Is Work Done by Gas or Gas and Spring Both?

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SUMMARY

The discussion centers on the application of the first law of thermodynamics to a system comprising a mono-atomic gas within a horizontal cylinder, equipped with a frictionless piston and a spring. The key conclusion is that when calculating work (W) in the equation W + delta(Q) = delta(E_int), it is essential to determine whether the focus is on the gas alone or the gas-spring system. If only the gas's internal energy change is considered, W represents the work done solely by the gas. Conversely, if the internal energy change of the gas-spring system is analyzed, W includes contributions from both the gas and the spring.

PREREQUISITES
  • Understanding of the first law of thermodynamics
  • Familiarity with mono-atomic gas behavior
  • Knowledge of work-energy theorem
  • Concept of spring constant (K) in mechanical systems
NEXT STEPS
  • Study the implications of the first law of thermodynamics in different thermodynamic systems
  • Explore the behavior of mono-atomic gases under various conditions
  • Learn about the work-energy theorem and its applications in thermodynamics
  • Investigate the role of springs in mechanical systems and their energy contributions
USEFUL FOR

This discussion is beneficial for physics students, thermodynamics enthusiasts, and engineers involved in mechanical systems analysis, particularly those interested in the interplay between gas behavior and mechanical components like springs.

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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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