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Adiabatic process derivation

by chuakoktong
Tags: adiabatic, derivation, process
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chuakoktong
#1
Mar26-13, 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?
Attached Thumbnails
pv gamma.jpg  
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Philip Wood
#2
Mar26-13, 12:09 PM
PF Gold
P: 945
The derivation in your attachment seems to start half way through. Where does your first line come from?
chuakoktong
#3
Mar26-13, 08:06 PM
P: 7
Please refer attachment it seem using this method, it able to change instaneneous equation to continuois equation.
Equation A
5/2P1V1-P2(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 gas-work 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/(γ-1)PV-P(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
Attached Thumbnails
integration.jpg  

Andrew Mason
#4
Mar27-13, 11:34 PM
Sci Advisor
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P: 6,654
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 = Cp-Cv and get (VdP + PdV)/dT = n(Cp-Cv) = nCv(γ-1) where γ = Cp/Cv

Then you apply the first law and substitute _____ for the nCvdT term (hint: nCvdT = dU). Can you do that?

AM
chuakoktong
#5
Mar28-13, 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/(γ-1)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/(γ-1)PV) correct?
Andrew Mason
#6
Mar28-13, 10:00 AM
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P: 6,654
Quote Quote by chuakoktong View Post
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/(γ-1)PV correct?
Correct for an ideal gas. This is just algebra ie:

The internal energy of an ideal gas is:

U = nCvT = n(Cp-R)T = nR(Cp/R-1)T = nR(Cp/Cv)(Cv/R)- 1)T = nR(γ(Cv/(Cp-Cv)-1)T= nR(γ(1/(γ-1))-1)T = 1/(γ-1)nRT

Since PV = nRT, U = 1/(γ-1)PV

B)when compress the gas by a small volume dv, the work done on gas is P*dv?
Not quite. The work done ON the gas is -PdV. The work done BY the gas is PdV.

C)The final internal energy is the sum of work done (P*dv)+initial internal energy(1/(γ-1)PV) correct?
Not quite. PdV is the work done BY the gas. The final internal energy is the sum of the work done ON the gas (-PdV) and the initial internal energy.

So:

U = Ui + ΔU = Ui + W = Ui - ∫PdV


AM
chuakoktong
#7
Mar28-13, 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(Vf-Vi)
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 eq-1
5/2P1V1-Pdv=5/2P2V2 eq-2

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.
Andrew Mason
#8
Mar29-13, 12:49 AM
Sci Advisor
HW Helper
P: 6,654
Quote Quote by chuakoktong View Post
Dear Mason,
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.
...

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
You appear to be trying to avoid using calculus by approximating. To get an exact answer you have to use calculus.

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.
You have to use the formula. I am not sure why you want do use an approximation. Just use PV^γ = constant.

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
chuakoktong
#9
Mar29-13, 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
5/2P1V1-Pdv=5/2P2V2

integrated become P1V1^γ
Andrew Mason
#10
Mar29-13, 02:22 PM
Sci Advisor
HW Helper
P: 6,654
Quote Quote by chuakoktong View Post
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
5/2P1V1-Pdv=5/2P2V2

integrated become P1V1^γ
But P2 = P1+dP and V2 = V1+dV. So:
your equation becomes:

(1/(γ-1))P1V1 - P1dV = (1/(γ-1))(P1+dP)(V1+dV) which is equivalent to:

P1dV = -(1/(γ-1))(P1dV + V1dP)

And all that says is that W = -ΔU (ie. it is adiabatic).

AM
chuakoktong
#11
Mar29-13, 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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