In thermo, i cannot figure out why the work done is not the same. I try to simplify my problem here. Considering again a cylinder with insulated wall is expand adiabatically and piston with surface area of 1m^2 . The cylinder have a volume of 1m^3 and massless piston. The piston is initially at the middle of the partition so volume= length of the end to the piston*surface area of piston=0.5*1=0.5m^3 The pressure inside the cylinder is 4atm. On the other side of the partition consist of a massless spring. I show my calculation in several step so someone can point out where is the mistake

Step 1

When the partition expand from it original position by distance x,(x<0.5) than the volume will be (0.5+x)A. So using P1V1^gamma=P2V2^gamma i can formulate it as 4(0.5A)^gamma=P2((0.5+x)A)^gamma. When it move a distance x than the spring will be compress by x disance

Step 2

The elastic energy of the spring will be 0.5kx^2. You can try giving any value of K as long as it make sense and will find that the work done by the gas is not equal. P1V1-P2V2/(gamma-1) is not equal to 0.5kx^2(using the formula above).

I been figuring since the internal energy of the gas is nCvdt, why cannot use 3/2RTi=1/2kx^2+3/2RTf where Ti is initial temperature and Tf is final temperature or 3/2P1V1=1/2kx^2+3/2P2V2. But just need to find the point when the force exert on the piston become equal to force exert by the spring. This also agree with in an adiabatic free expansion, since no work is done, it can be treated as an isothermal process.

I really can't figure out why it is not conserve. You can try another example considering an infinite long cylinder open end cylinder with insulated wall is expand until the pressure inside is same as the pressure outside. Threating that the pressure outside is atmospheric pressure and it is almost constant.

Let's break down your question about the work done during an adiabatic expansion in a cylinder with a piston and a spring. The confusion often arises from the interplay between the work done by the gas and the energy stored in the spring. To clarify this, we need to look at the principles of thermodynamics and how they apply to your scenario.

Understanding the System

In your setup, you have a cylinder with an insulated wall, a piston, and a spring on the other side. The gas inside the cylinder is expanding adiabatically, meaning no heat is exchanged with the surroundings. The initial conditions are:

• Initial volume, V1 = 0.5 m³
• Initial pressure, P1 = 4 atm
• Piston area, A = 1 m²
• Spring constant, k (arbitrary value)

Step-by-Step Analysis

Now, let's analyze the steps you've taken and identify where the misunderstanding may lie.

Step 1: Work Done by the Gas

When the piston moves a distance x, the new volume becomes (0.5 + x)A. Using the adiabatic condition, you correctly set up the equation:

P1V1^γ = P2V2^γ

However, remember that during this process, the pressure inside the cylinder will change as the volume changes. The work done by the gas during this expansion can be expressed as:

W = ∫ P dV

As the piston moves, the pressure is not constant, which complicates the calculation. You need to integrate the pressure over the changing volume to find the total work done.

Step 2: Energy in the Spring

The elastic potential energy stored in the spring when compressed by distance x is given by:

U_spring = 0.5kx²

At equilibrium, the force exerted by the gas on the piston must equal the force exerted by the spring:

P2A = kx

From this relationship, you can express x in terms of P2 and k:

x = (P2A)/k

Energy Conservation Considerations

Now, regarding your question about energy conservation, the first law of thermodynamics states:

ΔU = Q - W

In your adiabatic process, Q = 0, so:

ΔU = -W

This means that the change in internal energy of the gas is equal to the negative of the work done by the gas. The internal energy change can also be expressed in terms of temperature:

ΔU = nC_vΔT

Connecting the Dots

When you try to equate the work done by the gas (using the formula P1V1 - P2V2)/(γ - 1) with the spring's potential energy, you are overlooking the fact that the gas does work not only against the spring but also in changing its own internal energy. The work done by the gas is not solely converted into spring energy; some of it contributes to changing the internal energy of the gas as well.

Final Thoughts

In summary, the discrepancy arises because the work done by the gas is not just about the energy transferred to the spring. It also involves changes in the gas's internal energy due to the temperature change during the adiabatic process. In an adiabatic free expansion, no work is done, and thus the internal energy remains constant, which is why it can be treated as isothermal in that specific case.

To further clarify, consider your infinite cylinder example. As the gas expands to match the external pressure, it does work against that pressure, and the energy balance must account for both the work done and the internal energy changes. Always remember that in thermodynamic processes, energy conservation is key, and it often involves multiple forms of energy interacting with one another.