How can I find the deflation rate of a balloon?

You need the following information to calculate the deflation rate of a balloon.

1. The balloon is deflating because the air is escaping through an orifice. Are we talking about a large orifice and rapid deflation, or are we talking about a slow process? I think rapid deflation can be approximated as adiabatic, while the slow process can be isothermal.
2. You need an "equation of state" for your balloon. To expand a balloon we increase the pressure inside. How much it expands for a given pressure change depends on the properties of the elastomer it is made of. So in some sense, you need an equation that relates the volume of the balloon to the pressure inside. Let us call it the Equation of State (ES) for the balloon. It would look like $P=f(V)$ for some function $f$. In addition to the material the balloon is made of, $f$ also depends on whether this process is adiabatic or isothermal.
3. You need a description of the orifice that letting the air out. That is, we need an equation relating the pressure difference across the orifice to the flow rate, $dn/dt = g(Delta P)$. The function $g$ depends on the geometry of the orifice. It would look very deferent if the air is escaping from the opening of the balloon with a smooth entrance and a pipe-like extension compared to the case of having a hole in a sphere with a sharp boundary. Equations of this type can be found in some engineering fluid mechanics textbooks.
4. You also need an equation of state for the air inside. I think this one can be safely assumed to be an ideal gas, so if the process is isothermal then $V = nRT/P$ is your ES. If it is adiabatic, then you have something like $V=stuff/P^gamma$, with $gamma = C_v/C_p$.

Once you have all of the above information, this is how you could theoretically solve the problem:

• 2 and 4 give you two ES. The first one has two unknown variables $V$ and $P$, while the second one has three $V$, $P$, and $n$. Plug in the first in the second to eliminate $P$, then solve for $V$ as a function of $n$. Let's call this function $h$, i.e. $V = h(n)$.
• Then use chain rule: $$frac{dV}{dt} = frac{dh}{dn} frac{dn}{dt}$$
• Then use 3 to substitute $dn/dt$: $$frac{dV}{dt} = frac{dh}{dn} g(P-P_{out})$$
• Then you can write everything in terms of $V$ and solve the resulting differential equation: $$frac{dV}{dt} = left.frac{dh}{dn}right|_{h^{-1}(V)} g(f(V)-P_{out})$$

Linear Response Solution

Let us see if we can solve the problem in the simple case that all of the response functions are linear and the system stays isothermal. That is $f(V) = P_{out} + k V$, $g(Delta P) = -C Delta P$, and $PV = nRT$. This might be a good approximation for small orifice and low pressure (note: I think the linear form for $g$ is not very realistic, and $Delta P$ is more likely proportional to the second power of the flow rate)

If I haven't made an algebraic mistake, we have $$V = h(n) = frac{-P_{out}+sqrt{P_{out}^2+4k n R T}}{2k} n = h^{-1}(V) = frac{V(P_{out}+kV)}{RT} frac{dh}{dn} = frac{RT}{sqrt{P_{out}^2+4knRT}} left.frac{dh}{dn}right|_{h^{-1}(V)} = frac{RT}{P_{out}+2kV} frac{dV}{dt}=-RTCkfrac{V}{P_{out}+2kV} V(t) = frac{P_{out}}{2k},Wleft(frac{2kV_0}{P_{out}}e^{k/P_{out}(2V_0-CRT,t)}right) $$ where $V(0)=V_0$ and $W$ is the Lamber W function. Let us define the dimensionless time $tau = kCRT,t/P_{out}$ and the dimensionless volume $nu = 2kV/P_{out}$ with the initial condition $nu(0) = nu_0 = 2kV_0/P_{out}$. Then we have $$nu(tau) = Wleft(nu_0e^{nu_0-tau}right)$$ Here is what it looks like:

Is this realistic? I have no idea.

I am sure I have made assumptions that I am not aware of, so please help me improve this answer.