the definition of the strain tensor is not unique. There are many tensorial forms at large deformation that reduce to the same infinitesimal strain tensor in the limit of small deformations. For example $$\boldsymbol \epsilon=\frac{1}{2}(\bf G-\bf I)$$where G is the...

Thermal chemical decomposition of the rubber.

The more general form of the 1st law of thermodynamics is $$\Delta U+\Delta (PE)+\Delta (KE)=Q-W$$where PE is the gravitational potential energy and KE is the non-random kinetic energy.

At any point during the process, the pressure P is related to the mass m by $$PA=mg$$or $$m=\frac{PA}{g}$$Therefore, mass can be used as a surrogate for pressure. For an isothermal path, $$mV=m_1V_1=M_2V_2$$ For an adiabatic reversible path, $$mV^{\gamma}=m_1V_1^{\gamma}=m_2V_2^{\gamma}$$...

The total work is path independent, but the path for adiabatic reversible is different from the path for isothermal reversible, and the final states of the gas differ for the same change in hight. In other words, the lmass variation vs z is not the only measure of the path.

Initial pressure = ##P_0A=m_0g##, where ##m_0## is the finite amount of mass that must be added to the piston to hold it in equilibrium when the first stop is removed. We then have $$\delta P_n=g(\delta m_n)$$ and $$P_nA=g(m_0+\Sigma {\delta m_n})$$The work done by the piston on the gas is...

If your solution satisfies the differential equation, then it must be correct. (assuming y is the only other independent variable involved).

Here is an example. If you initially have 1 mole of H2 and 1 mole of I2, and you let the reaction to to equilibrium, forming x moles of HI, then the final amounts of H2 and I2 will be 1-x/2 and the final amount of HI will be x.

Are you saying that there is no effect of the fluid velocities at the two points?

There is no cycle involved. You simply heat the air and gases to cause them to expand and make the propeller above to rotate.

I assume that the heat flow is in the vertical direction and the bottom is hotter than the top, so we do not need to include natural convection. Is this correct?