Recent content by Pendleton

Woaaaahhh... yeah, I remember that from my other classes! It never occurred to me how common that property is, but now that you point it out, being able to transform the operator inside a squared binomial is downright amazing. So many problems become easy if you can do that. Thanks so much again!

(Record of my thoughts as I did this problem...) Really? I wish you'd told me more. Ugh, fine, I'll try, but I doubt it'll work... $$R = \left({1-\beta \over 1+\beta}\right)^2, \quad\quad T = {\epsilon_2n_1 \over \epsilon_1n_2}\left({2 \over 1+\beta}\right)^2$$ $$\therefore R+T=...

Consider polarized light crossing a sharp boundary between two volumes, each of a different but uniform refraction index ##n_1## or ##n_2##. Prove that the sum of the transmission and reflection coefficients of this light ##R+T=1##, where $$R \equiv {I_R \over I_I} = \left( {E_{0_R} \over...

Attempt at A Solution Problem 1 Reversible Process - A cylinder of ideal gas at pressure P is in mechanical equilibrium with a piston of area A driven by a spring of force F = PA and thermal equilibrium with a reservoir of temperature T. The piston is moved a small distance dx toward the...

Sorry for taking so long to reply! You are right—the chambers are in thermal contact with each other. Therefore, here is my analysis. For the system, $$dU_S = Q_S + W_S$$ $$dU_S = 0, ~ dW_S = 0$$ $$0 = dQ_S + 0$$ $$dQ_S = 0$$ For the reservoir, $$dQ_R = -dQ_S$$ $$dS_R = \frac{dQ_R}{T}$$...

The partition is unclamped, allowed to move to the right by the force of the higher-pressure gas against it, reclamped, and then allowed to equilibrate. Therefore, if F1 and F2 are the forces of the gases, $$F_{NET} = F_1 - F_2 = ma$$ where a is the rightward acceleration of the partition...

Yes, and here ##\delta## is small. Oh, ok! It is very inconvenient to type out By the above formula, $$\Delta S_1 = \frac{P_1 A\delta}{T}$$ $$\Delta S_2 = \frac{P_2 A (- \delta)}{T} = - \frac{P_2 A \delta}{T}$$ By the fundamental relation, $$\Delta S_R = \frac{dQ_R}{T}$$ The heat absorbed...

Yet this seems to contradict our previous agreement that the change of entropy of an ideal gas isothermally compressed a small distance by a piston of area A was $$dS = \frac{-PA}{T} δ$$ This problem, despite describing an infinitesimal change, mercifully does not require a path integral over...

Woaaaaah—viscous stresses in an ideal gas? I thought their particles never interacted. Tell me more. My lecture notes state that because T, S, and V are all state variables, ##dU = TdS - PdV## is true for all processes, reversible and irreversible.

That is not true because the pressures are different. The expanding gas in the left chamber, where the pressure is higher, does more work on the gas it is compressing than the latter does on the former.

First point, yes. I see now. The formula was right. Second point, I think you had it backward. Also, I edited in a question about why we don’t just say that $$dS = \frac{PdV}{T}$$ That means the overall process is both adiabatic and isothermal. How is this possible?

Hold on. The entropy change formula $$dS = nRln{\frac{V_f}{V_o}}$$ implies that $$dS_1 = nRln{\frac{V+Aδ}{V}}$$ $$= \frac{P_1}{T}ln(1+\frac{Aδ}{V})$$$$dS_2 = nRln{\frac{V-Aδ}{V}}$$ $$= \frac{P_2}{T}ln(1-\frac{Aδ}{V})$$ because, just to clarify, the volume of chamber 1, which is the one on...

The volume of chamber 1 increases by Aδ and volume of chamber 2 decreases by Aδ. The temperature of both chambers is T, the same as the initial temperature. Pressure changes only with the second order of δ.

My goodness, it’s you! Thanks for being here. The final state of the system is that the piston has been reclamped, having been moved δ rightward, and the system returned to equilibrium.

Attempted Solution: Gas Entropy This system is isothermal: the energy of each gas remains constant. $$dU = 0$$ By the combined statement of the first and second laws, $$dU = TdS - PdV$$ Therefore, $$0 = TdS - PdV$$ $$dS = \frac {PdV}{T}$$ Therefore, $$dS_1 = \frac {P_1 dV_1}{T} = \frac {P_1...