Pressure-Volume Work

[Kakashi]

Consider a gas contained in a rigid cylinder with a frictionless, weightless piston. The cylinder is in thermal contact with a thermostat at fixed temperature T. The space above the piston is evacuated. Initially, the piston is held by stops and the gas occupies a volume V1.

If the gas is compressed in a quasi-static matter do we envision the process as follows:
When the stops are removed, small masses are added infinitesmally on top of the piston so that it descends quasi-statically and compresses the gas to a final volume V2. At each step the piston is momentarily at rest and the gas pressure equals the external pressure.

At mechanical equilibrium, the force balance on the piston is

$$ \overrightarrow{F}=PA\hat{n}=mg\hat{n}$$

Where n is the unit vector perpendicular to the surface

$$ \int \overrightarrow{F} \cdot \overrightarrow{dz}=\int PA\hat{n} \cdot \hat{n} dz=\int PdV=\int m(z)gdz $$

Here, however, the mass m is not constant it changes as additional infinitesimal masses are added to maintain quasi-static equilibrium during compression and mass is the cause for the change in pressure.

I am confused because force of gravity is usually a conservative force field, implying that the work it does should be path independent.

 

[anuttarasammyak]

I am confused because force of gravity is usually a conservative force field, implying that the work it does should be path independent

What path dependency do you find in this case? I am not sure of it.

 

[sophiecentaur]

I am confused because force of gravity is usually a conservative force field, implying that the work it does should be path independent.

The change in gravitational potential would depend on the change of position of the CM of the body of gas. I think that would take care of your concern. If your integrals applied for a vertically symmetrical system (two pistons - one at the top and one at the bottom) would not conflict.

 

[Chestermiller]

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 $$W=-\int{PdV}=-\int{PAdz}=-g\int{m(z)dz}$$

 

[Kakashi]

What path dependency do you find in this case? I am not sure of it.

Appreciate everyones help.I didnt show path dependency. My question comes from recalling multivariable calculus, where the gravitational force is a conservative vector field and therefore its work is path independent. In the quasi-static piston setup, the external pressure is generated by gravity acting on the added mass, and the integral I wrot involves this gravitational force acting through a displacement of the piston.

I was trying to understand why the pressure–volume work is path dependent in thermodynamics. I now see that because mechanical equilibrium Pgas=Pext, and Pgas is a state-dependent quantity rather than a position-dependent force field. Different processes connecting the same states (isothermal, adiabatic, etc.) will lead to different p(V) and therefore different values of work.

 

[anuttarasammyak]

I was trying to understand why the pressure–volume work is path dependent in thermodynamics. I now see that because mechanical equilibrium Pgas=Pext, and Pgas is a state-dependent quantity rather than a position-dependent force field. Different processes connecting the same states (isothermal, adiabatic, etc.) will lead to different p(V) and therefore different values of work.

The first law of thermodynamics : dU=dQ-dW
Energy can be supplied not only as work dW, which may originate from conservative forces, but also as heat dQ which depends on procedure.

State equation of gas : f(p,V,T)=0
$f(p_1,V,T_1)=f(p_2,V,T_2)=0$
Same volume but different pressure means different temperature.

 

[Chestermiller]

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.

 

[Chestermiller]

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, $$mV^{\gamma}=m_1V_1^{\gamma}=m_2V_2^{\gamma}$$
Therefore, the final states for these two different paths must be different.