Constant pressure process (piston) on gas violates Newtons 2nd law?

[questionmonkey123]

I often see this set up in thermodynamic problems and need clarification on how Newton's Laws are involved for the piston:

Gas inside a piston cylinder (1) is heated expanding the gas and raising the piston (initially at rest) to a height (2) in a constant pressure quasi-equilibrium process. At (2) the piston is also at rest, thus the FBD at (1) and (2) are the same.

My understanding: In order for piston to travel to any height from initial rest, there is acceleration involved thus a net force. The only forces on the piston are the ones shown (for any point in the process), and the gas pressure is constant throughout the process so there is never a net force to move the piston up from rest.
How is this type of problem justified in terms of Newton's laws? What assumptions am I overlooking?

 

[anuttarasammyak]

When you think of NOT force BUT work or energy, gas inside the cylinder did work of $p_1 Ah$.
Constant pressure quasi-equilibrium process is not constant pressure equilibrium.

 

[questionmonkey123]

When you think of NOT force BUT work or energy, gas inside the cylinder did work of $p_1 Ah$.

But how does this clarify acceleration with no net force? I already know how to calculate work.

 

[anuttarasammyak]

When you heat gas its pressure goes up slightly $\triangle p$ which make force imbalance and the piston moves up. By gas inflation its pressure goes down back to $p_1$ and the piston oscillates and ceases oscillating in a long time transferring all kinetic energy to potential energy and stops at $\triangle h$ up for there is no loss of energy or heat generation for a reversible process. Such an infinitesimal process is repeated infinite times for the final piston height.

 

[jack action]

At state 1:
$$\left(P_1 - P_0\right)A = mg$$
After heating, the pressure increases by $\Delta P$:
$$\left(\left(P_1 + \Delta P\right) - P_0\right)A = m(g + a)$$
And the acceleration $a$ depends on $\Delta P$.

But as soon as the piston moves, the pressure decreases because the volume increases. Eventually, $\left(P_1 + \Delta P\right)$ will go back to $P_1$ and $a$ will go back to zero. The only difference is that the volume and temperature have increased.

 

[Lnewqban]

What happens in cases of abrupt increase of internal pressure?

 

[Drakkith]

The only forces on the piston are the ones shown (for any point in the process), and the gas pressure is constant throughout the process so there is never a net force to move the piston up from rest.
How is this type of problem justified in terms of Newton's laws? What assumptions am I overlooking?

Assuming a frictionless system, the piston still has inertia and resists moving when the gas is first heated. Thus the gas is initially at a higher pressure than it will be later, as jack action explained in post #5.

 

[boneh3ad]

The simple answer here is that the given example is a simplified abstraction of what would actually occur in such a situation. Of course in the real world you would need some initial (and final) acceleration and this whole process is transient, but the purposes of problems like this are perfectly well-served by neglecting those details.