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OK. I'm going to try to do it using ehild's condition, rather than vacuum on the right hand side and see where that takes us. The force balance on the piston in this situation during the change is:$$P=P_1+\frac{k}{A}x$$where x is the displacement of the piston to the right (equal to the absolute change in length relative to the unextended state). In this situation, x is related to the volume change of the right hand chamber by:$$x=\frac{V-V_1}{A}$$ So if we combine these two equations, we get:$$P=P_1+\frac{k}{A^2}(V-V_1)$$ This equation applies at (V2, P2). So, here again, we get the exact same relationship between pressure and volume that we obtained with the assumption of vacuum on the right side of the piston:$$P=P_1+\frac{(P_2-P_1)}{(V_2-V_1)}(V-V_1)$$So the question is, why did ehild get different results from us for case A if we have the same P-V relation applying.
The we showed that the work done on the spring is: $$W=\frac{(P_2+P_1)}{2}(V_2-V_1)$$ This is also the energy stored in the spring. From the ideal gas law,$$P_2=\frac{V_1}{V_2}\frac{T_2}{T_1}P_1$$
In case A, ##P_2=1.5P_1##, so ##W=\frac{5}{4}P_1V_1##. This is not what ehild got. We need to resolve this.
See next post for resolution.
Chet
The we showed that the work done on the spring is: $$W=\frac{(P_2+P_1)}{2}(V_2-V_1)$$ This is also the energy stored in the spring. From the ideal gas law,$$P_2=\frac{V_1}{V_2}\frac{T_2}{T_1}P_1$$
In case A, ##P_2=1.5P_1##, so ##W=\frac{5}{4}P_1V_1##. This is not what ehild got. We need to resolve this.
See next post for resolution.
Chet
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