Irreversible adiabatic expansion in which P_ext=0

[Raghav Gupta]

Homework Statement

In an adiabatic irreversible expansion of ideal gas , if P_ext = 0 then which is true,
A. T2=T1
B. Q=0,
C. P1V1= P2V2
D. P1V1^γ= P2V2^γ

Homework Equations

ΔU = Q - W
For ideal gas, PV= nRT[/B]
ΔU ∝ ΔT

The Attempt at a Solution

I know Q= 0,
Work done is zero,
So ΔU = 0, hence T2= T1,
So also P1V1 = P2V2,
But why option D is incorrect?
I now,
Also have a problem that why work done is zero. Doesn't the gas will do work in expanding?
Our work done will be zero.

 

[mooncrater]

For an irreversible adiabatic process the work done is equal to
W=P_ext[V₂-V₁]
And since P_external=0 (as you said) therefore work done is zero here.
NOTE: For a reversible adiabatic process you have a different formula.
You said if gas is expanding then work should be done, but here there is no resistance against the expansion of the gas as P_ext=0. Therefore no work is done here for its expansion.
Moreover the D option is a basic result of an adiabatic process therefore it is true here (thats what I think).

 

[Raghav Gupta]

Got it, that work done is zero.

Moreover the D option is a basic result of an adiabatic process therefore it is true here (thats what I think).

That d option is false, I have seen the answer key.
How they are getting is the main problem.

I think for derivation of PV^γ = constant for adiabatic expansion,
We take ΔU = -W,
⇒ nC_vdt = -P dv = - (nRT/V) dv
The fact we are taking P = nRT/V is incorrect in irreversible I guess. Is that correct?

 

[mooncrater]

The fact we are taking P = nRT/V is incorrect in irreversible I guess. Is that correct?

Yup... I think you are (absolutely) correct. If in the question its not given that the equilibrium pressure is equal to the external pressure then ideal gas equation can not be applied on this pressure. More over we have to equate external pressure to the internal pressure to derive the result of option D. Derivation (of my book ) is attached in which this step is performed.

Sorry for:

Moreover the D option is a basic result of an adiabatic process therefore it is true here (thats what I think).

 

[Chestermiller]

Got it, that work done is zero.

That d option is false, I have seen the answer key.
How they are getting is the main problem.

I think for derivation of PV^γ = constant for adiabatic expansion,
We take ΔU = -W,
⇒ nC_vdt = -P dv = - (nRT/V) dv
The fact we are taking P = nRT/V is incorrect in irreversible I guess. Is that correct?

d is true only for an adiabatic reversible expansion, where P_ext=P. It is not true for your irreversible adiabatic expansion. Solve the problem for the adiabatic irreversible expansion, and you will see that the answer is different.

Chet

 

[Raghav Gupta]

Thanks, the derivation part I know already.

 

[Raghav Gupta]

d is true only for an adiabatic reversible expansion, where P_ext=P. It is not true for your irreversible adiabatic expansion. Solve the problem for the adiabatic irreversible expansion, and you will see that the answer is different.

Chet

But how I will integrate in irreversible expansion, when I don't know how pext is changing as a function of something?

Wait all of you. For a moment it seemed I knew everything but now it is seeming that here PV^ gamma can be constant , where P is gas pressure.

 

[mooncrater]

But how I will integrate in irreversible expansion, when I don't know how pext is changing as a function of something?

P_ext is a constant... why will it change with time?

 

[Raghav Gupta]

P_ext is a constant... why will it change with time?

Irreversible expansion. See this link ChesterMiller's reply in which he is saying P ext can be varied.

 

[mooncrater]

For an irreversible process, we can control Pext as a function of time t during the deformation to be whatever we wish.

(Emphasis is mine)
You can control P_ext externally by yourself. You can change it to any value you want but if you have given it a constant value then it won't change by itself.
Consider an example: If you put some water in a bucket (say 20ml) then it is your choice that you have put 20 ml water in it. It won't change with time (no evaporation) until you change it by yourself.

 

[Raghav Gupta]

Getting expression after solving,
T2 - T1 = -P_ext(V2-V1)/C_v
Now, we don't know about gas pressure from all this but only about external pressure.
How then one can say d option is incorrect?

 

[Chestermiller]

Getting expression after solving,
T2 - T1 = -P_ext(V2-V1)/C_v
Now, we don't know about gas pressure from all this but only about external pressure.
How then one can say d option is incorrect?

You can't use the ideal gas law for all the states that the gas passes through as it expands, but the initial and final states of the gas are thermodynamic equilibrium states, so you can use the ideal gas law for those. Also, you've already been told in the problem statement that P_ext=0 during the expansion, so you know immediately that T2=T1. From this, you can determine P2, which is not the same as Pext, since, once the expansion is manually constrained to end at V2, the gas pressure re-equilibrates to P2. Pext=0 is present only during the expansion.

Chet

 

[Raghav Gupta]

Got it. Thank you to both of you.

 

[Chestermiller]

Here's a thread that you guys might find interesting and might want to help with: https://www.physicsforums.com/threa...ite-using-thermodynamics.808847/#post-5077211

Chet