Why is air pressure different between the inside and the outside of my house?

Question

I have two identical barometers. I made sure they were calibrated to each other in the same room. Then, weeks ago, I placed one outside.

Right now, the outside one reads $30^\circ\text{F}$ and $1030\,\text{hPa}$ whereas the inside one reads $66^\circ\text{F}$ and $1020\,\text{hPa}$. These values have been in this same ballpark for hours.

Knowing that ideal gas laws involve closed systems and this one is a decidedly open one, I can't figure out how to think about what's happening here. My house is a fairly tightly built modern house, but even so there are many cracks and crevices where air pressure from the outside can seep inside or vice versa.

What is maintaining this pressure difference? I thought that maybe it was because the house is warmer, but it is closer to a closed system, and shouldn't air warm up and expand as it enters the house, thus increasing the house pressure?

My prediction is that they should have been equalized, but a pressure difference has been persistent.

Perhaps posed another way — is this consistent with the behavior of gases in general in an open system with this temperature difference, or is my house unusual (in this case, likely an appliance of some kind is driving this)?

Answer 1

Open your front door and hold up a piece of paper. Does it get pushed outward? If so there is a net outflow of air from your home.

It does not surprise me too much that there would be this small pressure difference. You mentioned the outdoor temperature is 30 F, meaning you have the heat on. That means there is a net influx of energy into your home, and the air is constantly heating up and expanding, trying to flow outward. This would be consistent with a small $\Delta P$.

Might I suggest shutting the heat off for a short time and seeing if there is a difference.

As a footnote, I applaud you for doing real experiments, taking data, and trying to see physical principles in action. Physics is fundamentally experimental, and too often these days students and practitioners do nothing but read books about it and write out equations, without ever getting a real world physical experience.

Answer 2

If your internal pressure is 10 hPa (or 1000 N/m2) higher than outside then your outside door (area about 1.5 m2) will need a force of 1500 N to prevent it being blown out which is about 150 kg or 350 pounds. Similarly your windows and walls have even larger areas and couldn't possible withstand the pressure.
So I have to question your calibration and is it valid for different temperatures? It may be easiest to put your internal unit inside your refrigerator (or outside your house) for a while then, inside, quickly check its reading.
I expect you barometer is electronic and consumer grade. Electronic pressure sensors measure a tiny capacitance change as a minute silicon diaphragm deflects. It's so small that it's a miracle that it is actually possible. Many electronic properties change with temperature including reference voltages, oscillator frequencies, material properties, resistance, capacitance - the list goes on. While pressure sensors all have some form of temperature compensation I expect in a consumer grade device it is insufficient for the accuracy you are expecting.
As a comment on some of the other answers if a closed space is being heated the air inside will expand and leave via any openings but there won't be any movement if the inside temperature is stable.
Also I've obviously made approximations and ignored the effect of wind as the [stagnation pressure][1] is small at low velocities (because it has a v² term)

Answer 3

Assuming wind speed is >0 there will be a pressure differential between the windward side of a brick wall and the leeward side. I'm not saying it to harp on a technicality, it's a reminder that equilibrium is a construct of science. "Assuming all other things equal" is like "once upon a time, in a land far far away".

Answer 4

Assuming ideal gas law, the ~7% increase in temperature (in Kelvin) inside the house relative to outside, if perfectly isolated, would have lead to a change in pressure of ~+75hPa, a pressure difference equivalent to a ~600m change in elevation. Since the actual difference you see is a lot smaller (~1%) than this, the pressures difference inside and outside the house have indeed equalized, implying the presence of processes/passages that allow this. The small difference (in the "wrong" direction) could be due to local factors inside the house like humidity, environment around the thermometer, air currents etc. or more likely wind or a cold draft at the outside thermometer.

Answer 5

As has been pointed out in some other answers, it is not possible to maintain a pressure difference of 1000 Pa between the inside and outside of a house: small as the difference may seem, it would add up to 2000 N on a 2-square-meter front door, so you'd need to pull on an inward-opening door with something near 500 lb to open it. Very clearly, the instruments are in error (unless, perhaps, you have put your outdoor sensor on a tall pole--the pressure decreases as you go up, and near sea level, you could explain the observed difference with an 80-meter-tall pole).

The question is what is the source of the error? I understand you have swapped sensors (a clever suggestion) and obtained a (more or less) consistent result. This means that there is likely a temperature-dependent bias in your sensors: when you cool them, they read low.