Expansion wave in a hermetically sealed room Suppose there is a human in a hermetically sealed room who takes in a small amount of air without letting it out. The room is initially at atmospheric pressure. The suction is similar to the case in which a piston is withdrawn to generate an expansion wave in the surrounding media. If I measure the pressure in a region furthest away from the human, will it still be atmospheric? How will equilibrium in pressure be achieved in this case compared to the case when the room is not hermetically sealed (opened to atmosphere)? 
 A: When you breathe in you expand your lungs. This decreases the pressure inside your lungs, and if the room is sealed as you describe it will increase the pressure in the room.
Assuming you keep your mouth open, air from the room flows into your lungs through your mouth until the pressure of the room and your lungs has equalised. At this point there has been no change it the total volume of the air in the room so the pressure is unchanged i.e. the room pressure is still the same as it was before you breathed in.
If you close your mouth then expand or compress your chest the pressure in the room will be changed. The air has to flow between your lungs and the room for the pressure to equalise.
A: It would be easier to answer your question, if a human taking in some air is replaced by an expanding or contracting mechanical object, say a sphere. 
Either expansion or contraction will generate a spherical acoustic wave which will bounce off the walls (ceiling, floor, etc.) a couple of times, before it decays. 
In the new state the pressure everywhere in the room will increase (in case of the expanding sphere) or decrease (in case of the contracting sphere) by a tiny amount compared to the initial state.
A: Let's do a simple analysis. Let quantities with subscript "r" and "l" correspond to the room (excluding the lung) and the lung respectively. Since in a sealed room total mass of air $m_0$ must remain constant, we have: $$m_r+m_l=m_0$$
Assuming that temperature is uniform and steady, ideal gas law gives:
$$p_rV_r=m_rRT\\ p_lV_l=m_lRT\\ \therefore\quad p_rV_r+p_lV_l=(m_r+m_l)RT=m_0RT,~\textrm{a constant}\\$$
If initially the pressure was $p_0$ everywhere (same inside room and lung) then $p_0V_0=m_0RT$. Thus: $$p_rV_r+p_lV_l=p_0V_0$$
Now we assume $V_0=V_r+V_l$; this is a good assumption as far as breathing is concerned. Above equation becomes:
$$(p_r-p_0)V_0+(p_l-p_r)V_l=0$$
If $p_0$ is atmospheric pressure, the question whether room pressure is above or below atmospheric subsequent to inhalation concerns the sign of $(p_r-p_0)$. Its sign is opposite to that of $(p_l-p_r)$. If pressures inside the room and lung were equalized at the end of inhalation, i.e. $p_l=p_r$, then room pressure remains atmospheric, i.e. $p_r=p_o$. Maintaining unequal pressure requires external force on the lung, and a non-zero pressure difference can result if breathing was stopped before pressures were equalized. In this case, higher (lower) lung pressure compared to that of room coexists with below (above) atmospheric pressure in the room.
To obtain the result for a room which isn't sealed but open to atmosphere (or a sealed room which is very big compared to the lung), we take the limit $V_0\to\infty$. Since $(p_l-p_r)V_l$ is a finite quantity, we must have $p_r\to p_0$. Therefore in a room which isn't sealed pressure will practically always be atmospheric.
