1. Apr 24, 2005

Sorry about the repost, but I am frustrated with this problem.

An ideal monatomic gas, consisting of 2.47 mol of volume 0.0890 m3, expands adiabatically. The initial and final temperatures are 22.3oC and -64.3oC. What is the final volume of the gas?

I've tried solving for P in terms of T using PV = nRT and it didn't seem to work out. Am I even starting in the right place?

2. Apr 24, 2005

OlderDan

In your earlier discussion I think you got as far as understanding that $PV^\gamma$ is a constant. Try writing $V^\gamma$ as $V*V^{(\gamma-1)}$ and replacing the product PV with what you know from the ideal gas law. I think there is no escaping the $\gamma$ in this problem, so you will have to get a reasonable value from your tables

3. Apr 25, 2005

GCT

Did you use the proper units...convert celcius to kelvin?

4. Apr 25, 2005

ZapperZ

Staff Emeritus
The problem with using PV=nRT to solve this is that there's nothing in here that's a constant. So even while it is valid, it is not a very helpful form. However, from what I've gathered based on what Olderdan has said, you should already arrive at a very useful form for any adiabatic process, which is

$$PV^{\gamma} = constant$$

You should KNOW this cold whenever an adiabatic process is involved. Now, using this, and the PV=nRT relations, you can substitute P as nRT/V, which gives you another useful adiabatic relation, which is

$$TV^{\gamma -1} = constant$$

This is the form that is relevant to your question, because you are given the two values of the temperature, and the initial volume of the gas.

Moral of the story: you need to not only know what to do, but also be aware of WHY an approach doesn't work. PV=nRT is not a suitable starting point here because there's no quantity or relationship here that's constant. So you have to go hunt for a more appropriate form.

Zz.

5. Apr 25, 2005

GCT

I would imagine that P is constant. Internal energy is directly converted to $$P \Delta V$$ work, work done by expanding against a constant pressure atmosphere/constant pressure situation.

6. Apr 25, 2005

ZapperZ

Staff Emeritus
Come again?

If you look at an adiabatic curve on a PV diagram, P isn't a constant (it is not a flat, horizontal straight line). Look at the Carnot cycle, for example.

Zz.

7. Apr 25, 2005

Dr.Brain

tHE EXACT SOLUTION:

As per First law of Thermodynamics:

dQ=dU+dW

Therefore dU=-dW

dU=nC(dT)

W=p(dV)

where C= heat capactity at constant volume
for monoatomic gas

dT=change in temp.

dV=change in volume

Put the forms and do the calculations . a Correct answer is assured.

8. Apr 25, 2005

GCT

nevermind...I was unfamiliar with the concept, confused it with a vague recollection of another situation.

espnaddict, what course is this for?

9. Apr 26, 2005

Andrew Mason

This is a non-trivial calculation. Why not just use the resulting adiabatic relationship: $PV^\gamma$ per Dan and Zapperz rather than develop it from first principles every time you use it?

AM

10. Apr 26, 2005

Dr.Brain

^^^^
ANDREW MASON,

PV^ lamda will not give the correct answer to the above question for sure.Its by the basics that physics has developed. My method is correct.

PV^lamda will give an answer which is independent of the number of moles being used and change in temperature taking place.Therefore it is wrong method as it is independent of the constraints which can make a huge difference.

11. Apr 26, 2005

ZapperZ

Staff Emeritus
Eh?

Why would the "number of moles" matter in THIS case? The amount of gas doesn't change - it is a constant at both temperatures. And simply via substitution using the ideal gas law, one arrives at the OTHER form for an adiabat with a T-V relationship, which is ALL that is needed in solving the original problem.

.. unless of course you're claiming that this form is also "incorrect".

Zz.

12. Apr 26, 2005

Clausius2

Well, I think the initial question is not clear enough. The main question is:

-Is it that adiabatic expansion reversible?

If it is reversible, then Andrew, Zz and Dan are right, and the result is independant of mass.

If it is irreversible, then you're right, but you don't actually know how to solve the problem. Let's see:

$$nc_vdT=-PdV$$ is the equation for an irreverisible adiabatic curve as you have stated. Now try to integrate it. You don't know how pressure evolve with T nor V. So that, you cannot integrate it without additional information.

I recall have done similar problems with non-reversible adiabatic process, but all of them consisted in a cylinder with a piston in mechanical equilibrium with atmosphere, so that it can be considered that pressure remains constant across that adiabatic curve. Remind all that an irreversible adiabatic process can have the pressure constant.

13. Apr 26, 2005

Andrew Mason

I don't think that is correct. Work it out from $dU = -dW$ and $nC_v = dU/dT$:

$$dU = d(PV) = Pdv + Vdp= nRdT = n(C_p-C_v)dT = -dW = -Pdv = nC_vdT$$

$$\frac{Pdv + Vdp}{n(C_p-C_v)} = -\frac{Pdv}{nC_v}$$

$$Pdv + Vdp = - Pdv\frac{(C_p-C_v)}{C_v} = - Pdv({\frac{C_p}{C_v} - 1})$$

$$Vdp = - Pdv\frac{C_p}{C_v}}$$

$$\frac{dp}{P} + \frac{dv}{V}\frac{C_p}{C_v}} = 0$$

Integrating:

$$\int\frac{dp}{P} + \int \frac{dv}{V}\frac{C_p}{C_v}} = K$$

Integrating and letting $\gamma = C_p/C_v$:
$$lnP + \gamma lnV= K$$

$$PV^\gamma= e^K = constant$$

AM

14. Apr 26, 2005

Andrew Mason

How can it not be? The same amount of work is required to compress it back to its original state as the gas does in expanding. So if you store the work done (eg by lifting a weight) you can reverse the process without going outside of the system.

AM

15. Apr 26, 2005

siddharth

Adiabatic expansions need not be reversible. Consider this example. I have some moles of a gas in an insulated container .Now the gas adiabatically expands against a constant external pressure (p). In this case work done while expanding, that is, in the intergral pdv, is simply -p(v2-v1) as the pressure is constant and equal to external pressure. In this case the work done will not be Pv^gamma though the process is adiabatic but will be -p(v2-v1). The reason is that in a reversible adiabatic expansion, the system and surrondings are almost in equilibirium in every stage. But in an irrerversible expansion we first rapidly increase the pressure to the external pressure and then expand against that pressure.

16. Apr 27, 2005

Andrew Mason

I disagree, especially with your last two sentences. (BTW the work done is never $PV^\gamma$. That is just a constant).

If a gas expands adiabatically, this means that the work done by the gas is equal to the change in internal energy. If a gas expands from P1V1 to P2V2 and does work of only P2(V2-V1) then heat is lost to the system because:

$$P_2(V_2-V_1) < \int_{V_1}^{V_2}Pdv \implies Q \ne 0$$

So, by definition, this process cannot be adiabatic.

What happens in a case where a gas at P1V1 does work against an external pressure of P2 where P2<P1, is that the work done by the expanding gas produces kinetic energy as well as doing P2(V2-V1) of work:

$$\Delta U = P_2(V_2-V_1) + KE = \int_{V_1}^{V_2}Pdv$$

The work done by the gas is greater than P2(V2-V1) by the amount of this kinetic energy (that area in the PV diagram below the adiabatic gas curve and above the line P=P2).

If this kinetic energy is stored along with the P2(V2-V1) work (for example, by raising a weight and giving it KE and its kinetic energy allowing the weight to move higher after expansion ceases), the gas can be compressed with that stored work back to its original state.

So, I would argue, every truly adiabatic expansion/compression is reversible.

AM

17. Apr 27, 2005

siddharth

Oops. Well your absoulutley right there of course. My mistake.

What i should have said was that to calculate the final temperature using $PV^\gamma$ and then substituting P=nRT/V will not work in case of irreversible adiabatic process . This is because the gas law holds only when the gas is in equilibirium. In my example, the final temperature can be found by the dU=q+w. As the process is adiabatic, q=0. Therefore nCv(T2-T1)= -p(v2-v1) and T2 can be found

Because the value of the inital pressure P1 is rapidly changed to P2 before the expansion actually occurs the value of
$$\int_{V_1}^{V_2}Pdv = P_2(V_2-V_1)$$
so q=0 is still valid

Last edited: Apr 27, 2005
18. Apr 27, 2005

xanthym

.

The classic example of an IRReversible adiabatic process is that of "adiabatic free expansion". Let an Ideal Gas be confined in the Left compartment of a 2 compartment completely insulated container. The Right compartment is separated from the Left by a thin massless plate (currently held by a stop) and contains a vacuum.

The separator plate is now released from its stop. The gas in the Left compartment chaotically expands, filling the entire Right compartment (pushing the thin massless plate to the end of the Right compartment). The Ideal Gas now occupies the total volume of the Left and Right compartments.

By the FLT:
ΔU = ΔQ - ΔW

The process is adiabatic so that {ΔQ = 0}. Furthermore, since the gas expanded into a vacuum, no work was done and {ΔW = 0}. Thus:
ΔU = ΔQ - ΔW = (0) - (0) = (0)

Because we have an Ideal Gas:

$$1: \ \ \ \ \ \Delta U \ = \ 0 \ \ \, \ \Longrightarrow \ \ \, \ \color{red}T_{final} \ = \ T_{initial}$$

The above Eq #1 contradicts the following commonly used formula for "adiabatic processes":

$$2: \ \ \ \ T_{final} \, \ = \, \ T_{initial} \left ( \frac{V_{initial}}{V_{final}} \right )^{(\gamma \, - \, 1)} \ \ \color{red}\neq \ \ 1 \ \ \ \ \ \ \ \ \color{black}(\mbox{for} \, \ \ V_{initial} \ \neq \ V_{final})$$

where:

$$3: \ \ \ \ \gamma \, \ = \, \ \frac{C_{p}}{C_{v}} \, \ = \, \ 1 \, + \, \frac{nR}{C_{v}} \, \ > \, \ 1$$

The reason is that Eq #2 applies only to REVERSIBLE adiabatic processes and always predicts non-zero temperature changes dependent only on initial and final volumes (and γ). The case of adiabatic free expansion, where {ΔT = 0}, clearly indicates Eq #2 is NOT applicable to IRReversible adiabatic processes.

~~

Last edited: Apr 27, 2005
19. Apr 27, 2005

Clausius2

I think Siddharth has answered you yet. But anyway I'll try to clear it up.

Imagine there is a closed cylinder filled with an ideal gas at pressure $$P_1$$ with a free piston over it. The system is submerged into an atmosphere of pressure $$P_o<P_1$$. Besides there is a mass put onto the piston. The whole system is adiabatic. At some instant, the mass is instantaneously removed and the piston goes up. Assuming the mass of the piston is negligible, then the mechanical equilibrium must yield internal pressure equals atmospheric pressure. Due to the rapidity of the process, it must be considered irreversible. Such irreversible trajectory is represented by the equation $$\delta Q=0$$, or which is the same $$dU=\delta W$$. So that, the decreasing of internal energy is the same than the work done by the system over the atmosphere. Of course $$W=-P_o\Delta V$$. And $$PV^{\gamma}=const$$ has no sense for this problem because it represents an adiabatic reversible trajectory (isentropic).

In order to make this process be reversible, the mass would be removed very very slowly, such that internal pressure evolves yielding mechanical equilibrium in each infinitesimal variation.

Also it can be demonstrated that the temperature of internal gas is greater after irreversible adiabatic expansion than after a reversible adiabatic (isentropic) expansion.

Turning back to the original problem, I don't know which is the real case, by the way we have not obtained any feedback of the thread author as usually.

Last edited: Apr 27, 2005
20. Apr 27, 2005

Andrew Mason

But if the initial pressure is rapidly changed before the expansion, without doing work, heat must leave the system. There is no other way that can happen. By definition that would not be adiabatic.

$$P_2(V_2-V_1) < \int_{V_1}^{V_2}Pdv + \int_{P_1}^{P_2}Vdp = \Delta U \implies Q \ne 0$$

AM