# Using Ideal Gas Law to Calculate Vertical Pressure Gradient

• Ghost Repeater
In summary, the conversation discusses the use of the ideal gas law and the weight of a cylindrical parcel of air to derive the equation dP/dz = -mg/(kT)P. The conversation also includes a mistake in the algebra that was corrected by analyzing the units.
Ghost Repeater

## Homework Statement

Consider a cylindrical parcel of air of area A and infinitesimal height dz. If this air parcel is to remain stationary, the difference between the total pressure forces exerted on its top and bottom faces must be equal to its weight. Use this information and the ideal gas law to show that

$$\frac{dP}{dz} = - \frac{mg}{kT} P$$

where m is the mass of an air molecule, g is the gravitational field strength, k is Boltzmann's constant, T is the absolute temperature, and P is the air pressure at the vertical position z corresponding to the center of the particle.[/B]

## Homework Equations

$$PV = NkT$$

$$\rho = \frac{M}{V}$$[/B]

## The Attempt at a Solution

I'm calling the pressure difference between top and bottom dP. Since this has to equal the weight of the cylinder, I have

$$dP = -W_{cyl} = -Mg$$

where M is the weight of all the air in the cylinder. I can express this in terms of the density of air as

$$dP = -\rho_{air} V g$$

where V is the volume of the cylinder.

This in turn can be written in terms of the given mass m of a single molecule:

$$dP = -m \frac{N}{V} Vg$$

I can write the volume of the cylinder differentially as V = Adz, so now I have:

$$\frac{dP}{dz} = -m \frac{N}{V} A g$$

Finally, using the IGL, I can write

$$\frac{N}{V} = \frac{P}{kT}$$.

So I have

$$\frac{dP}{dz} = -m\frac{P}{kT}A g$$[/B]

So you see my problem: I have this factor of A still hanging around and I can't find a way to get rid of it. I don't see the flaw in my algebra, but physically it's obvious that the area of the cylinder can't make any difference to the pressure gradient!

Check all your equations for proper dimension left and right (and in between) of = signs !

And only use boldface sparingly

BvU said:
Check all your equations for proper dimension left and right (and in between) of = signs !

And only use boldface sparingly

Sorry about the boldface! Didn't even realize it was happening.

Thanks for your suggestion to analyze the units. I think I found my mistake right away. I should have started by writing

$$dP = \frac{Mg}{A}$$

where M is the mass of the entire cylinder. This gives me units of

$$\frac{[N]}{m^2} = \frac{[kg] [N]}{[kg] [m^2]}$$

And that gives me a factor of 1/A to cancel my leftover factor of A, besides just being true to the definition of pressure!

Thanks!

BvU

## 1. How do you calculate the vertical pressure gradient using the Ideal Gas Law?

The vertical pressure gradient can be calculated using the formula ΔP/Δz = -ρg, where ΔP is the change in pressure, Δz is the change in height, ρ is the density of the gas, and g is the acceleration due to gravity. This formula is derived from the Ideal Gas Law, which states that PV = nRT.

## 2. What are the units of measurement for the vertical pressure gradient?

The units of the vertical pressure gradient are typically in Pascals per meter (Pa/m) or kilopascals per kilometer (kPa/km). However, other units such as bar/meter or millibars/kilometer may also be used.

## 3. How does temperature affect the vertical pressure gradient?

According to the Ideal Gas Law, an increase in temperature will result in an increase in pressure if all other variables remain constant. This means that as temperature increases, the vertical pressure gradient will also increase. Similarly, a decrease in temperature will result in a decrease in pressure and a decrease in the vertical pressure gradient.

## 4. What is the significance of calculating the vertical pressure gradient?

The vertical pressure gradient is an important parameter in atmospheric science and meteorology. It helps us understand the distribution of air pressure in the atmosphere, which is a key factor in weather patterns and atmospheric circulation. It is also used to calculate the strength and direction of winds, as air moves from areas of high pressure to areas of low pressure.

## 5. Are there any limitations to using the Ideal Gas Law to calculate the vertical pressure gradient?

Yes, there are limitations to using the Ideal Gas Law to calculate the vertical pressure gradient. This formula assumes that the gas is in a state of thermodynamic equilibrium, which may not always be the case in dynamic atmospheric conditions. It also assumes that the gas is an ideal gas, which may not hold true for all gases. Additionally, the vertical pressure gradient may be affected by other factors such as humidity, which are not accounted for in this formula.

• Introductory Physics Homework Help
Replies
10
Views
1K
• Materials and Chemical Engineering
Replies
1
Views
888
• Mechanical Engineering
Replies
3
Views
1K
• Classical Physics
Replies
18
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
920
• Biology and Chemistry Homework Help
Replies
1
Views
235
• Thermodynamics
Replies
23
Views
1K
• Classical Physics
Replies
10
Views
802
• Introductory Physics Homework Help
Replies
3
Views
362
• Introductory Physics Homework Help
Replies
4
Views
1K