First, we have air density \(\rho(t)\), viewed as a function of time, and the mass of all the air \(M(t)=\rho(t)~V\) in the depressurizing compartment of volume \(V\). The air escaping the spacecraft is limited to the speed of sound, which we (simplistically) assume is a constant \(v_{sound}=343.15\text{ m/s}\). The mass loss is just the mass flow rate out the hole (of area \(A_{hole}\)). So, the air's mass is described by the following ODE:\[
\frac{d}{d t} M(t) = -\rho(t) ~ A_{hole} ~ v_{sound}
\]
Using the above relationship between \(M(t)\) and \(\rho(t)\), we can substitute in for one of the variables, see that this is a separable differential equation, and solve for the solution:
\begin{align}
M(t) &= \rho(0)~V~\text{exp}\!\left( \frac{-A_{hole} ~ v_{sound}}{V} ~ t \right)\\
\rho(t) &= \rho(0)~~~~~\text{exp}\!\left( \frac{-A_{hole} ~ v_{sound}}{V} ~ t \right)
\end{align}
For notational simplicity, define that with a constant \(c_2\):
\[
\begin{matrix}
M(t) = \rho(0)~V~\text{exp}\!\left( c_2 ~ t \right)\\
~\,\rho(t) = \rho(0)~~~~~\text{exp}\!\left( c_2 ~ t \right)
\end{matrix}
\text{,}\qquad\text{where}\qquad
c_2 := \frac{-A_{hole} ~ v_{sound}}{V}
\]
Far from the hole, the velocity of the flow (again, taken to be \(v_{sound}\) in the hole) is spread out by the ratio of the areas. That is, if the room has cross-sectional area \(A_{room}\), then the flow far from the hole will be \(v_{sound}~A_{hole}/A_{room}\), or just \(v_{sound}~A_{ratio}\) for short.
We're now ready to look at objects in this flow. An object traveling with speed \(v_{obj}(t)\) in the flow has, by the above, a relative velocity to the flow of:
\[
v_{rel}(t) = v_{sound}~A_{ratio} - v_{obj}(t)
\]
Now recall the drag equation. Here, the force \(f_{obj}(t)\) on an object with mass \(m\), viewed as a function of time \(t\), is computed from the relative velocity \(v_{rel}(t)\). The object has a drag coefficient \(c_d\) and cross-sectional area exposed to the wind \(A_{obj}\).
\[
f_{obj}(t) = \frac{1}{2} ~ \rho(t) ~ (v_{rel}(t))^2 ~ c_d ~ A_{obj}
\]
For notational simplicity, define that with a constant \(c_1\):
\[
f_{obj}(t) = m ~ c_1 ~ \text{exp}\!\left( c_2 ~ t \right) ~ (v_{rel}(t))^2
\text{,}\qquad\text{where}\qquad
c_1 := \frac{\rho(0) ~ c_d ~ A_{obj}}{2 m}
\]
This is a force, and we can convert it into an object's induced acceleration \(a_{obj}(t)\) by using Newton's second law (i.e., divide \(f_{obj}(t)\) by \(m\)).
\[
\frac{d}{dt} v_{obj}(t) = a_{obj}(t) = \frac{f_{obj}(t)}{m} = c_1 ~ \text{exp}\!\left( c_2 ~ t \right) ~ (v_{rel}(t))^2
\]
This is also a separable differential equation, and assuming that the object starts at-rest, the solution is:
\[
v_{obj}(t) = v_{sound}~A_{ratio} - \left( \frac{c_1}{c_2} \left( \text{exp}(c_2 ~ t)-1 \right) + \frac{1}{v_{sound}~A_{ratio}} \right)^{-1}
\]
For the case where the room's volume is taken to be infinity, the atmosphere's mass and density are not affected over time:
\begin{align}
M(t) &= \rho(0)~V\\
\rho(t) &= \rho(0)
\end{align}
The analysis proceeds exactly as-above, but with simpler equations. The final results of both cases are:
\[
\begin{matrix}
\text{Finite Volume} & \text{Infinite Volume}\\
\begin{aligned}
M(t) &= \rho(0)~V~\text{exp}\!\left( c_2 ~ t \right)\\
\rho(t) &= \rho(0)~~~~~\text{exp}\!\left( c_2 ~ t \right)\\
v_{obj}(t) &= v_{sound}~A_{ratio} - \left( \frac{c_1}{c_2} \left( \text{exp}(c_2 ~ t)-1 \right) + \frac{1}{v_{sound}~A_{ratio}} \right)^{-1}\\
v_{rel}(t) &= v_{sound}~A_{ratio} - v_{obj}(t)\\
a_{obj}(t) &= c_1 ~ \text{exp}\!\left( c_2 ~ t \right) ~ (v_{rel}(t))^2\\
f_{obj}(t) &= a_{obj}(t) ~ m
\end{aligned} \quad & \quad \begin{aligned}
M(t) &= \rho(0)~V\\
\rho(t) &= \rho(0)\\
v_{obj}(t) &= v_{sound}~A_{ratio} - \left( c_1~t + \frac{1}{v_{sound}~A_{ratio}} \right)^{-1}\\
v_{rel}(t) &= v_{sound}~A_{ratio} - v_{obj}(t)\\
a_{obj}(t) &= c_1 ~ (v_{rel}(t))^2\\
f_{obj}(t) &= a_{obj}(t) ~ m
\end{aligned}
\end{matrix}
\]
