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enter image description here

This problem has been bugging me for a long time now.

It states that given an insulated cylinder which contains a spring and an ideal gas, you must calculate the work required to compress the gas, work required to compress the spring, work done by the atmospheric pressure $p\ $ ($p=1\ bar$) the work done by the block $m$, and the tension force at the end of the process. The mass of the piston in negligible, the spring has a spring constant $k=200\ N/cm$ the initial state of the gas is $p_1 = 0.8 \ bar, \vartheta_1=20°C$, the block is held up by a tension force $F$, the block is lowered onto the piston such that is causes the piston to descend by $0.2\ m$, the weight of the block is $10000\ N$, the ideal gas has an isentropic exponent $\kappa =1,37$

Now, my question is:

If I decide to set the bondary of my system around the spring and the gas, so the only objects I am studying are the sping and the gas, applying the first law of thermodynamics should give me this: $$Q_{12}=W_{12}+\Delta U + \Delta E_{ep}$$ where $\Delta E_{ep}$ is the change in the elastic potential energy of the spring. What's bugging me is what exactly is $W_{12}$?

  1. the work done by the sping due to the spring force and the deformation of the spring:$$W_{spring}=-\Delta E_{ep}=-\frac{k}{2}(\delta_2-\delta_1)$$ where $\delta_2$ and $\delta_1$ are the final and initial deformations of the spring
  2. the work done by gravity on the block:$$W_{gravity}=-\Delta E_{gp}=-mg(z_2-z_1)$$ where $\Delta E_{gp}$ is the change of the gravitational potential energy of the block, $z_2$ the final height and $z_1$ the initial height using the bottom of the cylinder as reference
  3. the work done by the atmosphere:$$W_{atmosphere}=pA(z_2-z_1)$$ where $A$ is the cross-sectional area of the piston

Drawing an FBD of the piston in the initial and final position gives: enter image description here

the problem:

enter image description here

the translation into english:

Inside of an insulated upright cylinder from the image there is an ideal gas ($\kappa = 1,37$) with an initial state $0,8 \ bar$ and $\vartheta =20°C$ and a spring with a linear characteristic (spring constant $k=200\ N/cm$). From the outside of the piston there is air at a pressure of $1\ bar$. In equilibrium state the piston is $50 \ cm$ away from the bottom of the cylinder. Using a crane, a weight of $10000\ N$ is being placed onto the piston, in doing so the piston descends by $20 \ cm$. What is the remaining force in the crane rope at the end of the process? What is the work required to compress the gas? what is the work required to compress the sping? How much work was done by the atmosphere, and how much weight by pushing the rod?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Jun 6 '20 at 15:48
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Your system is the spring and the gas.

When the first law is written in this form, $\Delta U= Q-W$, then $\Delta U$ is the change in internal energy of the system which includes the spring ($U_{\rm final} -U_{\rm initial})$, $Q$ is the heat input to the system (positive if into the system and negative if out of the system) and $W$ is the work done by the system, (positive if the work is done by the system and negative if the work is done on the system).

So you will have to decide about the sign(s) of the $W$(s).

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  • $\begingroup$ since the cylinder is insulated then $Q_{12}=0$, what would be the $W$(s) you speak of? $\endgroup$ – herosai Jun 6 '20 at 11:43
  • $\begingroup$ If you are including change in elastic potential energy of the spring in $\Delta U$ then you can't also include it in $W$ $\endgroup$ – BioPhysicist Jun 6 '20 at 12:31
  • $\begingroup$ why is that the case, that always bugged me! $\endgroup$ – herosai Jun 6 '20 at 12:56
  • $\begingroup$ If you included the spring in both the system and the surroundings then you would be double counting its effect.. $\endgroup$ – Farcher Jun 13 '20 at 8:52
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The force balance on the piston at any time during the compression is given by: $$P_gA+F-mg-kx-P_{atm}A=0$$ where x is the upward displacement of the spring from its unextended length. If we multiply this by the differential (upward) displacement of the piston during the process $dx=\frac{1}{A}dV$, we obtain:$$P_gdV+Fdx-mgdx-kdx-P_{atm}dV=0$$Integrating this equation between the initial and final locations of the piston yields $$\int{P_gdV}+\int_{x_i}^{x_f}{Fdx}+mg(x_i-x_f)+\frac{k}{2}(x_i^2-x_f^2)+P_{atm}(V_i-V_f)=0$$The first term represents the work that the gas does on the piston, the second term represents the work done by the force F on the piston, the third term represents the work done by the mass m on the piston, the fourth term represents the work done by the spring on the piston, and the fifth term represents the work done by the atmosphere on the piston.

The gas undergoes an adiabatic reversible compression, and the work done by this can be obtained separately by applying the formula for work in an adiabatic reversible compression. The work from the variable force F can then be obtained by difference using the final equation above.

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