Spacecraft Decompression Times

References Consulted:
Vacuum Exposure: How long will it take a spacecraft to decompress? by Geoffrey A. Landis

The above link goes into much more detail on this subject; this is merely a simplification of the above link into a more general science text for rule of thumb use.

Time Required to Reach New Pressure:

Time = 0.086 (VolumeCraft / HoleArea) * Ln(PressStart / PressEnd) / SQRT (Temperature)

Where:

Time is in Seconds
Volume
Craft is in Cubic Meters
Hole
Area is in Square Meters
Press
Start is Dimensionless (See Note)
PressEnd is Dimensionless (See Note)
Temperature is in Kelvins.

NOTE: PressureBeginning and PressureFinal are dimensionless, they can be in either PSI or Pascals, as long as they are consistent; e.g. both sides use the same units.

Some rough values to be used in this formula are given below.

A wide range of variables are included for the temperature level; because some areas of the spacecraft may have air pressure in them, but are not heated to enable ease of repair, e.g. the spacecraft may have a double hull, in which only the inner hull is fully pressurized to sea level and 70F, while the space between the two hulls is only pressurized to 20,000 feet and 32F to save on weight. Additionally, some areas may reach high temperatures during combat if the life support system is overloaded and cannot keep up with temperatures from spacecraft equipment.

Earth Air Pressure (Sea Level): 14.7 PSI (101.34 kilopascals)
Earth Air Pressure (8,000 ft) 10.91 PSI (75.22 kilopascals) (Airliner Cabin Pressure)
Earth Air Pressure (20,000 ft): 6.4 PSI (44.12 kilopascals) (Oxygen Mask required)
Earth Air Pressure (50,000 ft): 1.6 PSI (11.03 kilopascals) (Pressure suit required)

-10 F (-23.3 C) = 249.8 Kelvin
0 F (-17.7 C) = 255.37 Kelvin
32 F (0 C) = 273.15 Kelvin
70 F (21.1 C) = 294.2 Kelvin
100 F (37.85 C) = 311 Kelvin
125 F (51.85 C) = 325 Kelvin

Example:

A spacecraft with a pressurized volume of 6.17 m3 is hit by a meteorite which punches a 2 cm diameter hole in it's hull. (1 cm radius * 3.14 = 3.14 cm2 = 0.000314 m2). The internal air pressure is 10.91 PSI; at 294.2 Kelvin. How long would it take before the air pressure dropped to a level where the crew would have to don pressure suits (1.6 PSI)?

0.086 (6.17 / 0.000314) * Ln(10.91 / 1.6) / SQRT (294.2) = 189.1 seconds (or 3.15 minutes)

As you can see; this is more than enough time to put a patch over the hole, or to don an emergency pressure suit. If you play with the temperature of the air inside; you find that the lower the temperature of the air, the longer it takes for the air to leak out. This leads one to the conclusion that part of the SOP for any Space Navy going into combat would be to lower the air temperature inside their ships; to not only add additional time to seal a leak; but also to provide a heat sink in the worst case scenario.