Sciencemadness Discussion Board - Calorimetry problem…

Calorimetry problem…

blogfast25 - 8-3-2011 at 10:36

The following is a bit of a thought experiment, the results of which could impact seriously on real life calorimetric experiments I’m conducting.

Imagine a bomb calorimeter closed system, perfectly thermally insulated and with an internal volume of V. It contains n moles of reactive gases, at pressure p1 and starting temperature T1.

By means of some device (a spark for instance) the gases are made to react in an exothermic reaction to n moles of reaction product gases (assume the reaction complete and reagent gases and reaction product gases to be close to ideal gases), for instance: H2 (g) + Cl2 (g) --- > 2 HCl (g) (n1 = n2 = 2 moles of gas).

During reaction an amount of reaction enthalpy ΔH is released. Since as no mechanical work W has been done (ΔV = 0), the internal energy U has been increased by ΔU = ΔH (1st Law of Thermodynamics for the case ΔW = 0).

It is known that ΔH = n Cv (T2- T1) with T2 the end (equilibrium) temperature, and Cv the molar heat capacity at constant volume for the reaction product gases.

The new equilibrium pressure p2 can now be calculated from p1/T1 = p2/T2 or p2 = T2/T1 x p1.

But in my thought experiment things go wrong: an alarm goes off to tell me the new pressure is dangerously high and to avoid disaster, I open a relief valve and allow gas to escape until the pressure inside the calorimeter has been reduced back to p1, the starting pressure.

My question is simple: how much of the internal energy U2 (ΔU = U2 – U1) is carried off by the escaping gas?

The first problem I can’t get my head around is: what is the nature of the transformation that takes place when the gas is released? It can’t be isobaric because obviously the pressure inside the calorimeter decreases. Then I was thinking ‘adiabatic’ because the calorimeter is perfectly insulated but of course heat does escape because with the valve open, it becomes an open system and hot gas escapes from the calorimeter. Can it be isothermal?

Answers on a postcard, please!

Ozone - 8-3-2011 at 11:25

None, so long as all of the heat has already been transferred...

OK. Where W = 0, DH=nCvDT = DU=mCvDT (and is only true for pure compounds). Standardize Cv for your bomb using a standard (benzoic acid, for example) and calculate Cv. Then bomb a known mass of your sample (we 1 g total sample under 19 atm O2 in an Parr 1103 oxygen bomb; not to exceed 8000cal and/or 40 atm O2).

We measure the DT in 2 L volume of pure water, and because our system (Parr 6200LE) is isoperibol, it also measures differential heat leakage in a temperature controlled jacket surrounding the water bucket. Once DT = 0 (dynamic mode) the experiment is concluded and the DU calculated. Escaping gas has nothing to do with this...unless the sample is burning while pressure is allowed to escape thus violating w=0.

If done correctly, in a bomb made for this purpose and in good repair, it *will not* explode. Measure DT and forget about it. Do the expermient in a hood or behind a shield if you are that concerned.

Cheers,

O3

[Edited on 8-3-2011 by Ozone]

blogfast25 - 8-3-2011 at 13:29

Thanks for the info, Ozone, but you seem to be answering your own question, not mine. You’re basically giving me the operating manual of a Parr bomb calorimeter.

Your very first point for instance: well, no; in my thought experiment the calorimeter is perfectly thermally insulated and so after the ‘burn’ and reaching of equilibrium the temperature inside the bomb is T2, at p2 (and would stay like that forever w/o intervention). Now I open the valve and bring the pressure back to p1. Hot gas escapes, taking with it heat. How much heat escapes?

Put slightly differently, what’s the end temperature of the gas remaining in the bomb after adjusting pressure from p2 to p1 by release of a sufficient amount of gas?

You will argue that that’s not how calorimetry is done but that’s beside the point.

watson.fawkes - 8-3-2011 at 19:45

Quote: Originally posted by blogfast25

My question is simple: how much of the internal energy U2 (ΔU = U2 – U1) is carried off by the escaping gas?

You've got two kinds of energy being carried off here: thermal (internal) energy and mechanical (external) energy. So to say a priori that you're only concerned with internal energy is going to lead you down the wrong conceptual path. The gas that escapes could move a piston (say), so it's doing mechanical work. There's also heat lost through the heat capacity of the gas released. You can consider your process iso-volumetric (by ignoring everything on the other side of the vent), so you can use the constant-volume heat capacity cV.

To get the right answer, start by writing down a differential equation for energy balance whose key variable is the molar flow dn/dt through the vent.

P.S. Constant volume processes are called "isochoric", a word I can never manage to remember.

[Edited on 9-3-2011 by watson.fawkes]

Ozone - 8-3-2011 at 21:26

If the escaping gas is hot, it is not at equilibrium...unless the water is also hot (which it most likely is not).

Good luck deconvoluting that over the 19 atm (in my case) of O2 that you added in the first place (which, granted, is do-able if you assume perfectly adiabatic conditons

).

Cheers,

O3

blogfast25 - 9-3-2011 at 08:25

Watson:

Funny that: mulling the problem over late at night I concluded that reducing the problem to internal energy alone would lead nowhere. I also concluded the starting point should be a differential equation based on matter escaping the system: dn/dt. But I haven’t had time to elaborate any further.

‘Isochores’ is what iso-volumetric transitions are called in Dutch. I wasn’t sure the term applies also in English.

Quote: Originally posted by Ozone

If the escaping gas is hot, it is not at equilibrium...unless the water is also hot (which it most likely is not).
O3

Firstly if the hot gas was at equilibrium to begin with, the only condition for the release to be reversible is to conduct the transition infinitesimally slowly. But I’m not sure reversibility/non-reversibility plays a part here. I still think you’re kind of looking at the wrong problem, frankly. Here, let me simplify it:

A perfectly thermally insulated gas cylinder of constant volume V is filled with n moles of hot, near-ideal gas. After equilibrium is achieved the conditions are thus p, V, T (and n). A valve is now opened and slowly the gas pressure is reduced to p’. Relevant questions are: how many moles of gas are released before pressure reaches p’? How much heat H was carried off by the released gas and how much mechanical work W did it carry out?

I find it rather amusing that this seemingly simple and potentially real world problem seems to defy easy solution somewhat!

[Edited on 9-3-2011 by blogfast25]

entropy51 - 9-3-2011 at 10:11

Quote: Originally posted by blogfast25

I also concluded the starting point should be a differential equation based on matter escaping the system: dn/dt. But I haven’t had time to elaborate any further.

When I worked in process dynamics modeling we would have used

M(dh/dt) = Win hin - Wout hout - V dP/dt - h dM/dt + Q

where for your problem Win = mass flow rate into system = 0 and assuming good insulation the heat transfer rate Q = 0

and you need an expression for presssure as a function of density and energy of the gas (the equation of state). We usuallly used tabular data or numerical fits instead of assuming ideal behavior.

You also solve for mass in the system as a function of time:

dM/dt = Win - Wout

Since volume is constant, the density as a function of time is

rho = M/V

And of course the flow Wout is some function of pressure P, depending on the hydraulic characteristics of the safety valve.

We solved large systems of these types of equations numerically using Runge-Kutta or similar integration.

[Edited on 9-3-2011 by entropy51]

[Edited on 9-3-2011 by entropy51]

blogfast25 - 9-3-2011 at 10:40

Could you please specify the definition and units? I assume everything is mass based, not molar based...

In one of my practical text books on industrial termodynamical processes ('heat machines' to put it bluntly), h (little H) is defined as mass specific enthalpy (h = H/M).

Also Win would be mechanical work delivered into the open system, not 'mass flow rate into the system'.

entropy51 - 9-3-2011 at 10:53

h = enthalpy, BTU/lb or cal /gm
W = mass flow rate, lb/sec or gm/sec
M = mass in system, lb or grams
P = pressure, needs converstion factor (144/778 for p in psi)
t = time
rho = density lb/ft3 or gm/cm3
V = volume, ft3 or cm3
Q = heat transfer rate, BTU/sec or cal/sec

A fairly complex application of this type of analysis to a real world problem is described in detail here.

[Edited on 9-3-2011 by entropy51]

blogfast25 - 9-3-2011 at 13:54

Thanks, entropy, that clears that up. Thanks also for the *.pdf. From Batelle, eh? Very prestigious indeed!

The equation:

M(dh/dt) = Win hin - Wout hout - V dP/dt - dM/dt + Q

… is not dimensionally consistent though. In SI units all terms would be J/s, except for dM/dT, which is in kg/s!

watson.fawkes - 9-3-2011 at 15:09

This is the sort of problem where you don't have to know the whole dynamics in order to get the desired answer. If you want to know how fast the gas escapes, that takes all the apparatus entropy51 mentions. This one, though, can be "solved" solely through conservation of energy.

The total energy in this kind of system is called (by convention) the enthalpy H = U + PV. For an ideal gas, its internal energy U = cVnRT. Its external energy is PV = nRT. Thus H = (1+cV

PV. Since V is constant, H is proportional to P. Thus the final energy H2 = H1 p2/p1. (You can do all this with dn/dt, which then cancels out.) From this analysis alone, you don't know how much gas escaped (Δn) or what the final temperature is (ΔT). As a result, the energy remaining in the vessel is partitioned in some as-yet unconstrained way between internal and external energy. All this is to say that the problem as stated is insufficient to provide the answer you were seeking. In other words, there are multiple values for n2 and T2 that satisfy all the stated premises of your question.

To get the full answer, you need all of the constitutive equations, not just the two I used above. This key here is the state equation entropy51 mentions.

entropy51 - 9-3-2011 at 15:26

Quote: Originally posted by blogfast25

Thanks, entropy, that clears that up. Thanks also for the *.pdf. From Batelle, eh? Very prestigious indeed!

The equation:

M(dh/dt) = Win hin - Wout hout - V dP/dt - dM/dt + Q

… is not dimensionally consistent though. In SI units all terms would be J/s, except for dM/dT, which is in kg/s!

Sorry, it's h(dM/dt) which I corrected above.

blogfast25 - 10-3-2011 at 08:10

Thanks both Watson and entropy.

I’ll be reading the Batelle paper tonight and will try and relate it to the equation you quoted (I can’t at this point see where it comes from, although I’m sure it’s based on conservation of mass and energy).

M(dh/dt) = Win hin - Wout hout - V dP/dt - h dM/dt + Q

Which can be simplified to:

M(dh/dt) = - Wout hout - V dP/dt - h dM/dt

Isn’t Wout (acc. entropy’s symbol definitions) simply dM/dt?

------------------------

Like Watson I’ve been revisiting the basics again to refresh understanding on H = U + pV and Joule’s "U is a function of T only" and thus it can be shown easily that:

ΔU = n Cv ΔT, no matter the nature of the transformation.

One thing that’s missing in Watson's reasoning (I think!) is the work that the system has to exert to expel the gas. Assume the surrounding air pressure is p0 (with p > p0) and that at t the gas flow is dn/dt, with the gas at pressure p and temperature T. During an interval dt, dn moles escape and the mechanical work needed for this is (correct me if I’m wrong):

dW = (p – p0) dV and from pV = n R T, dV = RT/p . dn, or dW = RT [(p – p0) / p] dn

As regards the equation of state, isn’t that complicated by the fact that this isn’t a steady state problem? If it was we could invoke Bernoulli.

The function dn/dt must be a continuous function with negative gradient over the range (dn/dt)start – (dn/dt)end, approaching dn/dt = 0 asymptotically, when p = p0 and dW = 0. An exponential or power function of sorts…

[Edited on 10-3-2011 by blogfast25]

blogfast25 - 11-3-2011 at 07:56

Going totally back to basics, I’ve developed a time-independent 1st order differential equation of the kind f(p) dp/dT = 1/T for this problem but I’m not sure if it’s correct. Here it comes:

1st Law of Thermodynamics: dQ + dW = dU (dQ and dW are positive if they are inputs into the system). For our case: dQ = 0, so dW = dU: all lowering of internal energy U is due to work performed by the system.

At a given moment in time we have n moles in the system, at pressure p and constant volume V. The pressure outside the system is a constant p0. Assume also that the flow rate is very slow so that equilibrium is maintained inside the system at all times.

During an infinitesimal interval dt, dn moles of gas are expelled at pressure p and temperature T. The volume expelled is dvout = RT/p dn and the work performed by the system to expel the gas is dW = (p – p0) dvout, so:

dW = (RT/p) (p – p0) dn (eq. 1)

Now with pV = nRT, differentiate both sides:

pdV + Vdp = RT dn + Rn dT and with dV = 0:

Vdp = RT dn + Rn dT and:

RT dn = Vdp – Rn dT, insert into (eq. 1):

dW = [(p – p0) / p] . (Vdp – Rn dT) or:

dW = [(p – p0) / p] .Vdp – [(p – p0) / p] . Rn dT (eq. 2)

With pV = nRT, n = pV/RT, insert into (eq. 2):

dW = [(p – p0) / p] . Vdp – [(p – p0) / p] . R (pV/RT) .dT or

dW = [(p – p0) / p] . Vdp – (p – p0) (V/T) .dT (eq. 3)

Subsequently with U = n Cv T, the gas must have cooled down very slightly to supply this work, so:

dU = (n – dn) Cv dT ≈ n Cv dT and with n = pV/RT:

dU = (pV/RT) Cv dT (eq. 4)

Since as dW = dU with (eq. 3) and (eq. 4):

[(p – p0) / p] Vdp - (p – p0) (V/T) dT = (pV/RT) Cv dT, all terms in Joule (J).

Drop V from both sides:

[(p – p0) / p] dp – (p – p0) (1/T) dT = (p/RT) Cv dT

Divide by dT, so dP/dT = p’:

[(p – p0) / p] p’ – (p – p0) (1/T) = (p/RT) Cv

[(p – p0) / p] p’ – (p – p0) (1/T) = (Cv/R) p (1/T)

[(p – p0) / p] p’ = (Cv/R) p (1/T) + (p – p0) (1/T)

[(p – p0) / p] p’ = [(Cv/R) p + (p – p0)] (1/T)

[(p – p0) / {p [(Cv/R) p + (p – p0)]}] p’ = (1/T)

Separation of variables has been achieved.

For clarity, make (eq. 5) equal to (N/D) p’ = 1/T

With N = p – p0

And D = p [(Cv/R) p + (p – p0)]

Clearly this can be integrated mathematically, albeit not without some tedium. If the boundary conditions were p1 -- > p0 and T1 -- > T2, T2 could be calculated by determined integration and the drop in internal energy ΔU could be determined because the end state is unambiguously defined by p0, V and T2.

What say you?

[Edited on 11-3-2011 by blogfast25]

watson.fawkes - 11-3-2011 at 10:29

Quote: Originally posted by blogfast25

[...] and the work performed by the system to expel the gas is dW = (p – p0) dvout

You've got a basic conceptual error here: internal and external energies are not conserved separately, but rather it's their total that's the conserved quantity. In the present example, you've really got two systems: the interior of the calorimetric bomb (an isochoric system) and its exterior. These are connected, by assumption, through an orifice. Now the work done by the inside (its change in external energy) is not the same as the work done to the outside. The total decrease in energy on the inside is equal to the total increase in energy on the outside, but it's not the case that the work done by/to is necessarily the same, no more than the thermal energy transferred from/to is the same.

The upshot is that you can assume all you want about the outside system, but you still need the full constitutive equations of the inside gas to know how it partitions its external vs. internal energy.

blogfast25 - 11-3-2011 at 13:04

Well, my hopes weren’t high nut have now been cruelly dashed!

Quote: Originally posted by watson.fawkes

The upshot is that you can assume all you want about the outside system, but you still need the full constitutive equations of the inside gas to know how it partitions its external vs. internal energy.

And unless someone can point to ‘how it partitions its external vs. internal energy’, I’m lost…

Which is a shame because the obtained function T = F(p, pi, Ti, p0) after integration is neat and simple and seems to have the right shape. I’ll post it here tomorrow after one final proof reading… It would appear I've found the solution to 'a' problem, it just isn't this one!

watson.fawkes - 11-3-2011 at 14:51

Quote: Originally posted by blogfast25

And unless someone can point to ‘how it partitions its external vs. internal energy’, I’m lost…

The partition I'm talking about is in the equation H = U + PV, namely, how much of H is in U and how much is in PV. It boils down to what the final temperature inside is. That gives you the internal energy, the moles of gas left inside, etc. It's also generally path-dependent and depends on the constitutive equations (a.k.a. equations of state).

For example, one relatively common constitutive model is that the pressure is a function of density only, i.e. P &equiv; P( ρ ). In the present case, the interior has ρ = n / V, and this is enough to give you a solution. If you want to work out the differential equation, start with this model. Differentiating the function P gives you a (state-dependent) relationship between dP/dt and dn/dt.

blogfast25 - 12-3-2011 at 06:55

OK, back to the drawing board with that!

For what it’s worth, the previous DE integrated as:

ln {p [α p – p0]^γ} + C = ln T

For pi -- > p, Ti -- > T where (pi, Ti) is the initial state. Determined integral, reworked:

T = Ti (p / pi) {[α p – p0]/ [α pi – p0]}^γ

With α = Cv/R + 1 and γ = - Cv / (Cv + R)

For a monoatomic ideal gas:

Cv = 3/2 R

So α = 5/2 and γ = - 3/5

T declined smoothly with declining p, as a plot showed.

[Edited on 12-3-2011 by blogfast25]