The following problem occurred to me today:
Suppose a $100\mathrm{cfm}$ fan is pushing air out of a large room which is airtight except for a $10 \mathrm{cm}^2$ hole. The air pressure outside the room is $101.3\mathrm{kPa}$, and the atmosphere consists of $80\%$ nitrogen and $20\%$ oxygen. All temperatures are $300\mathrm{K}$. What is the equilibrium pressure inside the room?
I believe I have given enough information to determine the resulting pressure. My inclination is to take a naïve statistical mechanics approach to the problem. This leads me down the following train of thought:
- Particles are distributed evenly inside and outside the room, with a cubic meter containing $N_i=\frac{P_i}{k_BT}$ and $N_o=\frac{P_o}{k_BT}$ molecules respectively, where $P_i$ is the pressure inside and $P_o$ the pressure outside.
- Velocities are distributed symmetrically and speeds satisfy a Boltzmann distribution, so for a fixed type of molecule $M$ the distribution of $v_z$, the $z$-coordinate of velocity squared, is $$D_M(v_z)=\int_{S^2}\mathrm{exp}\left(\frac{-m_Mv_z^2}{2k_BT\cos\theta^2}\right)\mathrm{d}\theta\mathrm{d}\phi$$ where $m_M$ is the mass of the molecule $M$.
- The number of molecules of $M$ that strike an area $A$ over time $t$ from inside is $$\int_0^\infty Ax_MN_i\int_{d/t}^{\infty} D_M(v_z)\mathrm{d}v_z\mathrm{d}d$$ where $d$ denotes the distance along the $z$-axis of the molecule from the area and $x_M$ denotes the fraction of all molecules which are $M$. Similarly, the number of molecules of $M$ that strike an area $A$ over time $t$ from outside is $$\int_0^\infty Ax_MN_o\int_{d/t}^{\infty} D_M(v_z)\mathrm{d}v_z\mathrm{d}d.$$
- The difference between these two, summed for oxygen and nitrogen, must be equal to the number of molecules of $M$ removed by the fan, which is $.0472\cdot N_it$.
Thus we have $$.0472\cdot N_it=A(N_i-N_o)\int_0^\infty \int_{d/t}^{\infty} \left(x_{N^2}D_{N^2}+x_{O^2} D_{O^2}(v_z)\right)\mathrm{d}v_z\mathrm{d}d$$ so solving for $N_i$ we get $$\begin{align} N_i &=\frac{AN_o\int_0^\infty \int_{d/t}^{\infty} \left(x_{N^2}D_{N^2}+x_{O^2} D_{O^2}(v_z)\right)\mathrm{d}v_z\mathrm{d}d}{A\int_0^\infty \int_{d/t}^{\infty} \left(x_{N^2}D_{N^2}+x_{O^2} D_{O^2}(v_z)\right)\mathrm{d}v_z\mathrm{d}d-.0472t}\\ &=\frac{.1\mathrm{m}^2\cdot\frac{101.3\mathrm{kPa}\cdot \mathrm{m}^3}{1.38e-23 \mathrm{J}/\mathrm{K}\cdot 300\mathrm{K}}\int_0^\infty \int_{d/1\mathrm{s}}^{\infty} \left(.8D_{N^2}+.2 D_{O^2}(v_z)\right)\mathrm{d}v_z\mathrm{d}d}{.1\mathrm{m}^2\int_0^\infty \int_{d/1\mathrm{s}}^{\infty} \left(.8D_{N^2}+.2 D_{O^2}(v_z)\right)\mathrm{d}v_z\mathrm{d}d-.0472\mathrm{s}}\\ &=\frac{2.45e24 \mathrm{m}^{2}\int_0^\infty \int_{d/1\mathrm{s}}^{\infty} \left(.8D_{N^2}+.2 D_{O^2}(v_z)\right)\mathrm{d}v_z\mathrm{d}d}{.1\mathrm{m}^2\int_0^\infty \int_{d/1\mathrm{s}}^{\infty} \left(.8D_{N^2}+.2 D_{O^2}(v_z)\right)\mathrm{d}v_z\mathrm{d}d-.0472\mathrm{s}}\\ \end{align}$$ which I could probably evaluate using Mathematica if my desktop were working, but sadly it is not. From there it would be trivial to calculate $P_i$.
My question is whether my reasoning up to this point is correct, and whether there are any factors I have left out. Additionally, is there an easier way to calculate $P_i$?
