Number of moles necessary to get piston back to initial position

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SUMMARY

The discussion focuses on calculating the number of moles required to return a piston to its initial position in a thermodynamic system. Key equations include the ideal gas law and entropy change calculations, specifically using the formulas for temperature, pressure, and volume relationships. The final result indicates that 3 moles of gas are necessary to achieve this state, derived from the equations involving initial and final conditions of the gas. The work done by the gas on the spring is also calculated, confirming the energy transfer involved in the process.

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lorenz0
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Homework Statement
In a cylindrical piston of section S a piston can move without friction. A mole of ideal monoatomic gas is confined to the left, the right chamber is empty and contains a spring with elastic constant k which would be at rest if the piston were all the way to the left.
Initially the gas is at rest in an "A" state (##p_A = 10^5 Pa,
V_A = 25 dm^3##). It takes place through an exchange of heat with the outside
a slow transformation until reaching a "B" state with volume
halved. Calculate :
a) the heat ##Q## exchanged by the gas with the external environment to complete the transformation,
b) the entropy change ##\Delta S_G## of the gas,
c) how many moles ##n_{agg}## of gas must be added to bring the piston back to the initial position assuming that the temperature remains that of state "B"?
Relevant Equations
##F_{spring}=-kx, \Delta U=Q-L, W=\int_{a}^{b} \vec{F}\cdot\vec{dx}, \Delta S=nC_V\ln(\frac{T_f}{T_i})+nR\ln(\frac{V_f}{V_i}), PV=nRT##
a) ##T_A=\frac{p_AV_A}{nR}=300.7K, P_A V_A=kL^2=nRT_A##, ##P_B S=k\frac{L}{2}\Rightarrow P_B V_B=k(\frac{L}{2})^2 \Rightarrow P_B=\frac{kL^2}{2V_A}=\frac{P_AV_A}{2V_A}=\frac{P_A}{2}##, ##W_{spring\to gas}=\int_{L}^{L/2}kxdx=-\frac{3}{8}kL^2=-\frac{3}{8}nRT_A####\Rightarrow Q=L+\Delta U=-\frac{3}{8}nRT_A+n\cdot\frac{3}{2}R (\frac{T_A}{4}-T_A)\simeq 3750 J##

b) ##\Delta S_{gas}=nC_V\ln(\frac{T_f}{T_i})+nR\ln(\frac{V_f}{V_i})=nC_V\ln(\frac{T_A /4}{T_A})+nR\ln(\frac{V_A/2}{V_A})=-4nR\ln(2)=-23 J/K##

EDIT: I think I have managed to solve also part c)

c) ##T_B=\frac{P_BV_B}{nR}=\frac{1}{nR}\cdot\frac{kL^2}{2V_A}\cdot\frac{V_A}{2}=\frac{1}{4}\frac{kL^2}{nR}=\frac{1}{4}\cdot\frac{P_AV_A}{nR}=\frac{1}{4}T_A## so ##n_f=\frac{P_AV_A}{RT_B}=\frac{10^5\cdot 25\cdot 10^{-3}}{8.314\cdot\frac{300.7}{4}}mol=4mol## so ##\Delta n=n_f-n_i=(4-1)mol=3mol##
 

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In part (a), your W is the work done by the gas on the springbecause ##Q=W+\Delta U##. So, for Q, I get $$Q=-1.5RT_A=-3750\ J$$

The rest looks good. Very nice job.
 
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