Recent content by jrklx250s

Ah wonderful. Thanks for your help here now I get it. So the INSULATOR actually increases its entropy which is done by its temperature change which is T(t) = 300K + (10A^2*25Ω)/(8.36 J/K) = 599K Now the Entropy change of the resistor inside the insulator is ΔS = ∫CdT/T = 8.36*ln(599K/300K) =...

Yea exactly, that's why I'm confused...being thermally insulated Q=0 and being held at a constant temperature it is isothermal. Therefore this is adiabatic and isothermal so...if the ΔS=ΔQ/T and there is no heat change well then there is no entropy change? Clearly I'm wrong though because the...

Homework Statement A current of 10A is maintained for 1s in a resistor of 25Ω while the temperature of the resistor is kept constant at 27°C. This resistor is thermally insulated with a mass of 10g. If it has a specific heat of 836 J/kg*K. What is the entropy change of the resistor? Universe...

Haha wow...thank you serena I was making this so much more complicated than it was. Yea of course you can just conclude that since P1=P2 V2=V3 so therefore... P1V1=nRT1 P3V3=nRT3 P2V1=nRT1 P3V2=nRT3 solving for both T's T1=P2V1/nR T3=P3V2/nR sub this in my previous...

Hi Serena, Yes I believe so when i calculated the adiabatic processes for the power stroke... which i concluded that they were T1V1^(γ-1) = T3V2^(γ-1) T1P2^((1-γ)/y)=T3P3^((1-γ)/y) And Since I need to make T1/T3 = (V1/V2)/(P3/P2) This means that T1 = (V1*P2) and T3 = (V2*P3)...

Since no one has replied I'm assuming some are confused as to what I'm talking about so here is the ideal gas cycle that I need to calculate the thermal efficiency from. Here is the link to the picture of the cycle http://imageshack.us/photo/my-images/411/img1048u.jpg/

Homework Statement Given an imaginary ideal-gas cycle. Assuming constant heat capacities, show that the thermal efficiency is η = 1 - γ[((V1/V2)-1)/((P3/P2)-1)] Since i can't show you the cycle we are shown that l Qh l = which is absolute value of the heat at high temperature =...

Well since density is number of moles over volume σ = n/v and V = nRT/P σ= P/RT but I am unsure how this helps me because now I have the density at x=0 which is P/RTo and the density at x=L which is P/RTL

Indeed, PV=nRT But my main question is how to derive T = To + [(TL - To)/L]x To eventually make it into T = [(TL-To)/ln(TL/To)] I'm lost as to where to even start in this derivation

No I didnt show my formulas because I didn't think I was even on the right track for this question. I mean I know you must use PV=nRT, PiVi^γ=PfVf^γ, and P1V1/T1 = P2V2/T2 and for the last one Q = ΔU - W...im just unsure if these are appropriate for each question

Thanks for your reply, so I'm a little confused, the density is changing therefore we can write σ(x) = Pdx ∫σ(x) = P∫dx ∫σ(x) = P(xL-x0) But how does this help with the temperature equation?

Homework Statement A horizontal, insulated cylinder contains a frictionless non conducting piston. On each side of the piston is 54L of and inert monatomic ideal gas at 1 atm and 273K. Heat is slowly applied to the gas on the left side until the piston has compressed the gas on the right to...

Homework Statement Show that the heat transferred during an infinitesimal quasi static process of an ideal gas can be written as dQ = (Cv/nR)(VdP) + (Cp/nR)(PdV) where dQ = change in heat Cv= heat capacity while volume is constant n= number of moles of gas R= ideal gas constant Cp=...

Homework Statement Prove that the work done by an ideal gas with constant heat capacities during a quasi-static adiabatic expansion is equal to W= (PfVf)/(Y-1)[1 - (Pi/Pf)^((Y-1)/Y)] where Y = gamma, which is heat capacity at constant pressure over heat capacity at constant volume...

Homework Statement Given the temperature of an ideal gas in a tub where it varies linearly from (x=0) to (x=L) T = To + [(TL - To)/L]x where To is temperature at 0 length and TL is the temperature at the end of the tube We are suppose to show that the equation of state for this...