**Short answer:
** ** You can even calculate how things will be affected
using only High School level Physics or Chemistry.
**

We know how to calculate how the air pressure in the ball will change with temperature, using the Ideal Gas Law. The one subtly in the calculation is knowing that the Ideal gas Law uses Absolute Pressure & Temperature while pressure gauges measure "gauge pressure". Below is a table showing the final ball gauge pressure paired with a final ball temperature for a football inflated to 13 psi at 70°F as well as the pressure change from temperature alone. NFL rules require the ball to be between 12.5 psi & 13.5 psi. In our scenario, the temperature would take the ball "out of NFL spec" when its temperature dropped below 60°F or rose above 80°F. I've highlighted this "out of NFL spec" in orange:

Final ball Temperature (°F) |
Final Ball Gauge Pressure
Reading (psi) |
Ball Gauge Pressure Change
(psi) |

0 |
9.3 |
-3.7 |

5 |
9.6 |
-3.4 |

10 |
9.9 |
-3.1 |

15 |
10.1 |
-2.9 |

20 |
10.4 |
-2.6 |

25 |
10.6 |
-2.4 |

30 |
10.9 |
-2.1 |

35 |
11.2 |
-1.8 |

40 |
11.4 |
-1.6 |

45 |
11.7 |
-1.3 |

50 |
12.0 |
-1.0 |

55 |
12.2 |
-0.8 |

60 |
12.5 |
-0.5 |

65 |
12.7 |
-0.3 |

70 |
13.0 |
0.0 |

75 |
13.3 |
+0.3 |

80 |
13.5 |
+0.5 |

85 |
13.8 |
+0.8 |

90 |
14.0 |
+1.0 |

95 |
14.3 |
+1.3 |

100 |
14.6 |
+1.6 |

105 |
14.8 |
+1.8 |

*As
one can easily see, temperature alone can affect the pressure
in a football significantly. A football inflated to 13psi indoors at 70*°*F,
then taken outside in 45*°*F
weather and allowed to cool to 45*°*F,
will have it's pressure drop to 11.7psi, outside of NFL specifications. It
should be noted however, that in this example, if the football is brought back
indoors & allowed to warm back to 70*°*F**,
it's gauge pressure would return it's original pressure, which in this example,
is 13psi.*

You can learn about the Ideal Gas Law here:

**For parts of the calculations below, the equations all require that
we work in absolute units. Absolute units are units where the "zero" isn't
arbitrary, but is a true zero of the phenomenon being measured. Absolute
temperature scales, for example, have their zero at **
**
absolute zero****,
which is the coldest temperature even theoretically possible, as it is the
temperature at which all atomic motion stops. Absolute pressure is the gas
pressure compared to the pressure of a perfect vacuum. **

**Scientists normally work with the International System of Units
(SI), which is the system that all modern Scientific and Engineering
calculations are made. Most Americans are more familiar with
English units,
such as the foot, pound and degree Fahrenheit for temperature. Ironically,
even the English stopped using "English Units" decades ago, and use SI
units. The United States is almost the only nation still using English Units, which
in the US, are also called
United
States customary units. So for the calculations
below, we will use the
SI
Kelvin temperature scale and
occasionally the
Pascal pressure scale (1 pascal = 0.000145037738
pounds per square inch or psi).**

**A perfect ("ideal") theoretical gas has a relationship
between it's volume, absolute pressure, and absolute temperature according
to the relationship below, often refereed to as the
ideal gas law.**

*PV* = nR*T*

**where P is the absolute pressure (i.e. the pressure
compared to **

**A logical question to ask is "How well does the Ideal Gas
Law work in real life?". We'll answer that, but I want to address that
question from
the opposite direction, by asking "Can you modify the Ideal Gas Law to
work for real gases rather than ideal gases?" The answer is "yes", by introducing a factor
called the
Compressibility Factor**** ( Z)**

*PV* = nR*T Z*

**The intuitive way to think of the compressibility factor
is that it is the ratio of the true volume of a real gas to the same number
of molecules of an "ideal gas" at that same absolute temperature and
absolute pressure. The compressibility factor depends on the gas (or gas
mixture), as well as temperature & pressure. In other words, Z isn't a
constant, but is really a function Z(P,T) that can be quite complex.
So the more general form for realistic gases is: **

*PV* = nR*T Z(P,T)*

**Below we will show that for this football problem, the
ideal gas law & the gas law for real gases differ negligible
amounts.**

**Normal air is composed of roughly 80% nitrogen (N2) and
20% oxygen (O2). Not surprisingly, air is an extremely well studied gas
mixture, and it's compressibility factor is well known out to very high
pressures over a vast temperature range. The x-axis of the graph below is
labeled in
bars, which 1 bar is 100,000 Pascals or 14.5 psi. "Room temperature" is
about 300°K. (70°F
is 294.3°K)**

**Note that down below about 10 bars (145 psi, or almost 10 times
atmospheric pressure), the compressibility factor is close to 1
regardless of temperature. So what is Z for air around where we need to do
this calculation? Well, at T= 300°K
(80.3°F), Z_{air} = 0.9999 at 1
bar (14.5 psi), and when T= 300°K (80.3°F),
Z_{air} = 0.9987 at 5 bars (72.5 psi). So we can see that Z_{air}
changes by less than **

**Constant volume of football in pressure range of interest (isochoric
process)**

**One other question to address before we get down to
business is "Does the football's volume change significantly in the pressure
range we are talking about, which is 10-15 psi gauge pressure?" The short
answer is that before the pressure reaches that range, the
rubber air bladder inside the leather cover has fully expanded, and the
leather keeps the bladder's volume from expanding further, with negligible
volume change. So setting V = constant in the Ideal Gas Law will have a
negligible effect on the calculation.
**

Notice that in the above discussions, that just as I kept
referring to absolute temperature, I also referred to the pressure in terms
of absolute pressure. The gas laws above only work when one uses absolute
temperature and absolute pressure. Absolute pressure is the gas pressure compared to the
pressure of a perfect vacuum. But the gauges that are used to measure air
pressure in footballs, car & bicycle tires, and many other pressure gauges,
don't measure absolute pressure. They measure gauge pressure, which is the
pressure difference between the gas of interest in the ball or tire, and the
outside air pressure that the gauge is sitting in. In other words, the when
you measure the football air pressure to be, say, 13 psi, that is the gauge
pressure. The absolute pressure is that number *plus* the air pressure
*outside* the ball.
Standard atmospheric pressure is 14.7 psi, so to
convert that 13 psi gauge pressure to absolute pressure, we need to add 14.7
psi to it, which means that 13 psi gauge pressure is 27.7 psi absolute
pressure, or in Si units, 190,984.8 Pascals. *The failure to use absolute
pressure when performing calculations with the Ideal Gas Law is a common
error students make. It is also a common error made by Scientists who don't
deal with gases in their professional lives, and so are relying on what they can recall from
their student days. They, like inexperienced students, also often fail to
distinguish
between gauge and absolute pressure.*

So let's ask ourselves, if we pump up a football to say, 13 psi in a 70°F room, and then take the balls outside into a 45°F stadium and let them sit until they too are at 45°F, what would be the pressure drop in them? In order to use the Ideal gas Law, we need to convert those temperatures from Fahrenheit to Kelvin so that the temperatures are absolute temperatures. The formula for converting Kelvin to Fahrenheit is

T_{Kelvin} = (T_{Fahrenheit} + 459.67) * 5 /
9

which means that 70°F is 294.3°K, and 45°F is 280.4°K.

So we have the Ideal Gas Law

*PV* = nR*T*

which we want to do some algebra on to put all the variables that change on one side of the equals sign and all the things that don't change on the other. We know that the ideal gas constant R, amount of gas n, and the volume V won't change, and that the pressure P & temperature T will. So with a little algebra, we have

So now the variables that change are on the left side of the equals sign, and the variables that do not change are on the right hand side. This means that any pressure-temperature ratio for the football must be equal to any other pressure-temperature ratio for the football, or

or

Since we are interested in the final pressure when we change the temperature, we can do a little more algebra to get

So let's plug in our numbers! But first, I want
to point out that since we have our pressures in absolute pressure, and our
temperatures in absolute temperature, *and we are looking at ratios*, we
can just plug in even though our pressure is in English units and our
temperature is in SI units. So let's plug in *P*_{1}
= 27.7 psi absolute pressure, *T*_{1} = 294.3°K
(70°F), and *T*_{2} = 280.4°K
(45°F), which means

So once the ball cooled down to 45°F, it would be at an absolute pressure of 26.4 psi. But what would the pressure gauge read? For that, we need to convert from absolute pressure back to gauge pressure, which means subtract 14.7 psi, which means our gauge pressure will be 11.7 psi. Since we started at 70°F with a pressure of 13 psi, this means that temperature alone would drop the pressure by 1.3 psi.

Below is a table showing the final ball gauge pressure paired with a final ball temperature for a football inflated to 13 psi at 70°F as well as the pressure change from temperature alone. NFL rules require the ball to be between 12.5 psi & 13.5 psi. In our scenario, the temperature would take the ball "out of NFL spec" when its temperature dropped below 60°F or rose above 80°F. I've highlighted this "out of NFL spec" in orange:

Final ball Temperature (°F) |
Final Ball Gauge Pressure
Reading (psi) |
Ball Gauge Pressure Change
(psi) |

0 |
9.3 |
-3.7 |

5 |
9.6 |
-3.4 |

10 |
9.9 |
-3.1 |

15 |
10.1 |
-2.9 |

20 |
10.4 |
-2.6 |

25 |
10.6 |
-2.4 |

30 |
10.9 |
-2.1 |

35 |
11.2 |
-1.8 |

40 |
11.4 |
-1.6 |

45 |
11.7 |
-1.3 |

50 |
12.0 |
-1.0 |

55 |
12.2 |
-0.8 |

60 |
12.5 |
-0.5 |

65 |
12.7 |
-0.3 |

70 |
13.0 |
0.0 |

75 |
13.3 |
+0.3 |

80 |
13.5 |
+0.5 |

85 |
13.8 |
+0.8 |

90 |
14.0 |
+1.0 |

95 |
14.3 |
+1.3 |

100 |
14.6 |
+1.6 |

105 |
14.8 |
+1.8 |

In addition to temperature, the atmospheric pressure (barometric pressure) in the stadium itself depends on the weather. During storms for example, atmospheric pressure is lower, and during nice clear days, atmospheric pressure is higher. Below is a graph of the last few days of weather data (including barometric pressure) at Boston's Logan Airport. If you click the graph below, you'll be taken to a website that allows you to see similar graphs for major cities in the US.

Meteorologists usually measure barometric pressure in units
of "inches of
mercury" or "inches of
Hg"
or "inHg", where Hg is the
Periodic Table symbol for mercury. Standard atmospheric pressure of 14.7
psi corresponds to 29.9 inHg. But the reality is that it is not uncommon for weather to
vary the air pressure, in a matter of hours, by amounts on the order of 1 inHg. 1 inHg = 0.491 psi,
and remember the pressure gauge is reading the difference between the
absolute pressure in the ball and the absolute pressure outside the ball. So
in addition to the temperature effects computed above affecting the absolute
pressure *inside* the ball, depending on other weather, the absolute
pressure *outside* the ball can change in one direction or another on
the order of 0.5 psi, thus also affecting the ball's gauge pressure.

NOAA National Severe
Storms Laboratory

NOAA Severe
Thunderstorm Climatology

NOAA Severe Weather Potpourri

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