Pressure versus Volume system problem

A rigid air tight container of volume 1.5 m$^3$ is filled with a gas of density 8 kg/m$^3$. The container has a density meter that allows us to watch the changes in density in the gas in the container. Inside the container there is an rubber ball (example, spherical balloon) containing a different gas. We don't know the density or the volume of the inner ball. The membrane of the ball is perfectly insulated. Cold temperature is injected remotely into the inner ball. While this is happening, after taking a number of measurements from the density meter, we can see that density is dropping at the rate of 0.1 kg/m$^3$/sec. When the meter stops we know that the density inside the inner ball has reached 1000 kg/m$^3$ with a volume of $5*10^{-4}$. How long does the entire event last?

This is what I worked out so far, but I'm not sure if I am off track:

Let:

• $\rho_{b,0}$ and $\rho_{b,1}=1000 kg/m^3$ the initial and final density of the inner ball, respectively.

• $\rho_{c,0} = 8 kgr/m^3$ and $\rho_{c,1}$ the initial and final density of the outer container, respectively.

• $V_{c}=1.5m^3$ (the total volume in the large container) and $V_{c,0}$ and $V_{c,1}$ the gas volumes in the large container respectively.

• $V_{b,0}$ and $V_{b,1}= 0.0005 m^3$ the initial and ending volumes in the inner ball.

• $\frac {d\rho_c}{ dt} = 0.1kg/m^3/sec$ the rate pressure changes in the large container.

As the inner ball temperature drops, the density increases and the volume of the ball decreases. As the inner ball shrinks, that creates a reverse pressure (or vacuum) in the outer gas, causing the density of the outer gas to drop, and thus we see the density dropping in the meter.

Here are other facts that I think we know:

$$\rho_{b,o} V_{b,0} = \rho_{b,1} V_{b,1}\tag{1}$$ since the mass would be constant.

$$\rho_{c,o} V_{c,0} = \rho_{c,1} V_{c,1}\tag{2}$$

$$V_{c,0} = V_c - V_{b,0}\tag{3}$$

$$V_{c,1} = V_c - V_{b,1}\tag{4}$$

Also the rate of increase in the inner ball causes a rate of decrease in the outer ball (not sure if its expressed correctly),

$$\frac {d\rho_b}{ dt} (V_{b,0}-V_{b,1}) = - \frac {d\rho_c}{ dt} ( V_{c,1}-V_{c,0}) \tag{5}$$

I suppose that the only thing that matters here, in order to know "how long", is the final state of the system as a function of $t$. So somehow I need to take the expressions to the final state.

We now do substitutions from (3) and (4) into (5):

$$\frac {d\rho_b}{ dt} ( ( V_c- V_{c,0} ) -V_{b,1}) = - \frac {d\rho_c}{ dt} (V_{c,1}-(V_c - V_{b,0})) \tag{6}$$

We now substitute (1) and (2) and the value for the rate of change:

$$\frac {d\rho_b}{ dt} ( V_c- V_{c,0} - V_{b,1}) = - 0.1 ((\frac {\rho_{c,o} V_{c,0}}{ \rho_{c,1}})-(V_c - (\frac {\rho_{b,1} V_{b,1}} {\rho_{b,0} }))) \tag{7}$$

$$\frac {d\rho_b}{ dt} = - 0.1 \frac { ((\frac {\rho_{c,o} V_{c,0}}{ \rho_{c,1}})-(V_c - (\frac {\rho_{b,1} V_{b,1}} {\rho_{b,0} })))} {( V_c- V_{c,0} - V_{b,1}) }\tag{8}$$

$$\rho_{b,0}-\rho_{b,1} = - 0.1 \int_0^t {\frac { ((\frac {\rho_{c,o} V_{c,0}}{ \rho_{c,1}})-(V_c - (\frac {\rho_{b,1} V_{b,1}} {\rho_{b,0} })))} {( V_c- V_{c,0} - V_{b,1}) } } dt$$

And here is where I am stuck. First of all I am not sure if I had enough information to begin with in order to find all the answers required. But that is not so important. The most important issue is whether the expression I have so far is correct or not. Something tells me that I got myself in trouble.

Many thanks for any help that may be provided.

1. In your 5th "Let" statement "$\frac {d\rho_c}{ dt} = 0.1kg/m^3/sec$ the rate pressure changes in the large container", "pressure" should be "density" and there is a sign error.

2. Your equations 1-4 are correct, but it is important to determine what is known and what is unknown and analyze how many independent equations and unknowns you have. Also, mass of the gas in the ball is 0.5kg.

3. You should add to your list of equations: $\rho_{c,1} = 8kg/m^3 - (t)0.1kg/m^3/sec$. This is very important because you are trying to solve for time. Here you can immediately see that the solution set is limited to values between 0 and 80 seconds since density cannot be negative.

4. I don't see any basis for your equation 5. Instead, the rate of change of the volume of the gas outside the ball equals the negative of the rate of change of the volume of the ball.

5. At equilibrium, the pressure in the container and ball will only differ by any elastic force of the ball.

• Thank you for your help. I've been waiting for some help on this for quite some time. Let me make sure I understand your point 4. Are you suggesting, $\frac {dV_a} {dt} = -\frac {dV_b} {dt}$ ? If the density of A increases as the volume decreases and that causes the volume of B to increase causing the density to B to decrease, why not then: $\frac {dV_a} {dt} \frac {d\rho_a} {dt} = -\frac {dV_b} {dt} \frac {d\rho_b} {dt}$ – Luis Feb 25 '14 at 7:10
• "Are you suggesting, \frac {dV_a} {dt} = -\frac {dV_b} {dt}", yes exactly (although the subscripts should be b and c in terms of your orginal question) because the total volume is constant (rigid container). – DavePhD Feb 25 '14 at 12:59
• "why not then: \frac {dV_a} {dt} \frac {d\rho_a} {dt} = -\frac {dV_b} {dt} \frac {d\rho_b} {dt}" because that is saying that the densities in each container change at equal rates, which isn't true. Instead, use that fact that mass is constant for each of the two gasses, and (volume)(density)=(mass), if you want to solve for rate of change of density in terms of rate of change of volume, or visa versa. – DavePhD Feb 25 '14 at 13:17
• I have posted the answer based on the tips you gave me. Could you please let me know if it right? Many thanks. – Luis Feb 26 '14 at 1:27
• Also, I'm curious. I know the Boyle's law can be applied for one of the containers as the container changes volume and pressure: $VP$ = $V'P'$. However, if the gas was the same in both containers could we make the case that $V_cP_c$ = $V_bP_c$? – Luis Feb 26 '14 at 3:13

We start from point (5) above where things started to go wrong:

$\frac {dV_c}{ dt}= - \frac {dV_b}{ dt} \tag{5}$

$\frac {dV_c} {dt} = m_c / (\frac {d\rho_c}{dt} ) = - \frac {m_c } {0.1} \tag{6}$

\begin{align} \int_{V_c,0}^{V_c,1} {dV_c} = - \frac {m_c } {0.1} \int_0^t dt \tag{7} \end{align}

\begin{align} V_{c,1}-V_{c,0} = - \frac {m_c } {0.1} t \tag{8} \end{align}

$\frac {dV_b} {dt} = m_b / (\frac {d\rho_b}{dt} ) = 0.5 / (\frac {d\rho_b}{dt} ) \tag{9}$

From (5), (6) and (9):

$- \frac {m_c } {0.1} = - 0.5 / (\frac {d\rho_b}{dt}) \tag{10}$

\begin{align} \int_{\rho_{b,0}}^{\rho_{b,1}} {d\rho_b} = 20 m_c \int_0^t dt \tag{11} \end{align}

\begin{align} {\rho_{b,1}}-{\rho_{b,0}} = 20 m_c t \tag{12} \end{align}

And finally with:

$\rho_{c,1} = 8kg/m^3 - (t)0.1kg/m^3/sec \tag{13}$

We now have 3 expressions as a function of time $t$, namely (8), (12), (13). From here on, it should be just algebra.

• How did you go from 5 to 6? It looks like a calculus error. $V_c = m_c/\rho_c$ so $\frac {dV_c} {dt} = -(m_c /\rho_c^2) (\frac {d\rho_c}{dt})$ – DavePhD Feb 26 '14 at 12:01