Challenge Questions

Would that not simply be two equations and two unknowns?

I am assuming the mass and temperature of the vessel walls are not to be considered.

However, I would simply look at the final state of the two vessels and the initial state of the vessel with the gas.

Vessel A will have the same volume as the initial state, so the final temperature should be simple once you determine how much of the gas is left.

Vessel B will contain the remaining gas so knowing its volume and using the original gas temperature/pressure you should be able to arrive at the final temperature and pressure for vessel B.

Then again, I might be full of hot air myself.

I agree with BobD, using PV/T = P'V'/T' since P' = ½P and V' = 2V with Temp in absolute 27C ≈ (273 +27)K = 300K

Then the final temp will be 150K = -123C at steady state

The temp will be even lower when the first bit of gas expands into the vacuum, but this is way to hard for me to work out.

if pV/T is constant, when you increase V and decrease P by the factor of 2 the temperature must remain constant, because PV is constant in that case.

Hi ffeg,

How do you know that it is "1/2" P? No where in the question it says the volumes of the tanks are equal. Your statement will be true only if the volunes are equal. Immagine that Vb = 10cc and Va = 1000,000 cc (Internal volume of a spherical vessal having an I.D. of 1.2407 meters, which is not that big).

Job Thykkoottathil

Yes, good point. I am talking about the special case mentioned.

GA

Isn't that "adiabatic expansion" with the gas doing work on a moving surface?

Hi

I do not quite agree. I see temperature as the rate at which molecules collide against each other. In the second vessel after everything has settled down, there are now less than the total original molecules. Hence a lot less collisions, hence a drop in temperature....or am I barking up the wrong tree.

Bob

My previous mail may indicate that I do not agree with you. I meant I also do not agree with the increase in temperature in the question posted

Carel

If you assume that the gas that escapes initially doesn't change temperature, then the subsequent gas that escapes from A has the back pressure of gas already in B to work against. That means that the gas in A is doing work while it expands, and its temperature must go down, and B is being compressed, so its temperature rises. Paradoxical, but that's life

That means that the gas in A is doing work while it expands

No, the gas can do work only against external pressure, that means when the volume of the vessel is changing by moving, for example a piston.

Imagine, that in vessel A is gas A and in vessel B is gas B (no matter at what pressure). Opening the valve causes expansion of two gases into the whole (A+B) vessel. And none of them is doing any work, nevertheless there is some gas the opposite vessel.

Ideal gas is built of independent molecules which DO NOT "see" other molecules (no interactions). Thus mixing of ideal gases is like expansion of both of them into vacuum.

"Ideal gas" contains a number of assumptions that can be seen as contradictory. The critical ones here are negligible size molecules and frequent collisions. These are useful assumptions, because although the molecules in a quasi-ideal gas are small compared with the available space, the linear dimensions are generally sufficient multiples of the mean free path that collisions are indeed frequent. When there is an interface between two bodies of gas at similar but unequal pressures and temperatures, in the short term the frequent collisions allow you to treat the interface region as if it was a solid boundary, with only modest mixing across the boundary.
Note that even the most basic aerodynamic equations require frequent collisions.

In addition, in the absence of collisions between gas molecules the velocity of each molecule would be constant (ideal container), so the only heating we would see in container B would be due to the greater frequency with which the faster molecules pass through the orifice. In that case it is straightforward to see that the temperature in B would peak immediately following the opening of the valve. At that point, the mean energy of particles passing through the valve may be found by integrating
k.T.(4 + probability.velocity_component²/rms_velocity)
The velocity distribution is usually taken to be "normal". I haven't done the calculation explicitly, but I would expect the temperature rise for this case to be in the order of 20-degrees.

In addition, in the absence of collisions between gas molecules the velocity of each molecule would be constant (ideal container), so the only heating we would see in container B would be due to the greater frequency with which the faster molecules pass through the orifice.

I think, that you can be right, but the faster molecules also more ferquently would go back to the vessel A. "Colder" molecules go to B rarely, but stay there for a longer time. I still suspect, that the temperature in B will be the same as in A.

Consider only the initial phase, when the pressure in B is very low. The rate of gas movement from A to B would swamp the return, so the initial peak temperature would exactly satisfy this behaviour. In no way does this represent what happens, of course.

This sounds reasonable.

The dynamics of the process described here may be (well, certainly it is) very complicated. And it can not be solved with simple, macroscopic, gas models. What we can easily calculate are the conditions at the beginning and at the end of the whole expansion.

Here are my summarised thoughts:

1. Do we really have here isoenthalpic process? Such expansion (like in Joule-Thomson experiment) requires porous plug and constant pressures on both sides of a valve. It is not a case because we have constant volumes and changing pressures on both sides of a valve

2. What we have here (in my opinion) is free expansion of a gas (Joule effect), in which the internal energy remains constant (no heat exchanged, no work done)

3. If the gas is ideal, because its internal energy is only fucntion of temperature, the temperature remains constant.

4. For non-ideal gas the temperature will drop. It can be easy calculated for Van der Waals gas, in which the change of the temperature is always negative.
Delta(T)=-a*n^2*(k-1)/(k*Va*Cv)
where a is always positive coefficent, n is quantity of gas (in moles) and k is ratio Vb/Va.

Such cooling can be observed while using e.g. "compressed air" which gets very cold or any spray.

5. Do we really have zero pressure in vessel B? I would rather say that it is equal to 1bar, after a shot. In such a case for ideal gas the temperature will again be constant. For real gas the drop will be smaller (but I can not calculate it quickly, unless we assume the process as two independent expansions - from A to B and in the opposite direction - but I can not say if this estimation is reasonable)

OK, thats all.
Kajot

Having thought about it a bit more:
If we consider only the gas in enclosure A: apart from a small section near the valve, the behaviour is spatially uniform (pressure travels and equilibrates much faster than diffusion or anything else). So we can consider what is happening in a small fraction of this uniform region and use the results for the total. That portion is expanding against an identical pressure applied by the remainder of the gas in enclosure A. That means that we can apply the standard "piston-based" adiabatic law to determine the temperature of the gas in enclosure A. That in turn tells us how much kinetic energy hes been transferred to the gas in chamber B, which determines the temperature once the valve is closed and the gas in B is allowed to mix. That sounds very similar to what slide-ruler was suggesting all along - but unfortunately I haven't time right now to check that I have exactly the same in detail.

So, it's an elegant problem for which we do not need to know the detail of the process near the valve or in container B - provided that the valve orifice and the mean-free path are both small compared to the sizes of both enclosures. (I already gave it 5-stars - maybe someone else could help bring it up to an appropriate rating?)

Given my previous comments I will try to answer this from basic principles. I will define temperature as the average kinetic energy of the gas particles and assume that no energy is lost in collisions with the container walls (perfectly elastic collisions). Once the valve is opened the gas particles in container A will start to move into container B. The rate at which this will happen will depend on the size of the pipe between them and the pressure in A. Regardless of these factors the average kinetic energy of the particles will not change as no work is done by the gas such as would happen if container A was to expand over a period of time. Since the gas does no work and no work is done on the system and no heat flows in or out of the system the average kinetic energy remains the same and so the temperature remains the same, ie, 27 deg C regardless of how much gas is transfered or how big container B is.

I'm with bob on the 111 degree thing! If container B is indeed a vaccuum and both vessels are the same size than the effective volume of A would double when the valve is opened, reducing the pressure which would necesitate a drop in temperature. I'd expect a temperature of 1/2 x 27 degrees or 13.5 degrees in both chambers, essentially to power the expansion of the gas.

Perhaps as all this stuff is in Kelvin the 111"C" is a typo. However, the basic rule is PV/T, (T in K), so if the volume doubles the P or T must adjust. Which one – who cares – as the "maximum temperature" of any gas in 'B' will be at the instant of release or before there is time to change energy. So it would be 27 C, if you assume a time dimensionless valve and the pipe volume is part of 'B'. As everything else is "perfect" – why not? Or the question is about time and inertia.

27 C... At least initially. I started thinking about ambient temperatures and gas expansion. Size of the vessels and duration of the valve being opened. The difference of mass and pressure of the gas being less, no matter how long the valve is actuated, would reduce the temp. Plus a vacuum should be a slightly lower temp, do to a lack of molecular activity. So I say 27 C or about 80.6 F is the greatest temperature to vessel B.

I agree with you MJ. Th problem doesn't say what the size of the second container is. Let's start by defining the size of the second container as zero. If it is zero then the temperature remains 27. As we increase the size of the second container, the gas has more room to expand, decreasing the average amount of kinetic energy and the temperature is less than 27.

Oops, havent read this one yet. Now I disagree with you.

Wait a minute here. Back when I was in navigator training in the USAF, we would take the primary altitude chamber to 12,000 ft pressure altitude, and another smaller chamber to 40,000 ft pressure altitude. Then the small diaphragm separating the two would be ruptured to simulate what would happen in the event of a sudden decompression at high altitude. The chamber cooled and it got very foggy for a short time. Where did the heating in this example come from?

Size of the container of air and of the container with the vacuum?

Internal and External Temp of the second and unknown container?

Ambient room temperature?

Location of this experiment?

Time of year?

Amount of time the temp measurement is allowed for reading?

Type of measruing device?

Age of measuring device?

Color of bra and panities the cute 24 year old brunette female is wearing in the apartment up the street?

We need a few details...

HOWEVER, But I'd say roughly 85 to 90 degrees farenheit simply because the friction of the air rushing to the second container as it warmed from 80.6 degree F as I GUESSED that the second unknown container and transient tube and valve are at least 20 degrees warmer because they are placed in the shade in Phoenix, Arizona in May.

Her panties and bra are white...

http://mealercompanies.com

We were supposed to be serious about the answers?

No one told me!

JL Mealer

http://mealercompanies.com

I'm with the "temp will be lower" crowd...

When you allow gas to escape to a lower pressure system, the result is cooling, not heating.

Tony

If temperature is proportional to the molecular kinetic energy then it is a function of the molecule's velocity. When the valve is first opened, the molecules fly down the pipe and into vessel B at the same velocity that they were before the valve was opened. Thus the first gas into vessel B has the velocity, kinetic energy, and temperature that it had in vessel A. The temperature will be 27 deg-C.

Don

OK here is my third attempt. Since no heat flows in or out of the system the only way for the temp in B to be higher than 27 deg is if the temp in A drops below 27 deg such that total energy in the two containers is the same. Clearly this situation can not be maintained for an indefinite period and so is a transient state. Here is how I think it might happen. When the valve is released the gas has a bulk flow into A which acts to compress the gas which has initially flowed into it so that it increases in temperature while at the same time the flow also reduces the temp in B (adiabatic cooling). At some point in time the pressures will be equal so the flow will stop however B will have a smaller number of molecules (because it has a higher temperature). Over time the temperature will even out between the two vessels until they are both at 27 deg with equal pressure and same number of molecules in each. So now if this is indeed the process what is the maximum temperature that could be achieved in B. Since the heating of the gas in B continues until the pressure is equal then if the valve is stopped before this time the gas does not reach its max temperature so the max temperature is reached when the pressures are first equal between A & B after the valve is opened. Now the only variable left is the size of B. If it is very large compared to A then most of the gas would be in B so its temperature increase could not be as high as when the sizes were the same. If it was a very small it would also be difficult to achieve a high temperature due to the large imbalance between the number of molecules in each vessel, so the most efficient temperature transfer would happen when the containers are of a size that results in equal molecules in both containers. Now if only I could work out what that was...

Well, I also do not know, how they can reach 111C expanding gas until pA=pB in case where Va=Vb.

If the gas is ideal, and enthalpy remains unchanged, then the temperature of the gas also remains unchanged, because the enthalpy of an ideal gas is only fucntion of temperature.

More details:
Lets start from the first law of thermodynamics reding:
dU = dW + dQ (of course dW and dQ are not differentials, they are "elementary" changes).
dW = 0 because the working pressure is 0 (vacuum)
dQ = 0 as it is stated

so dU = 0 = Cv dT while Cv<>0 dT must be 0

there is no assumption of Volumes at start and at the end of the process.

What about the case of closing the valve before pA=pB?
Because enthalpy and internal energy are the functions of state, their changes do not depend on the way, so we can devide our process into any number of simpler processes, giving the same start and end conditions.
We can devide the process into two stages. First one - expanding gas from vessel A to B with a smaller volume (we can devide vessel B into two parts with a removable wall) untill pA=pB. As a result we have the same pressure and temeperature (T is exactly the same as it was before expansion). In the second stage we can remove the wall in vessel B, making the second expansion. now pB will be lower than pA, but the temperature will again not change.

So my answer is : the temperature of an ideal gas expanding to vacuum in adiabatic conditions remains unchanged.

My question is: How to reach 111C in the example given???

Two more comments

We know from experience that rapid expansion of a gas cools it down (or cools the container). Rapid means that no heat can be transferred.

The problem is that ideal gas law is correct only for low pressures and not to high tempeartures. If we consider itermolecular forces like in Van der Waals law, using the same first law of thermodynamics we will obtaine decrese od the temperature even while expanding into vacuum.

The Cp/Cv ratio is used for quasistatic expansion of an ideal gas (Piossons equation) but it is not the case, because expansion is into vacuum.

The enthalpy of a gas is not just a function of temperature. The internal energy of an ideal gas is only a function of temperature. Enthalpy is equal to the internal energy plus pV.

Isn't it what I have written above? Maybe not too clearly:-)

The gas in Vessel A is expanding (doing work) and since no heat is being added, it must be cooling: Similarly the gas in Vessel B is being compressed and heated.

If the initial enthalpy in Vessel A is H, the H = HA + HB at all times. HA depends on both the massA and the temperatureA; same for HB.

For further explanation see #37

Gas expanding into a vacuum DOES NOT do any work.

dW = -P(external) dV, and Pext is equal to 0 in this case.
I also would be careful in stating, that gas in B is compressed. B is being filled with gas, gas amount is increasing, thus the pressure too, but it is not what we call compression, in my opinion.

IF enough pressure on the a side and enough vacuum on b side, it would be an air conditioner.

Or at least that is the Netflix(CKS) answer. (for those who haven't heard it is a commercial which asks non sense questions)

PV=mrT is fine for steady-state before-and-after conditions when any two of P, V or T are known. But they are not given here.

Even if they were known, the problem poses intermediate conditions where initially, sonic velocities are encountered, which brings in mach numbers and the rate-of-change of pressure and temperature, pipe diameters and orifice sizes.

If there is no exchange of heat or work with the external system, then the expansion is adiabatic. The work done is at the expense of internal energy so the temperature drops as does the associated pressure. The Joule-Thompson effect.

The temperature is linked to the pressure-ratio (r) by T₂=T₁ x r ^(n-1)/n

In adiabatic expansion the gas does work on a moving surface. That is not strictly a closed system. But I agree if you mean that the expanding gas in A is providing a compressive force on the gas already in B.

No, see my answer to #62.

See my answer to your #68

27°C

Hi Guys

The containers are elastic.

The vessels are not elastic. See #37

OK the temp. of vessel B will be higher and the temp. of vessel A will lower. I play paintball and when I fill my compressed air tank for my gun, the tank I am filling warms up and the tank that is losing the air get cold and sometimes frost on it. So yes the vessel B will warm up and Vessel A will get coller.

Scott "Paintball Wizard"

Maximum temperature in B is 195ºC

I have a similar problem on a design that I am working on it, but instead of elastic body I have a double action cylinder that will be emptied by a Liquid jet compressor.

The volume of the chamber that is being emptied is being reduced by the piston on the back stroke.

If in the future I make money with it, I will share it with the guys that kindly will help me on this project.

My posts at CR4 with info:

http://cr4.globalspec.com/thread/37678

http://cr4.globalspec.com/thread/38182

http://cr4.globalspec.com/thread/38177

Thanks

David

It's been a while since I studied thermo but it seems to me that, since the expansion of the gas into vessel B does not involve any work being performed, there is no change in the energy of the system and therefore no change in temperature.

Almost all of you guys are missing the clue in the posted question. What is the 1.4 all about? I think the equation is supposed to read PV=nRT

I teach vacuum so I will stay out of this one.

The heat capacity ratio of 1.4, shows that it is a diatomic gas with five degrees of freedom. It is governed by the ideal gas laws.

The material of the pressure vessels is not specified, and for example if is made of rubber, and vacuumed, the volume of B is zero, and when the gas moves into, it will increase the temperature due to friction.

If we let the system equalize in pressure and temperature, Pa=Pb and Va=Vb (only possible if the vessels are equal) the temperature is 111ºC.

Without transfer of heat with the surroundings, the higher temperature of B will be 195ºC. Then if we keep the valve open, the temperature of 27ºC of A (elastic walls the volume did reduce) will increase 84ºC and B will decrease 84ºC.

Can you explain in details how the temperature in B can be 111C?

We need information about the gas, the roughness of the walls, and the Reynolds number. They do not give that, but they assume that the temperature will be even at 111ºC in both vessels when it comes to rest and with the valve open.

If, as we empty the vessel A, the volume of A reduces because is elastic, then the remaining gas will be at 27ºC, just before the flow to B stops.

If you close the valve just when the flow from A to B stops, we will have the higher temp. in B, again due to friction.

If they even out at 111ºC, this means that B was at 195ºC.

I still can not understand it:-(

If the gas is ideal, how can it increase its temperature? Even if vessel A is elastic and B is not, the energy accumulated in the stretched vessel is much too small. Where is the source of energy absorbed by gas (it increases its internal energy)? There is no work, no heat transfer.

Can you calculate this 111C value with the conditions that Pa=Pb and Va=Vb (thats probably mean that Va and Vb are constant, vessels are not elastic)?


maybe I have to think more abut this:-)

Now we know that vessel A is at -27ºC when B is at 111ºC, then the vessels are not elastic and the solution is narrowed. The Joule-Thompson effect is taking place, and i am not sure if we can find the highest temperature with so litle info.

The challenge says no heat interchange with the walls. Assuming the valve orifice is small, interaction within the gas will randomise the motion - so why is the roughness of the walls relevant?

No, they don't even out at 111^OC, that is the peak temperature in B

I am the author of the challenge, which is based on a problem I had to solve in my work

The phenomenon taking place is seen in post #25 "paintball wizard"

The 111°C is correct for the simple case stated; the complete conditions are Tank A temperature -27°C, Tank B 111°C.

This is not the maximum possible in Tank B because it is easily(?) shown that with VA = VB if the valve is first opened and then closed when Pressure B is 25% of Pressure A initial, then Tank B temperature is 134°C.

Posting #21 by Horace40 deserves a GA since it is one of the key elements needed to solve the challenge.

The phenomenon taking place is seen in post #25 "paintball wizard"{br]
The gas in your paintball gun can not be treated as an ideal gas. The source of a gas is probably liquid (like in all sprays), so you can assume, that the pressure in the container A is high and constant. And we have almost constant increase of the amount of gas in our vessels Posting #21 by Horace40 deserves a GA since it is one of the key elements needed to solve the challenge.

I am not sure if we can use his equation to calculate the temperature, because the process is adiabatic, but not quasistatic. p*V^(gamma) remains constant only for quasistatic adiabatic process of an ideal gas.

I will think about this once again... later:-)

I do not know (yet) how to solve this problem but I do have the following insights. For the specific case given of equal volumes and equal pressures with a second vessel temperature of 111 C (384 K) with an initial temperature of 27 C (300 K), the temperature ratio is 1.28 which appears to be almost exactly equal to 2^(1/2k) where k is the specific heat ratio. The 2 might be the volume ratio. The second insight is that, while the pressures must be equal for this case, I'm not sure the temperature nor the mass of gas in each vessel needs to be the same. While there can be no net change in the total energy, the contents of the two vessels need not be identical.

I initially thought this was not a very good challenge question but now I've changed my mind.

Thanks,

Jim

My approach to the problem is as follows: There are two vessels with volumes V_A and V_B. These vessels are rigid. No work is done deforming them. Vessel V_A is intially at pressure P₁ and temperature T₁ = 27° C (300° K). Vessel V_B is initially empty. The mass of gas in V_A is m. The valve between the vessels is opened for a while leaving a fraction of the gas in V_A. This fraction is β. Once (1- β)*m has moved to V_B, the valve is closed. There is no heat transfer to or from the gas. The final pressure in V_A is P_A and the temperature is T_A. For vessel V_B, the the final values are P_B and T_B.

The two equations I use are PV = mRT and Pv^k = constant, where v = V/m and k = the specific heat ratio of 1.4 which I believe applies to a reversible, adiabatic process for an ideal gas.

Solving the first equation for R,

R = (P₁V_A)/(mT₁) = (P_AV_A)/( βmT_A) = (P_BV_B)/(1- β)/(mT_B)

For P_A = P_B and V_A = V_B, I get T_A/T_B = (1- β)/β

From the second equation,

P₁(V_A/m)^k = P_A(V_A/ βm)^k = P_B(V_B/(1- β)m)^k

For P_A = P_B and V_A = V_B, I get β = ½. This leads to T_A = T_B and then to P_A = P₁/(2^k) = 0.3789 P₁. T_A = T₁(P_A/P₁)/β or TA = 227° K (-46° C). This is conflict with the solution implied for the specific case.

For the more general case, if I let ε = the volume ratio V_B/V_A and note that P_B can not be greater than P_A. I get ε_max = (1- β)/β to avoid P_B being greater than P_A.

I now have two parameters that can be varied to explore the variations in the two vessel temperatures. These parametera are β and ε. When I put in values to β and ε, the largest value I get for T_B/T_A is 1. The highest value for T_B is T1 or 27° C (300° K). This does not seem like a reasonable solution in light of the value given for the specific case but unless my selected equations are wrong or I've made yet another math error, I don't seem how I can get anything else.

Thanks,

Jim

OK, this is my 5th post today, and for the 5th time I have to say NO :-(

The equation with k coefficent is valid only for adiabatic AND QUASISTATIC process of an ideal gas.

Now I have to sit and calculate something to write some more constructive comments :-)

Consider the figures below. There is cylinder with two pistons, one on each end. In the middle somewhere is a rigid wall that separates the cylinder into to separate spaces. The total of the two volumes is the same as volume of the first vessel in the challenge questions (V_A). Initially the pressures on both sides of the wall are the same and the temperature on both sides is the same, namely T1. The mass fraction of the gas on the left side is β and on the right side (1 - β). Since the temperatures and pressures on both sides of the wall are the same, the gas density is the same and the length of the enclosed volumes follow the same ratio. Next, the left piston is moved to the left until the volume on the left is V_A. If I do this very slowly and there is no heat transfer to the cylinder or wall, isn't this a reversible, adiabatic process? If so, then my approach above is valid. The piston on the right side moves very slowly until the trapped volume is V_B. Again, this would appear to be reversible, adiabatic process and the second half of my appoach above would seem to be valid. If I slowly move the pistons back to their original locations, I would end up with the same initial conditions.

Where did I go wrong?

Thanks,

Jim

In our case we have expansion into vacuum, that means there is vacuum on the outer side of a piston. If there is no friction gas will expand not doing any work until something will stop the piston.

In your explanation, holding the piston in a way, that it moves very slowly means, that you have to apply a force equal to the one applied by a gas (pressure*piston surface). So the gas do work against the pressure which is always equal to its pressure.

summarising, the process you described is quasistatic, but it is not the process described in this challenge.

Hope this helps,
Kajot

So what you are saying is the process is adiabtic but not reversible?

Thanks,

Jim

So what you are saying is the process is adiabtic but not reversible? exactly!

Jim

At the equilibrium condition, the fraction β in tank A is a constant mass which has expanded adiabatically following the law PV^k = constant: The temperature of the gas moving into B is a has been affected by compression, and mixing with the gas already present in B. The compression in tankB is therefore not adiabatic

For this reason, you need to use the fact that the enthalpy remains constant which is PA1*VA = PA2*VA + PB2*VB.

And again I can not agree:-(

PV^k = constant is not true because process is not quasistatic. See my comments above replying to jim's question. Kajot

Let a be the subscript for vessel 1 and b be the subscript for vessel 2

If no energy is entering or leaving the vessels then looking at the case where Pa=Pb and Va=Vb and the valve is closed just after Pa=Pb then we surely can say that

MaTa = MbTb where Ma + Mb = M and taking V=1 and M=1 and we may because this won't matter we get P=2490

Giving us MaTa = 150 and MbTb = 150

But we still do not know how many moles ( Ma ) is in Va or how many moles ( Mb ) is in Vb

If we kept open the valve and waited until everything settles down we have Ma = 1/2 and Mb = 1/2 so that Ta = Tb = 300K = 27 degrees C

(This post is by the author of this challenge)

Everyone seems to be having trouble with the simple(?) case so I'll show how it is solved and hope that there is enough interest left for the general case.

Let subscripts a,b apply to the vessels; and 1,2 denote the initial and final states.

Since no heat is lost or gained (constant enthalpy H)

Ha1=Ha2+Hb2 and as H=M*Cp*T for an ideal gas then

Ma1*Cp*Ta1 = Ma2*Cp*Ta2 + Mb2*Cp*Tb2

writing the M as P*V/(R*T) reduces this to Pa1Va=Pa2*Va+Pb2*Vb which for the simple case gives Pa2 (= Pb2) = Pa1/2

Adiabatic expansion of gas in vessel A

The gas present in Vessel A at state 2 has expanded from Pa1 to Pa2 adiabatically. Its temperature has therefore fallen to the value given by

Ta2=Ta1*(Pa2/Pa1)^((y-1)/y) where Ta1=300K; Pa2/Pa1=.5 from above; and y=1.4 (given)

Ta2=300*(.5)^.286 Ta2 = 246°K = -27°C

Constant mass

Ma1 = Ma2+Mb2 from which Pa1*Va/(R*Ta1) = Pa2*Va/(R*Tb2) + Pb2*Vb/(R*Tb2)

Remember in the simple case Pa2/Pa1=.5; Pb2/Pa1=.5; and Va=Vb so this equation reduces to

1/Ta1 = .5/Ta2 + .5/Tb2 and putting in the known values 1/300 = .5/246 + .5/Tb2

From which Tb2 = 384°K = 111°C

Note that the temperature rise in vessel B is greater than the temperature fall in vessel A. This is because there is less mass of gas in vessel B and so it raises to a higher temperature to absorb the heat lost by the greater mass in vessel A

I have to point out few mistakes in your solution:

1.
Ha1=Ha2+Hb2 and as H=M*Cp*T for an ideal gas then
That is not true, enthalpy has no absolute scale, you can not say, that it is 0 (or any other value). The equations above should read: Delta(H)=M*Cp*Delta(T)

Ma1*Cp*Ta1 = Ma2*Cp*Ta2 + Mb2*Cp*Tb2
writing the M as P*V/(R*T) reduces this to Pa1Va=Pa2*Va+Pb2*Vb which for the simple case gives Pa2 (= Pb2) = Pa1/2

You can have this solution only if Ta1=Ta2=Tb2, bot how do you know that? (It's true, beacuse it is ideal gas but you have to show it)

If you want to calculate changes of enthalpy separately for both vessels you have to remember, that they are changing mainly due to mass transfer (which is not included in my above equation for calculating Delta(H)), because they separately ARE NOT closed systems.


2.
Adiabatic expansion of gas in vessel A
Ta2=Ta1*(Pa2/Pa1)^((y-1)/y) where Ta1=300K; Pa2/Pa1=.5 from above; and y=1.4 (given)

NO, NO, NO!
The above (Piosson's) equation (I wrote it before) is only true for quasistatic processes. Expansion into vacuum is not quasistatic!

3.
Constant mass
OK, mass is constant, but point 2 is wrongly calculated, thus Ta2 is false, so Tb2 is also false

Quite destructive post:-) I will try to find a solution and post it tommorrow.
Because I have never played paintball, I was wrong stating earlier, that the gun gas conatiner is filled from a vessel with condensed gas. The ga is only highly compressed.

Hi Slideruler,

I am stuck at your first section where you state that H=M*Cp*T. Cp applies to the case where the gas is at constant pressure when the heat is applied (baloon or expandable container), while Cv applies to the case where the volume is constant and the presure varies (solid container). Given that in this scenario the pressure is not constant and the volume occupied by the gas also changes I am not sure that either of the C values can be used directly. Can you please explain why you used Cp in this case?

Bob

Sorry for my lateness in replying, but I have been computerless for a few days. I hope this helps.

(specific) Enthalpy (sometimes called "total heat") is the energy contained in a (unit) mass of gas. It comprises pure thermal energy U, and potential mechanical energy P*V. (both measured above Tabs=0) Thus enthalpy H is can be defined as U + PV

For an Ideal gas U=Cv*T and P*V = R*T; H=U+P*V; H=Cv*T + R*T: H=(Cv + R)*T

But for an Ideal gas R = Cp - Cv which leads to H = Cp*T

Hi Slideruler,

OK. Given your equations the max temperature that can be reached is 240 kelvin or 147 deg celsius. The temperature of B increases as the pressure of B is reduced, ie, the smaller the amount of gas that is allowed into B, however, this increase tends to a limit of 420 kelvin. I calculated this by an iterative process using your equations above. This limit seems to be independent of the volume of B.

420K?

Yes 420K I transposed the numbers and accidentally said 240K in my earlier post.

You wrote:

It is easy to find the Temperature in B for a specific case. For example If V_A = V_B and the valve is left open until P_A = P_B then T_B will be 111°C.

Now, suppose instead of 2 vessels connected like A and B, let's say you have a series of identical vessels connected in the same manner. According to your example, having opened the valve between A and B, the gas in vessel B is now at 111 C.

OK, now close the valve between A and B and open the valve between B and C. According to your logic, some of the gas from B should enter in vessel C and become hotter by doing so. If I followed your math correctly, the gas that has entered C should now be at 126 C.

So keep going. Close the valve between B and C and open the valve between C and D. Now some of the gas will enter D and become hotter by doing so.

This can go on as long as you like. Each time the gas gets hotter, even though no work has been performed and there are no external sources of heat. You start with gas at 27 C and in just a couple gas transfers, with no work done, have gas at a temperature nearly 100 C hotter.

Unless you have Maxwell's Demon working for you, this defies logic. I think there is a flaw in your reasoning, centered on what happens to the gas when it enters the vacuum in the adjoining vessel.

Coming to Slideruler's defence, (though he may not need assistance from lesser folks like us) I understand Maxwell's demon working for Slideruler is the mass of gas getting hotter gets smaller at each stage.

Btw Slideruler is it not time now that you upgrade your status to Calculator!!

From Wikipedia's explanation of the term "Maxwell's Demon" (quoting J C Maxwell):

"... if we conceive of a being whose faculties are so sharpened that he can follow every molecule in its course, such a being, whose attributes are as essentially finite as our own, would be able to do what is impossible to us. For we have seen that molecules in a vessel full of air at uniform temperature are moving with velocities by no means uniform, though the mean velocity of any great number of them, arbitrarily selected, is almost exactly uniform. Now let us suppose that such a vessel is divided into two portions, A and B, by a division in which there is a small hole, and that a being, who can see the individual molecules, opens and closes this hole, so as to allow only the swifter molecules to pass from A to B, and only the slower molecules to pass from B to A. He will thus, without expenditure of work, raise the temperature of B and lower that of A, in contradiction to the second law of thermodynamics."

Dear Guest,

Thanks for introducing me to an area I had never heard of before. When i made my comment I did not have any idea of JC Maxwells thought experiment.

Anyway I never had any illusion that 2nd law of thermodynamics can be contradicted.

Still I do not agree with your logic. I would like to align myself with the crowd who promote the idea that Joule–Thomson effect is more applicable in the scenario.

BTW why are you acting guest. CR4 needs more persons of intellect like you!

So my point in mentioning Maxwell's Demon is the idea that the gas in vessel A has a certain average velocity, and then the valve is opened allowing some of the gas to flow into vessel B. Since vessel B contained a vacuum, the gas from A flowed into B without doing any work. (Had there already been a small amount of gas there, the gas in B would have been compressed and this would have required work being performed.)

Since the gas from A flowed into B without doing any work, the average velocity is unchanged. So both A and B contain gas with the same average velocity, and thus both A and B are at the same temperature.

There is no Maxwell's Demon allowing only the fast-moving gas particles to pass through the valve. Once the pressures in A and B have come to equilibrium the two vessels will contain gas at the same temperature.

First - congratulations on a good problem - I gave it a five-star rating.

However, I agree with Kajot that the gas in vessel A is not expanding adiabatically. Neither is the gas in vessel B contracting adiabatically. (Note: entropy is not conserved => not adiabatic)

I'll try to explain my thinking:
Imagine that, instead of two vessels you have a single cylinder with a piston, if you move the piston slowly there will be mechanical work exchanged between the piston and the gas, so the adiabatic rules will apply. If you were to withdraw the piston substantially faster than the speed of sound and then stop it instantaneously, the gas would not be doing any work against the piston, and the temperature after settling would be the same as it was originally. Of course that will also apply to this system if the gases in the two cylinders are allowed to mix after the process is complete. Intermediate piston velocities will give intermediate results. Calculations for different cylinder velocities are relatively straightforward. Yours is a more difficult problem - I think it may be done by using a gas-jet velocity as a substitute for the cylinder velocity - but even there there is space for errors in the energy of the escaping gas.

It's an interesting enough problem that I doubt I shall be able to resist giving it my best shot. But as I'm starting from nearly nowhere and I'm short of time, that won't be for a few weeks yet.

However, I agree with Kajot that the gas in vessel A is not expanding adiabatically

I wrote, that the expansion is not QUASISTATIC. It is adiabatic if we assume no heat exchange.

OK. What I was trying to say was merely that the fact that it is "adiabatic" in the literal sense does does not necessarily mean that we can use the standard equations for adiabatic gas expansion.

How do you get the temperature of a perfect vacuum inside a perfect insulator, from which to measure a rise? There are no atoms/molecules to have energy levels.

We are calculating the temperature of gas from vessel A which has moved to B. Of course you can not measure the B temp at the begining if there is vacuum.

So, if there is no starting temperature, then there is no rise, just a final temperature in both less the A's starting absoute temperature due to pressure drop, assuming no frictional heating and perfect insulation.

that is what I tried to write here few times. For ideal gas there will be no drop, for more real models there will be drop of temperature.

Now I am writing answer to the answer which is wrong in my opinion. I want to strongly justify it, so it will take some time, but all my objections were already mentioned here (non quasistatic process, wrong equation for anthalpy).

Maybe this will help: The Joule expansion

What we have here is exactly the Joule expansion process, beacuse no work is done.

I really hope this will help,
Kajot

Oh,yea,I forgot to answer the question.All that can be said given the information of the OP is "lower" than "A" starting temp.

The reason for my answer:1= Boyles law. Expanding gas has a lower temperature than when it is compressed.The heat required(since this is a closed system) is eventually surrendered by the temperature in the gas in B.,thereby spreading the temperature over a greater area, resulting in less temperature overall.

Incidentally, the coldest place in our universe is an expanding gas cloud thrown off by a supernova explosion.It is barely above absolute zero.

The second tank is going from almost no pressure to more pressure. The temp will go up.

Is this a reasonable point of view?

Indeed silvCrow, totally reasonable, as the "vacuum" in B is spurious, or it's a remarkably novel gun.

I'd suggest what is happening is the new gas injection is compressing the residual of the last shot. The new problem is to quantify the stage of the propellant canister between liquid and gas, including an agitated mix, and the kinematics of a fast mass flow. Even so, and I think kojot will concur, it's under stated temps, so; I'd consider partial combustion of the mix. E.g. O2 and CO contamination in CO2 propellant, or O2 and NO in a nitrogen propellant. As triggers; consider the extremely high Reynolds number in a closing valve, or reactions with evolved plasticizer in the seals, or reactive lubricants, or blow-back products. What is certain is a paint ball gun is not a perfect adiabatic or isothermal, or uncontaminated apparatus – nor is there a vacuum.

OK, I will concure:-)

The problem is, that in the real case the gas IS NOT ideal. If the gas is ideal the temperature can not change, the joule-thomson coefficent is equal to 0.

I have to find some time to calculate the process using one of more "real" state equations for gas.

Can you tell me that really when you are filling a gun it is going burning hot???
I know that canister is going cold (of course, again, the gas it is containing can not be treated as ideal), but heating of a gun is still problematic for me...

Hi kajot, this link may help to illustrate the expansions, inter-cooling and re-compressions that take place - http://www.zdspb.com/media/tech/animations/ion_superrender3.gif

This is a second generation gun, first generation ones have more bits and perhaps more friction/heat, but I can't find a schematic or animation. Bolt friction is a "new" factor.

Perhaps the bolt return chamber outer sleeve gets hotish?

Good luck with the sums.

I believe that heating of B and cooling of A will be seen with an ideal gas. Gas in A doing work on gas in B will happen with ideal gas just as much as in non-ideal. The main effect of the diatomic gas postulated by slide-ruler will be to reduce the temperature differential (only one degree out of five contributes to the effect, whereas in a fully ideal gas it would be one degree out of three).

We could extend this to a two-stage process:
Open valve, allow gas to flow from A to B, but no mixing: entropy increases (the situation is not reversible), B is hotter than A, but total kinetic energy in the gas is unchanged.
Allow mixing: entropy increases further, temperature reverts to original.

I have to stress it once again.

Gas A CAN NOT do any work on gas B, until they are separated by a movable piston.
How can you compress gas B while its molecules can freely move to vessel A?

Stressing does not make it correct. This is a problem generated by the terminology of ideal gases. But the proper definition includes "no interaction other than by collisions". A region of gas that is longer than a few mean-free-paths can be treated as if it were a separating membrane). See also my reply to your number 68.

hmmm... I don't get it. This membrane. Can it prevent two gases from mixing? Or separate two regions (in one container) with different pressures?

I will try to use arguments, not capital letters in my future comments:-)

This idea is only an artefact to demonstrate a possible process. The fixed location of the vent means that for this process it doesn't happen like this everywhere (though slide-ruler may well be correct that the end result is similar).

Significant mixing occurs within the 'membrane'. But diffusion is a slow process compared with pressure transmission (sound). (You will recall that the velocity of sound is the uniderectional kinetic velocity, whereas diffusion is a random walk process whose individual steps are at the kinetic velocity).

Re: Your comment about the gas cloud from a supernova being the coldest place in the universe. While that may be true for a few SNRs like the Boomerang Nebula, most are quite hot. As described below.

What are the Stages of a Super Nova Remnant's Life?

The stages of a SNR's life represent an area of current study; however, basic theories yield a three-phase analysis of SNR evolution.

	In the first phase, free expansion, the front of the expansion is formed from the shock wave interacting with the ambient InterStellar Medium. This phase is characterized by constant temperature within the SNR and constant expansion velocity of the shell. It lasts a couple hundred years.
	During the second phase, known as the Sedov or Adiabatic Phase, the SNR material slowly begins to decelerate by 1/r(3/2) and cool by 1/r3 (r being the radius of the SNR). In this phase, the main shell of the SNR is Rayleigh-Taylor unstable, and the SNR's ejecta becomes mixed up with the gas that was just shocked by the initial shock wave. This mixing also enhances the magnetic field inside the SNR shell. This phase lasts 10,000 - 20,000 years.
	The third phase, the Snow-plow or Radiative phase, begins after the shell has cooled down to about 106 K. At this stage, electrons begin recombining with the heavier atoms (like oxygen) so the shell can more efficiently radiate energy. This, in turn, cools the shell faster, making it shrink and become more dense. The more the shell cools, the more atoms can recombine, creating a snowball effect. Because of this snowball effect, the SNR quickly develops a thin shell and radiates most of its energy away as optical light. The velocity now decreases as 1/r3. Outward expansion stops and the SNR starts to collapse under its own gravity. This lasts a few hundreds of thousands of years. After millions of years, the SNR will be absorbed into the interstellar medium due to Rayleigh-Taylor instabilities breaking material away from the SNR's outer shell.

Dozens of possible answers already given leave us in some doubt as to what the max temperature can be. You posed the question with the promise of an answer that now given seems to be in dispute. So who do I believe.

I appreciate you have to 'invent' intermediate conditions in order to arrive at an answer that in most cases will be somewhat academic.

In my line of business (or was - I've retired - but still take an active interest) which is to do with the everyday application of compressed air drying - where water vapour condenses due to the drop in temperature - mainly in the system from a hot compressor - but once cooled and in equilibrium with ambient temperature - there is a need to take account of additional cooling as air expands in the pneumatic devices - such that there is a risk of the device cooling enough for vapour to condense, or if already liquid, for it to freeze. Thus causing the device to fail.

From a drying point of view, the dryer must give a pressure-dew-point that is below the lowest likely temperature.

The similarity here, with this challenge, is that we have vessel A (tank, receiver, bottle) discharging into vessel B which is the atmosphere of infinite volume (by comparison) at 1 bar.

In my experience, calculating the discharge temperature is no better than forecasting the weather.

I can assure you that this challenge was based on experience of what happens in practice.

When the official answer is finally published, I will post my final comments. These may help you decide between the various answers posted.

My uinderstanding was that slide-ruler did not purport to calculate the discharge temperature (which can depend on all sorts of flow issues), but the equilibrium temperature after the valve was closed. That can be far more predictable - the requirement for predictability is that the linear dimensions of the orifice for discharge from A are small compared with dimensions of A.