# A thermodynamic problem

1. Apr 14, 2015

### mooncrater

1. The problem statement, all variables and given/known data
Two moles of an ideal mono atomic gas (C2=12. 55 JK-1mol-1) expands irreversibly and adiabatically from an initial pressure of 10 atm against a constant external pressure of 1 atm until the temperature drops from the initial value of 325 K to a final temperature of 275 K. Calculate the work done by the gas and its final volume.

2. Relevant equations
W=-Pext[Vf-Vi]---(1)
PV=nRT -----(2)

3. The attempt at a solution
work done can be found easily
W=-pext (ΔV)
=-1255 J
And V 1 can be found by the ideal gas equation.
BUT, there are two ways to find V2 : that is by ideal gas equation and by the work equation ( by putting the value of V 1 in it ).
Both these method give different values of final volume. Which method is wrong here? And why?

2. Apr 14, 2015

### Delta²

Nope that is not the work done by the gas, because the pressure of the gas does not remain constant and equal to pext as you assume. Use the given fact that the expansion is adiabatic to calculate the work from the 1st law.

3. Apr 14, 2015

### mooncrater

BUT for an IRREVERSIBLE process there is a same formula W=-Pext[ΔV] for both irreversible isothermic and irreversible adiabatic processes. From what I know you must be talking about the formula used in the reversible adiabatic process.

4. Apr 14, 2015

### Delta²

I am not sure if this formula is correct for all irreversible adiabatic processes. Well if your book mentions clearly that this is the case then i guess you should use it to get the final volume, now that i think of it it is unclear what the final pressure of the gas is , so you just cant use the ideal gas eq.

5. Apr 14, 2015

### Staff: Mentor

Actually, your result for the work done is correct. Since the process is adiabatic, the change in internal energy is equal to minus the work done on the surroundings:
$$ΔU=(2)(12.55)(275-325)=-1255 J$$
The work done on the surroundings is, as you correctly calculated, $W=P_{ext}(V_f-V_i)=1255 J$. (I assume you are using the sign convention where the work done by the surroundings on the system is W, while I am using the sign convention where the work done by the system on the surroundings is W).

So, there is no inconsistency. The work you calculated from each of the two equations matches.

Chet

6. Apr 14, 2015

### Staff: Mentor

Not only is this equation correct for all irreversible adiabatic processes. It is correct for all processes.

Chet

Edit: Actually, what I meant to say is the dW=PextdV for all processes. For reversible processes, Pext=P, where P is the equilibrium pressure of the gas.

Last edited: Apr 15, 2015
7. Apr 14, 2015

### Delta²

Ok maybe i am gonna sound dumb but why is this. Is it just the definition of work on the boundary and that the fact that external pressure remains constant?

8. Apr 15, 2015

### mooncrater

Yes it matches BUT the problem I am facing is with the final volume(as I said in the question). Do I have to use this work method or the ideal gas equation to find it cause they both are giving different values of final volume.

9. Apr 15, 2015

### Delta²

When you use the ideal gas equation what final pressure of gas you use ? do you use $p_f=p_{ext}=1atm$? I think the problem doesnt imply that the final pressure of gas equals $p_{ext}$

10. Apr 15, 2015

### mooncrater

Yeah..... it never said that final pressure is equal to pext..... but many a times we use ideal gas equation inside this work equation to get temperatures. And there we consider pext
as the final pressure of the gas. Like:
W=Pext[V2-V1]
Which is many times written as:
W=P2[nR T2/P2 -nRT1 /P1]
Which I have seen and used in many questions.

11. Apr 15, 2015

### Delta²

i just dont think in this case the final pressure is 1atm.

Because if it is then if you use it in conjuction with the ideal gas equation (and we know the ideal gas equation is correct no doubt about that) you get different value for the final volume than that of the value you get from the equation for work. So either the final pressure is not 1atm, or the equation for work is not correct. But Chet strongly agrees that your equation for work is correct, which leave us only one option that the final pressure is not 1atm.

12. Apr 15, 2015

### mooncrater

You are right. It is the final pressure that can be wrong here as no other thing can be wrong. But why?
I am uploading pics of this question and the next one in which they've put final pressure equal to the external pressure.

13. Apr 15, 2015

### Delta²

yes in the example (33) it says among others "until it is in equilibrium with a constant external pressure". This statement essentialy means that $P_f=P_{ext}$. In the example (32) there is no such statement, it just says "till the temperature drop to 275K".

14. Apr 15, 2015

### mooncrater

Okay....!!! Now it's clear to me.... in the example 32 the pressure may lie between anything from 10 atm to 1 atm.... whereas in the 33 rd example it is given that the equilibrium pressure is equal to external pressure therefore in that we can say that external pressure = final pressure but not for the 32nd example. Now I get it. Thanks for your guidance :).

15. Apr 15, 2015

### Staff: Mentor

It's not what I would call the definition. It's just the force exerted by the gas on the surroundings times the displacement. That's what we always call the mechanical work. dW = PextdV=(F/A)(Adx)

16. Apr 15, 2015

### Staff: Mentor

Good point. During the time that the expansion is occurring, the external force per unit area is held at 1 atm. After the piston is stopped, if the gas volume does not correspond to the volume that would be obtained from the ideal gas law at 1 atm and the final specified temperature, then the gas will re-equilibrate at a different final pressure, corresponding to the final volume and the specified final temperature.

Chet

17. Apr 15, 2015

### Staff: Mentor

Yes. This is what you get when the gas is allowed to expand all the way to the point where its final equilibrium pressure equals the constant pressure imposed during the irreversible expansion.

Chet