What I can tell you is that you'd have to be billionths of a degree above AZ in order for it to occur.

The pressure, or molecular movement, of an environment decreases and increases when the temperature is adjusted.

We can use the

Gay-Lussac Gas Law to see that temperature is tied to pressure in the most intimate of ways. Read the link. As temperature

*increases*, pressure

*increases*. Subsequently, as temperature

*decreases*, pressure

*decreases*.

Summarily, the pressure of a fixed amount of gas at fixed volume is directly proportional to its temperature in kelvins. As a gas expands, the average distance between molecules grows. Because of intermolecular attractive forces, expansion causes an increase in the potential energy of the gas. If no external work is extracted in the process and no heat is transferred, the total energy of the gas remains the same because of the conservation of energy. The increase in potential energy thus means a decrease in kinetic energy and therefore in temperature. This relation can be expressed mathematically as p/T= constant or p

_{1}/T

_{1} = p

_{2}/T

_{2}.

---

**Example 1 from the link:****Question:** Consider an environment with a volume of 22.4 L filled with a gas at 1.00 atm at 273 K. What will be the new pressure if the temperature increases to 298 K?

**Solution:** Using Gay-Lussac's Law and solving for p2 we get:

p

_{1}T

_{2} (1.00 atm)(298 K)

p

_{2} = ----- p

_{2} = -----------------

T

_{1} (273 K)

Or,

**p**_{2} = 1.09 atmNote: When the temperature increases, the pressure increases!

Also note that it is essential to use temperature on an absolute scale (i.e. use Kelvin instead of

^{o}C!)

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Using this formula lets make our own

**Example 2**:

**Question:** Consider an environment with a volume of 22.4 L filled with a gas at 1.00 atm at 273 K. What will be the new pressure if the temperature is decreased to 0.0000000001 K?

**Solution:** Using Gay-Lussac's Law and solving for p2 we get:

p

_{1}T

_{2} (1.00 atm)(0.0000000001 K)

p

_{2} = ----- p

_{2} = -----------------

T

_{1} (273 K)

**p**_{2}= 0.00000000000037 atmThus we see that an environment with a temperature of 0.0000000001 K has a standard atmosphere of 0.00000000000037 atm

Ergo we see that the amount of atmosphere is hardly anything at all; that if the temperature at the edges of the earth were sufficiently low there would be hardly an atmosphere which could be lost at the edges of the atmosphere gradient. According to an online pressure converter, atoms at the edge of the earth would be pushed out at a force of of 0.000000037 newtons per square meter [N/M

^{2}]. This is hardly significant; perhaps a few atoms per hour. And this is assuming that space is a perfect vacuum. It's not.