WD on the gas and by the gas

[i_island0]

A horizontal cylinder is fit with a frictionless and massless piston. Inside the cylinder there is a mono-atomic gas. Outside pressure is Po. The piston is connected with a spring (of spring constant K) the other end of which is connected with the walls of the cylinder. (the spring lies withing the cylinder).
My question is when we use first law of thermodynamics, W + delta(Q) = delta(E_int).

then shall we calculate the 'W' as work done by the gas, or work done by the gas and spring both. Please explain the reason as well.

To do this problem, I am using the equation, that I learned in work energy theorem; W + delta(Q) = delta(K) + delta(U) + delta(E_int).
Here, i am using delta(K) = 0 as i am not considering velocity of COM of the system.

 

[Hootenanny]

The first law of thermodynamics is simply a statement of the principle of conservation of energy. See http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/firlaw.html#c1" for more information. So if you are calculating the change in internal energy of the gas, then dW should be the work done by the gas, but if you are considering the change in internal energy of the gas-spring system then dW would be the work done by both the spring and the gas.

 

[charng]

I would like to clarify that the first law of thermodynamics is a fundamental principle that states that energy cannot be created or destroyed, only transferred or converted from one form to another. In the context of your question, this means that the total change in energy (delta(E)) of the system must be equal to the sum of the work done (W) and the heat transferred (delta(Q)).

In this particular scenario, the work done by the gas is the only relevant factor, as the spring and the walls of the cylinder are not external forces acting on the gas. Therefore, the work done (W) in this case would be the work done by the gas. The reason for this is that the spring is internal to the system and does not contribute to the transfer of energy from the gas to the surroundings.

Additionally, the equation you have mentioned from the work-energy theorem, W + delta(Q) = delta(K) + delta(U) + delta(E_int), is not applicable in this scenario. This equation is used for systems where there is a change in kinetic energy (delta(K)) and potential energy (delta(U)). In the case of a gas, there is no change in kinetic energy and the potential energy is negligible. Therefore, the equation simplifies to W + delta(Q) = delta(E_int).

In conclusion, when using the first law of thermodynamics to calculate the work done by a gas, only the work done by the gas itself should be considered. The spring and the walls of the cylinder do not contribute to the work done by the gas and should not be included in the calculation.