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Insanely vague thermodynamics question

  1. Jan 7, 2005 #1
    I'm seriously struggling with the following question:

    Ideal gas is in the container with a piston as one of the walls. In another wall of the container, there is a little hole, through which gas leaks out of the container. As a result of both heating and compressing the gas, its temperature increases 4 times, and pressure increases 8 times. How does the rate of the gas outflow through the hole change?

    Is there enough information here to solve the problem? I'm thinking no. However, I'm not asking for this problem to be solved. I just need to know if it can be solved with the given information, and any ideas on how I may approach this.

  2. jcsd
  3. Jan 7, 2005 #2
    no way you can solve this problem by the information given above
  4. Jan 7, 2005 #3


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    Yes, you can do it. It is in principal calculable. By applying the first law, you can model all of the phenomena which occur inside this container. Remember the change in internal energy of the gas inside the container equals the sum of (where I'll leave it to the reader to determine sign of the variable) equals the sum of the enthalpy coming in, plus the enthalpy leaving, plus the work coming in, plus the work going out, plus the heat going in, plus the heat going out.

    You can assume there is no heat transfer, and no gas coming in. You are left with the work and enthalpy leaving equal to the internal energy change of the gas inside the container.
  5. Jan 7, 2005 #4
    yes, you can do it. ( if you figure out how to solve an 1 equation, 3 unknown problem...)
  6. Jan 7, 2005 #5


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    Good point... there are 2 ways to do it.
    1) You model it on a computer. You have work in, enthalpy out. Take a guess at both. Does the guess you provided end up with ...

    2) You can create two equations such that:
    version 1 has
    1T (temp)
    1P (pressure)

    version 2 has
    4T (temp)
    8P (pressure)

    Now you have two unknowns and two equations. The unknowns are T and P.
  7. Jan 7, 2005 #6
    you still dont get the question, the unknowns are 1) heat goes in, 2) volume decrease, 3) the time which the action taken place, 4) the original volume, and finally, 5) the gas out flow rate.

    the first four are independent variables and can't be derive from the information given by the question.
  8. Jan 7, 2005 #7


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    vincent. you're wrong. keep thinking about it.
  9. Jan 7, 2005 #8
    Q goest, if you are right, why don't you show me the answer...
  10. Jan 7, 2005 #9
    Allright, here's what I did:

    Since we must find the rate of change of the rate of gas outflow, we must first derive an expression for the amount of gas in the cylinder in terms of the pressure, volume and temperature. Hence:

    [tex] N(t) = \frac{p(t)V(t)}{kT(t)} [/tex] where [tex] p(t) = 8t , T(t) = 4t [/tex].

    This reduces to [tex] N(t) = {\frac {2 V \left( t \right) }{k}} [/tex]

    Thus the general solution is [tex] \frac{2}{k}{\frac {d^{2}V(t)}{d{t}^{2}} [/tex].

    Where have I gone wrong? (if at all)

    PS: Q_Goest, I firmly believe that computer modelling is not needed for this question, considering the fact that it appeared on a physics contest preparation paper.
    Last edited by a moderator: Jan 7, 2005
  11. Jan 7, 2005 #10
    First of all, the question didn't state the temperature and presure increase LINEARLY, so, P(t) is not equal to 8Pt and T(t) is not 4Tt. The question is saying the FINAL TEMPERATURE and FINAL PRESURE is increased by 4 & 8 times. P(t) and T(t) is depend on the heat flows in and volume decrease, which may or may not be constant. Also the outfow rate is not a constant because the tempeture and presure and density of gas inside the box keep changing.. Secondly, you didn't say you know the original volume, temperature and presure of the box.
  12. Jan 7, 2005 #11
    darn, vincentchan you do have valid points there. I have tried looking at the enthalpy of the system and the energy equations, but I still can't connect the dots. :grumpy:
  13. Jan 7, 2005 #12
    As I told you at the very begining, unless you moldify your question, this question has no solution....
  14. Jan 7, 2005 #13
    I agree with vincentchan... but if we assumed the number of moles leaked out of the tiny hole is insignificant, we might be able to solve this using Grahm's law of Gas effusion.

    When uncompressed, and regular temperature...
    PV = (2*nKE)/3

    v = root of 3RT/m

    After compressed and temperature goes up
    V = nR*4T/(8*P)

    8P * nR*4T/(8*P) = 2/3 * 1/2 * mvo^2

    vo = root of 12RT/m

    According to Grahm's law of Gas effusion the rates are proportional to their average speed of the gas particles:

    ro/r = root of (12RT/3RT) <---------- masses cancel...

    so ro = 2 times faster then the initial rate... That's all i can say about this...
  15. Jan 7, 2005 #14
    apchemstudent, your calculations seem correct, but as you stated, they depend on the assumption of gas effusion. The question asks for 'how the rate of gas outflow changes' (essentially a double derivative is involved here), so i'm thinking that there is always a non-zero gas outflow. Ah well. I'll just write down "The rate of gas outflow increases."

    Tks to everyone who posted, for all the help.
  16. Jan 7, 2005 #15
    Here is just a stab at it, but im probably wrong:

    We know that

    [tex] Pv_1=n_1RT [/tex] (at initial conditions.)

    and that:

    [tex] 8Pv_2 = n_2R4T [/tex] (at final conditions.)
    [tex] 2Pv_2=n_2RT [/tex]

    If we subtract these two states:

    [tex] 2Pv_2 -pv_1 = n_2RT-n_1RT [/tex]
    [tex] 2P(v_2-v_1)=RT(n_2 - n_1) [/tex]
    but if we divide by the amount of time t,
    [tex] \frac {2P(v_2-v_1)}{t} = \frac {RT(n_2 - n_1)}{t} [/tex]
    and take the limit of the amount of change to be very small, and the time to be very small, it becomes a derivative and we get that:

    [tex] \frac {2P dv}{dt}= \frac {RT dn}{dt} [/tex]
    which is equal to:

    [tex] \frac {2P dv}{RT dt} = \frac{dn}{dt} [/tex]

    which means that the amount of change that leakes through any sized hole is proportional to the change in volume times some constant, which is 2P/RT.

    But its just a guess. Hope it might help. Oh, I assumed that the pressure and temperature changed Linearly with time so the cancelation was possible, but like vincentchan said, this may not be true. (lets cross our fingers that it is :smile: )
    Last edited: Jan 7, 2005
  17. Jan 8, 2005 #16


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    The only assumption to be made is that, since the hole is said to be "small" compared to area of the wall, the total number of molecules in the container before and after the change of pressure/temp is the same.

    Imagine a small portion of the container wall, area = A. In some short time [itex]\Delta t [/itex] let's calculate how many molecules hit the wall. If the mean velocity of the molecules in the +x direction is [itex]<v_x>[/itex], then all the molecules contained within a volume adjacent to this wall, of depth [itex]<v_x> \Delta t [/itex] will hit the wall. This number is simply the product of this volume with the number density of the molecules, [itex]n[/itex]. So,

    [tex]\Delta N = nA <v_x> \Delta t [/tex]

    But from the Ideal Gas Law, there's

    [tex] P = nkT => n = P/kT [/tex]

    And the average molecular speed is simply

    [tex] <v_x> = \int _0 ^{\infty} v_xf(v_x)dv_x [/tex]

    where f(v) is the Maxwell-Boltzmann distribution function, given by

    [tex] f(v_x) = [ m/2 \pi kT]^{3/2} e^{-mv_x^2/2kT} [/tex]

    Solving the integral gives the well known result (and you can probably use the result without having to derive it)

    [tex] <v_x> = \sqrt{\frac{8kT}{\pi m}} [/tex]

    So substituting for n, <v> into the first equation gives

    [tex] \frac{\Delta N}{\Delta t} = \frac{PA}{kT} \sqrt{\frac{8kT}{\pi m}} ~~ \alpha ~~ \frac{P}{\sqrt{T}} [/tex]

    So this is a measure of the rate of collision against the wall. Clearly if this small portion of the wall were removed (creating a small opening), the above number will give you a measure of the rate of escape, or the leak rate.

    So, if the P--->8P, T---> 4T, the leak rate will increase by a factor of [itex]8/ \sqrt{4} [/itex] or 4 times.
  18. Jan 8, 2005 #17


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    Gokul, it looks as if you're trying to calculate the mass flow out of the cylinder by calculating the number of molecules which will collide with the given orifice area. Although I really am impressed by that method, it's not very accurate. For example, the pressure inside and outside might be equal, such that there are equal numbers of molecules going into the cylinder as there are going out. As internal pressure increases, more go out than in. To calculate the flow rate out, one uses orifice equations. Would you like those?

    I'm not absolutely sure the below is correct, I haven't gone through the numbers, but I believe it should work, if not then some minor adjustments may be required. Also, I'm not going into the integration which should provide a more accurate answer, this is a strictly algebraic solution. Note that iterating as I explain below will give you the integrated answer. Given the amount of time I'm spending here, I'd rather not resolve the entire problem, but this should at least provide a start for you, Freemind.

    The problem states "As a result of both heating and compressing the gas, ... " which implies the piston is compressing the gas in an undefined way. Also, there is no mention of heat flux, so I assume the "heating" they are refering to is the heating due to work being done on the gas.

    This is a first law calculation, I'll use energy leaving the system as negative, and energy going into the system as positive.

    U-U1 = Wi - Ho

    Where U1 = initial internal energy
    U2 = final internal energy
    Wi = Work going into the system performed by the piston
    Ho = Enthalpy out

    Note there is no work done by the system, so no Wo (work out). There is no mass coming in, so no Hi. And no heat transfer so Qi and Qo are zero.

    Now U = m*cv*T
    where m = mass
    cv = Specific heat at constant volume
    T = temperature

    H = m*cp*T
    where m = mass
    cp = Specific heat at constant pressure
    T = temperature

    W = P*A*s
    Where P = pressure
    A = area
    s = distance piston moves through

    so we combine these and obtain:

    m2*cv*T2 - m1*cv*T1 = P(avg)*A*s - mo*cp*T(avg)

    Here I'm going to make the assumption the change in pressure and temp are linear, which is not exact. That's one reason modeling this would provide a more accurate description (ie: integrating or iterating using a computer model). Since mass flow rate out changes as pressure and temperature changes, we can make iterations with small changes in pressure and temperature such that the errors on any iteration are small. If iterating, these equations are used and the pressure and temperature are the pressure and temperature at the beginning and end of each iteration. Note also that if iterating, the mass flow rate out is calculated with each iteration. I haven't provided the mass flow equations for the orifice, and if you really want them I'll provide.

    We also use the following:
    m1 - m2 = mo
    Where m1 = the mass in the cylinder at the beginning of each step
    m2 = the mass in the cylinder at the end of each step
    mo = mass leaving the cylinder, calculated by orifice flow equations

    T2 = 4*T1 (or 4T1)
    P2 = 8*P1 (or 8P1)

    Now these are put into the equation above

    (m1-mo)*cv*4T1 - m1*cv*T1 = ((P1+8P1)/2)*A*s - mo*cp*((T1+4T1)/2)

    Use ideal gas law to give the initial conditions of pressure and temperature (ie: P1, T1)

    You can now determine the equation for the mass out as a function of distance through which the piston moves.

    mo(s) which will result in a 4T and 8P rise.

    Note: This doesn't guarantee a given gas can give you the given rise in pressure and temperature, there may be no solution for some gasses. This simply gives you the basics of how to go about calculating the answer if in fact it's possible.
  19. Jan 8, 2005 #18

    Andrew Mason

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    I would say that this is a mechanical energy question, not a thermodynamic one. It doesn't matter why the pressure increases.

    The rate of gas outflow is dm/dt where [itex]dm = \rho dV[/itex]

    Think of pressure as an energy density (energy/unit volume):
    [tex]P = \frac{dE}{dV}[/tex]

    In order to accelerate an element [itex]dm = \rho dV[/itex] of gas to speed v, you need an amount of energy:

    [tex]dE = \frac{1}{2}dmv^2[/tex]

    where [itex]dm = \rho dV[/itex] So in terms of energy density or pressure:

    (1) [tex]\frac{dE}{dV} = \frac{1}{2}\rho v^2 = P[/tex]

    Now, the density of the gas is a function of its mass and volume, but you don't have to know that because the hole is small so the mass does not change rapidly. So:

    [tex]v \propto \sqrt{P}[/tex]

    Since [itex]dm/dt = \rho dV/dt = \rho Adl/dt = \rho A v[/itex], this means that: [itex]dm/dt \propto vA \propto A \sqrt{P} [/itex]]

    So the rate of gas outflow increases by [itex]\sqrt{8} = 2.83[/itex]

    Correction: This would be correct if the volume did not change. I forgot that the increase in pressure was due to compression as well as adding heat. So there is a thermodynamic aspect here.

    [itex]\rho = m/V[/itex] so [itex]\rho = mP/nRT \propto P/T[/itex] since PV = nRT

    so from (1)

    [tex]\frac{1}{2}\rho v^2 = P[/tex]


    [tex]v^2 \propto P/\rho \propto P/(P/T) \propto T[/tex] !!

    [Further Correction: Since [tex]dm/dt \propto \rho v \propto P\sqrt{T}/T = P/\sqrt{T}[/tex] ]

    So the rate of gas outflow increases by [itex]8/\sqrt{4} = 4[/itex]

    So Gokul4321 was right!

    Last edited: Jan 8, 2005
  20. Jan 8, 2005 #19
    Since this questoin is come from a high school physics contest, I WILL MAKE IT SIMPLE (use the assumtion you guys made, the hole is small....)

    Frist, compress the box into half of its original volume thru an isothermal process (without changing its temperature)
    [tex] P_{a} = 2P_{i} [/tex]
    [tex] T_{a} = T_{i} [/tex]
    [tex] V_{a} = 1/2V_{i} [/tex]
    and most important [tex] \rho_{f} = 2\rho_{i} [/tex]

    Then, increase the temperature of the box by 4 times(Fixed the volume). Using
    [tex]PV = nRT n[/tex] is constant since the hole is small, We have
    [tex] P_{f} = 4P_{a} = 8P_{i}[/tex]
    [tex] T_{f} = 4T_{a} = 4T_{i}[/tex]
    [tex] V_{f} = V_{a} = 1/2V_{i}[/tex]
    [tex] \rho_{f} = \rho_{a} = 2\rho_{i} [/tex]
    [tex] v \alpha \sqrt{P} [/tex] however, the density has double
    the final outflow rate [tex] = 2*\sqrt{8}[/tex] initial outflow rate
    [tex] = 4*\sqrt{2}[/tex] initial outflow rate

    The spirit of a homework helper is not posting out the answer, but you guys made this question too complicate and the answer is the only way to clarify everything.......sorry
    Last edited: Jan 8, 2005
  21. Jan 8, 2005 #20
    where does density*root Pressure come from? Can you really say that density is directly proportional to flow rate? I mean, if the gas is thicker, shouldn't it be harder for it to go out not easier...

    If we're saying that the density does not change, just when it enters the hole ( what vincentchan is assuming)... then:

    v is proportional to sqrt(P/density) = sqrt(8/2) = 2 times initial flow rate...
    Last edited: Jan 8, 2005
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