# Problem in Thermodynamics

1. Sep 11, 2005

### siddharth

Here is the question.
100 g of N2 at 300K were held by a piston at 30 atm. Pressure was released suddenly to become 10 atm, adiabatically. Calculate $$\delta S$$. Cv=20.8 J/(K*mol)

Since this is adiabatic , q=0. So,

$$\delta U = w$$

Now, as the pressure was released suddenly to become 10 atm, this must be an irreversible adiabatic expansion.
Also, from the statement, what I understand is that the external pressure was rapidly changed to 10 atm and the piston was allowed to expand till the pressure of the gas inside was also 10atm.
So the work done in this case must be
$$-10(V_2 - V_1)$$

Also,

$$V_1 = \frac{nRT_1}{P_1} = 2.93 L$$

So,

$$(3.57)(20.8)(T_2-300)=-10(V_2 - 2.93)*10^2$$

and

$$P_1V_1/T_1 = P_2V_2/T_2$$

=> $$(30)(2.93)/300 = (10)(V_2)/(T_2)$$

Solving these two equations, I get

$T_2 = 243.44 K$ and $V_2 = 7.13 L$

As Entropy is a state function, considering the reversible path,

$$\delta S = nCvln(T_2/T_1) + nRln(V2/V1)$$

$$\delta S = 0.854$$

The answer which is given is -0.692
Can you check my work and tell me if I made any calculation or conceptual mistakes?

2. Sep 11, 2005

### marlon

What are you doing here ?

In the first term of this sum, shouldn't you loose the n. I think the formula should be $$\delta S = C_V ln(T_2/T_1) + nRln(V2/V1)$$

marlon

3. Sep 12, 2005

### Andrew Mason

This is an interesting problem. But I think the question leaves a few things unresolved. By making the change in pressure instantaneous, you have a dynamic system and you lose equilibrium. Where does all that kinetic energy go?

I may be taking too techinical a view of the problem, but I don't think it is quite as simple as subtracting the work done as $W = P_2\Delta V$. This does not take into account the kinetic energy transferred to the piston and load, which (presumably) leaves the system. This kinetic energy is the difference between the area under the curve of a reversible adiabatic change and the area represented by $W = P_2\Delta V$.

So, I would say that the adiabatic condition still applies:

$$P_1V_1^\gamma = P_2V_2^{\gamma}$$ where $\gamma=C_P/C_V = 7/5 = 1.4$

Also, using 1 mole at STP = 22.4 litres, I get the initial volume to be 2.73 litres.

Of course that gives you zero entropy change, so it is not what the question appears to be asking.

AM

4. Sep 12, 2005

### siddharth

For this process $q=0$
So $$\delta U = w$$
Also, for an ideal gas, for any process, the change in Internal Energy can be written as $$nC_v(T_2-T_1)$$ provided that the value of C_v remains constant in this interval.
Also, The external pressure against which the gas is expanding is Pext.
So, work done is $$P(V_2 - v_1)$$

This is how I got the formula.
For a reversible process, I can write
$$dq = \delta U - w$$

From that, writing $$\delta U = nC_vdt$$ and using PV=nRT and then ntegrating, I get the result. I consider the reversible reaction from P1,V1,T1 to P2,V2,T2 here as entropy is a state function and depends on the inital and final state only.

What is the load here? Since the mass of the piston is not given, I think that there is no kinetic energy transferred to the piston.
Also, I don't think you can use PV^gamma as constant.

5. Sep 12, 2005

### gnpatterson

I think you are right because the entropy change you have calculated is positive and that is what it should be for an irreversible process. I don't get how the book could be giving a negative number.

I follow your maths up to the last step, what do you get for each of the terms in delta S? I end up with an bigger positive number!

6. Sep 14, 2005

### siddharth

I made a mistake in my final calculation of $$\delta S$$

I get $$\delta S = -15.51 + 26.40 = 10.89$$

Is that what you got?

I don't understand what you mean by kinetic energy. Is it the kinetic energy of the piston or the kinetic energy of the gas molecules?
And why are you finding the difference in area between the curve of a reversible adiabatic change and P(V2-V1)?

Also, in the derivation of $$PV^\gamma = c$$, the ideal gas equation PV=nRT is used. Is that applicable in this process?
I learnt that PV=nRT is not applicable in adiabatic irreversible process (such as this one), and hence $$PV^\gamma = c$$ can't be used.

7. Sep 14, 2005

### Andrew Mason

My point was that it is a gross oversimplification to treat this as an instantaneous expansion of gas and ignore the kinetic energy of the expanding gas.

In any event, if you use 101325 Pa for 1 atm, I get T2 to be 251K and V2 to be 6.85 Litres.

I share gnp's confusion about your calculation of dS. Your figures do not seem to work for 100 g. of N2 (n=3.57 moles). Perhaps you could show what numbers you used.

AM

8. Sep 14, 2005

### siddharth

In my calculation, I used 1 atm as 10^5 Pa. From that, I get T2 to be 243.44K and V2 to be 7.13 L.

To calculate $$\delta S$$
$$\delta S = (3.57)(20.8)(ln(243.44/300)) + (3.57)(8.314)(ln(7.13/2.93))$$

There is something I don't understand. Isn't the kinetic energy of the gas molecules is accounted in Internal Energy?
From wikipedia,
"The internal energy of a system (abbreviated E or U) is the total kinetic energy due to the motion of molecules (translational, rotational, vibrational) and the total potential energy associated with the vibrational and electric energy of atoms within molecules or crystals."
So, what do you mean when you say that the kinetic energy of the expanding gas is ignored?
I really don't understand what you are saying in
It would be useful if you could explain what you mean in the above paragraph in more detail.

9. Sep 15, 2005

### Andrew Mason

But 1 atm is 1.01 x 10^5 Pa. The resulting difference is more than 1%.

The problem is with the assumptions in the question. It is not a real scenario. I also don't think you can call it truly adiabatic.

By the first law, energy cannot simply disappear. If it is converted into translational kinetic energy of the gas (which is not heat) and the load that is exerting external pressure on it, it does work. That work is in addition to external pressure x volume change.

Consider the following:

You have 3 L of an ideal gas at 30 atm. at temperature T1. It is contained by a piston of area .01 m^2 by a force of 30,000 N supplied by a 3,000 kg. weight. All of a sudden, the weight is reduced to a force of 10,000 N, reducing the pressure to 10 atm, by removing 2,000 kg reducing the weight to 1,000 kg. At the instant the weight is removed, the gas pressure is still 30 atm. The net force pushing the 1,000 kg is initially 20,000 N. So the weight accelerates upward at the rate of 20 m/sec^2. As the gas expands, the pressure decreases. When it reaches 10 atm, the weight has done P_ext x dV of work + the work required to accelerate the load and the centre of mass of the gas.

If you factor this work into the equation, it is equivalent to the work done in a reversible adiabatic process (ie. the kinetic energy is the area below the adiabatic curve and Pext x dv). If you simply let that kinetic energy do work back on the gas (eg. the piston eventually stops and falls back down, compressing the gas until it stabilises at a constant volume and pressure of 10 atm), then you are in effect allowing 'heat' to flow into the gas and it is not correct to call it an adiabatic process.

AM

10. Sep 15, 2005

### gnpatterson

I got the 10.9 that you now get.

I spotted your use of 10^5 Pa and was a little concerned with it but I decided that it wouldn't really matter that much.

I really think that worrying about piston KE is a red herring, it is after all a book question about an ideal gas. The whole question is essentially a fiction, but we all got used to point masses and weightless strings, so I think b*tching about massless pistons that achieve an impossible decompression is not going to be productive. In the end we have to get our head round the idea of entropy and doing it without some "magical" pistons is going to be even harder.

I think you might be able to visualise the piston as having a perfectly thin membrane infront of it that is perfectly light and only allows exact the right amount of pressure through it to push the piston at the final pressure. I am not going to discuse that issue further.

The amount/mass of gas is not relevant to the _sign_ of the entropy change.

I am stuned that the difference in estimate for 1atm produces such a big change in the results.

Last edited: Sep 15, 2005
11. Sep 15, 2005

### gnpatterson

Thinking about the principals of the situation, I cant get away from my perception that the entropy change has to be positive, as the process was an irreversible one, the amount of entropy must have increased.

The system could have got into the same state by having some heat dumped into and then being decompressed reversibly, the reversible decompression would have got more work out of the system to balance the heat (energy) we dumped into system, but the entropy of the equivalent heat transfer would remain unaltered.

perhaps playing with the numbers can change the magnitude of the entropy change but it can't change the sign.

12. Sep 15, 2005

### Andrew Mason

Well, thermodynamics is difficult enough. When one complicates it by ignoring real changes, it is that much more difficult to grasp.

The bottom line here is that you cannot change the pressure of an isolated volume of a gas without expanding it. So the gas does more than 10dV atm/L of work. I cannot conceive of a physical scenario in which this can occur suddenly without producing kinetic energy in excess of the PdV work.

Even if you allow the gas to leak slowly out of a cylinder through a small hole into an external pressure of 10 atm (which is not what the problem says - it says the pressure on the gas is released suddenly), the gas itself will gain kinetic energy (applying Bernoulli's formula). The gas in effect does work on itself. The problem suggests that you ignore the fact that this energy is kinetic and treat it as heat energy, which is what we have done in working out the answers.

AM

13. Sep 16, 2005

### gnpatterson

AM I know you wont let go this but I'll say it anyway, GIVE IT UP, you are archived as fighting this point from way back, are you try to take on thermodynamics?

If you cant accept the definitions of "sudden" as used in thermodynamics then you are just being obscurantist. No-body is asking about real gases in real pistons, the idea is to work with a concept that is defined in terms of mechanisms that are impossible, so using impossible tools is the name of the game.

As the red queen said "Why! I can imagine six impossible things before breakfast"

14. Sep 16, 2005

### siddharth

Just to make sure I understood, what Andrew is essentially saying is that for any process the total energy of an isolated system is
$$E = U + \frac{mv^2}{2} + mgh$$ and this energy is what is conserved.

What I initially thought was that the kinetic energy of the gas molecules during expansion of the piston was included in the Internal Energy.

But what I infer from Andrew's previous post is that, only the kinetic energy from the random thermal motion of the gas molecules is included in the Internal Energy and the kinetic energy of the gas molecules during expansion in the irreversible process is not included in the Internal Energy term.
So for an irreversible process such as this, the correct equation would be
$$\delta E = q + w$$
and not
$$\delta U= q + w$$ as we have to conserve the kinetic energy of the gas molecules also.
Is my above reasoning correct Andrew?
Also, I did make a serious blunder in approximating the value of 1atm. Thanks for pointing that out!

15. Sep 16, 2005

### gnpatterson

It think that you will have a problem if you continue to follow AMs line of thought because at the end of it entropy ceases to exist.

16. Sep 17, 2005

### Andrew Mason

Not quite. But you are thinking along the right lines. The essential point that important here is that the work done by the gas is more that simply the external pressure x change in volume (the difference being the kinetic energy produced by the rapidly expanding gas). You cannot possibly have a 3 L volume of gas at 30 atm rapidly expand by 4L at 10 atm external pressure and do only 10 atm x 4 L of work. The PV graph will not be flat from 3 L to 7 L. It will go down as a curve from 30 atm during that time.

So for a dynamic expansion:

$$dU = dQ + dW = dQ + P_{ext}dV + (P_{dyn} - P_{ext})dV = dQ + P_{dyn}dV$$

The difference is very close to the difference between the reverible adiabatic curve and the flat P_ext dV graph.

AM

Last edited: Sep 17, 2005