
#1
Mar2613, 10:02 AM

P: 7

Most of the book using PV=nRT for PV^gamma constant for adiabatic process.
However, please refer attachment, proving using conservation of energy will also make sense? 



#2
Mar2613, 12:09 PM

P: 864

The derivation in your attachment seems to start half way through. Where does your first line come from?




#3
Mar2613, 08:06 PM

P: 7

Please refer attachment it seem using this method, it able to change instaneneous equation to continuois equation.
Equation A 5/2P1V1P2(dv)=5/2(P2)(V1+(dv)) will be transform to Equation B 10.4/V1=P1+2/5P2 However the "V1" in equation B will represent the value of "V1+dv" in equation A and dv is eliminated both equation A and B showing same charateristic. I consider adiabatic derivation using conservation of energy and the same way as above Intial internal enegy of gaswork done by gas=Final energy of gas the P2 dv is change to Pdv since the pressure being compress of expand is the instataneous gas pressure in cylinder 1/(γ1)PVP(dv)=5/2P2(V+(dv)) However, dv can be +dv or dv depend the volume expand or compress After tranform it will ge PV^γ 



#4
Mar2713, 11:34 PM

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P: 6,591

Adiabatic process derivation
Hi Chuakoktong. Welcome to PF.
I am having some difficulty following your derivation. Start with the ideal gas law PV=nRT, which means that for any change: Δ(PV) = nRΔT, if n is constant. In other words d(PV)/dT = nR This means: (VdP + PdV)/dT = nR Now we can subsitute R = CpCv and get (VdP + PdV)/dT = n(CpCv) = nCv(γ1) where γ = Cp/Cv Then you apply the first law and substitute _____ for the nCvdT term (hint: nCvdT = dU). Can you do that? AM 



#5
Mar2813, 04:20 AM

P: 7

For the derivation of pv gamma using PV=nRT, i do know how to get it. However, i would like to try other concept to get it as well.
Sorry for my poor explanation. At the moment, I would like to understand the below question to know whether i on the right track. A)the internal energy of gas in a cylinder with volume V and pressure P can be written as 1/(γ1)PV correct? B)when compress the gas by a small volume dv, the work done on gas is P*dv? C)The final internal energy is the sum of work done (P*dv)+initial internal energy(1/(γ1)PV) correct? 



#6
Mar2813, 10:00 AM

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P: 6,591

The internal energy of an ideal gas is: U = nCvT = n(CpR)T = nR(Cp/R1)T = nR(Cp/Cv)(Cv/R) 1)T = nR(γ(Cv/(CpCv)1)T= nR(γ(1/(γ1))1)T = 1/(γ1)nRT Since PV = nRT, U = 1/(γ1)PV So: U = Ui + ΔU = Ui + W = Ui  ∫PdV AM 



#7
Mar2813, 08:42 PM

P: 7

Dear Mason,
Thank you for spending time for my post and answering my doubt. After checking, the work done is always defined by ∫Pdv or work done=P(VfVi) Vf=final volume and Vi=initial volume Thank for pointing out my mistake Consider a diatomic gas with pressure 1pascal undergo adiabatic compression by 0.1m^3 Using P1V1^γ=P2V2^γ (1)(1)^1.4=P2(0.9)^1.4 P2=1.1589pascal This is what cause of my confusion I been thinking may be is possible to get the same answer as above using the formula as mention below I obtain it by using approxumation method letting dv=0.05 and i believe using a small dv ie.0.01 could obtain a nearer value. Please compare the calculation using equation 1 and 2 5/2P1V1+Pdv=5/2P2V2 eq1 5/2P1V1Pdv=5/2P2V2 eq2 since dv is 0.05^m3 gas is compress initially by 0.05m^3 and undergo another compression by 0.05m^3 The formula below using eq 2 5/2(1)(1)(1)(0.05)=5/2(p2)(0.95) P2=98/95pascal 5/2(98/95)(0.95)(98/95)(0.05)=5/2(p3)(0.9) P3=1.0659pascal If using eq 1, 5/2(1)(1)+(1)(0.05)=5/2(p2)(0.95) P2=102/95 5/2(102/95)(0.95)+(102/95)(0.05)=5/2(p3)(0.9) P3=1.1571pascal Compare value obtain from eq 1 and eq2, eq1 is closer to value of pressure obtain using PV^γ approximation method always accompanied by error when comparing to the integral form. However, using a formula does not match the original equation result in a bigger error? At the moment i havent try using smaller dv Also in adiabatic free expansion, most of it mention P1V1/T1=P2V2/T2 and T1=T2 lead to P1V1=P2V2 I am wondering also consider the case of there is no change in the internal energy of gas in adiabatic free expansion 5/2P1V1=5/2P2V2 which lead to P1V1=P2V2 also I been thinking by using conservation of energy it should be able to hold a formula to predict any process but still the Pdv sign and +pdv sign is confusing me At the moment i will try to use a smaller dv to check the result for eq 1. 



#8
Mar2913, 12:49 AM

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P: 6,591

As far as the PdV and PdV, all you have to remember is that the work done BY the gas is always ∫PdV. When a gas expands dV is positive so work done by the gas = PdV is positive. AM 



#9
Mar2913, 01:35 AM

P: 7

Dear Mason,
Thank for your fast reply. The reason i advoid using calculus is because i trying to break down thing part by part for my further understanding. I been thinking as far as things are interrelated, maybe same result/answer can also be obtained by using different way(i.e taking different path leading the same destination) I am trying by means of calculus can the equation as below 5/2P1V1+Pdv=5/2P2V2 or 5/2P1V1Pdv=5/2P2V2 integrated become P1V1^γ 



#10
Mar2913, 02:22 PM

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P: 6,591

your equation becomes: (1/(γ1))P1V1  P1dV = (1/(γ1))(P1+dP)(V1+dV) which is equivalent to: P1dV = (1/(γ1))(P1dV + V1dP) And all that says is that W = ΔU (ie. it is adiabatic). AM 



#11
Mar2913, 06:05 PM

P: 7

Yes, from there i get the point.
Thank alot for all the work and explanation you show to me. I really appreciate it a lot. 


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