# Does pressure of a closed box with air in it change as a function of the gravitational acceleration?

If I took a hypothetical box of fixed volume and filled it with 1 mol of N2 at home, measured the pressure, then teleported this box to increasingly distant points from the Earth, would I see the pressure drop?

Since Pressure = Force / Area and Force = mass x acceleration

Decreasing the gravitational acceleration should decrease the pressure? This in turn should decrease the temperature based on PV=nRT. Though maybe ideal gas law does not apply in this context?

Where am I going wrong with this?

• Consider: the atmosphere in the ISS is practically weightless, yet it still has positive pressure. Also, in PV = nRT there's no term for the gravitational acceleration. – PM 2Ring Feb 16 at 13:35

In the presence if gravity there will be a pressure gradient in the box, with the pressure at the bottom of the box greater than that at the top. As you move the box away, or reduce gravity, then the pressure gradient will decrease, with the pressure in the box becoming uniform as gravity disappears.

EDIT: Let's be a bit more quantitative. The pressure gradient in the box will be,

$$\frac{dP}{dz}=-\rho g$$

where $$z$$ is measured upward from the bottom of the box. The ideal gas law can be recast as:

$$P=\frac{n}{V} RT=\frac{\rho}{M}RT$$

where $$M$$ is the molar mass.

Then we have, $$\frac{d}{dz}\left(\frac{\rho}{M}RT\right)=-\rho g$$ or

$$\frac{d\rho}{dz}+\left(\frac{Mg}{RT}\right)\rho=0$$

So, the solution for $$\rho$$ is $$\rho (z)=\rho_0 \exp\left(\frac{-zMg}{RT}\right)$$, where $$\rho_0$$ is the density at the bottom of the box. If the height of the box is small such that $$H<, then this can be linearized as, $$\rho (z)=\rho_0\left(1- \frac{zMg}{RT}\right)$$.

Now, if the volume of your box is $$V$$ and the total mass of the gas that you place in the box is $$m$$, then the denisty of the gas in the box in the absence of gravity is $$\rho=\frac{m}{V}$$. If this same mass of gas, in the same volume, is now exposed to gravity then we must have

$$\frac{1}{H}\int_{0}^{H}\rho_0\left(1- \frac{zMg}{RT}\right)dz=\frac{m}{V}=\frac{m}{HA}$$

Where $$H$$ is the height of the box, and $$A$$ is the area of its base. This gives:

$$\rho_0=\frac{m}{V}\frac{1}{\left(1-\frac{MgH}{2RT}\right)}$$

Then the pressure in the box is:

$$P(z)=\frac{\rho_0 RT}{M}-\rho_0 gz$$

Note that this analysis assumes that the temperature of the box remains constant.