Is Work Done by Gas or Gas and Spring Both?

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In the scenario of a horizontal cylinder with a frictionless piston containing a mono-atomic gas, the application of the first law of thermodynamics requires careful consideration of work (W). When calculating W, it is essential to determine whether the focus is on the gas alone or the combined gas-spring system. If only the gas's internal energy change is considered, W should reflect the work done by the gas. Conversely, if the internal energy change of the entire gas-spring system is analyzed, W must include contributions from both the gas and the spring. This distinction is crucial for accurately applying the first law of thermodynamics.
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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.
 
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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.
 
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