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WD on the gas and by the gas

  1. Jan 26, 2008 #1
    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.
    Last edited: Jan 26, 2008
  2. jcsd
  3. Jan 26, 2008 #2


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    The first law of thermodynamics is simply a statement of the principle of conservation of energy. See hyperphysics 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.
    Last edited: Jan 26, 2008
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