The Mass of Earth's Atmosphere

[lmannoia]

Homework Statement

The Mass of an Atmosphere: What is the total mass of Earth's atmosphere? You may use the fact that 1 bar is the pressure exerted by about 10,000kg pushing down on a square meter in Earth's gravity. Remember that every square meter of Earth experiences this pressure from the atmosphere above it.

Homework Equations

V for a sphere = 4(pi)R^2
Mass = density x volume

The Attempt at a Solution

I know that the Earth's atmosphere is approximately 10 km thick. What I don't know is how to go about figuring out the density of the atmosphere by using pressure and the fact that every square meter of Earth experiences this pressure from the atmosphere above it.
Any push in the right direction would be greatly appreciated. Thanks!

 

[Chi Meson]

It's a lot easier than the route you are taking.

The problem already states that for every square meter of Earth's surface, there is the weight of 10,000kg of atmosphere above it.

So, what is measured in units of "kg"?

[And you should know that "1 bar" is almost exactly "1 atm." Standard atmospheric pressure is actually more closely 1.013 bar]

 

[lmannoia]

Oh wow, that went completely over my head. Thank you!

 

[Chi Meson]

Oh wow, that went completely over my head.

like the atmosphere

 

[rwisz]

I would approach this problem by first identifying the variables and equations that are relevant to finding the mass of Earth's atmosphere. The given information states that 1 bar (unit of pressure) is equivalent to 10,000 kg/m^2 (unit of force/area). This provides a starting point for calculating the mass of the atmosphere.

The relevant equation here is the ideal gas law, which relates the pressure, volume, and temperature of a gas. It is given by PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. In this case, we can assume that the temperature of the atmosphere remains relatively constant, so we can focus on the relationship between pressure and volume.

We can rearrange the ideal gas law to solve for n (number of moles), which is equal to the mass of the gas divided by its molar mass. In this case, the gas is Earth's atmosphere, and its molar mass is approximately 28.97 g/mol. Therefore, we can write the equation as:

n = (mass of atmosphere) / (molar mass of atmosphere)

To find the mass of the atmosphere, we need to calculate the volume of the atmosphere. We can use the given information that the Earth's atmosphere is approximately 10 km (or 10,000 m) thick. We can assume that the atmosphere forms a spherical shell around the Earth, with a radius of 6,371 km (the radius of the Earth). Therefore, the volume of the atmosphere can be calculated using the equation for the volume of a sphere:

V = 4/3 * pi * (r_outer^3 - r_inner^3)

In this case, the outer radius (r_outer) is equal to the radius of the Earth plus the thickness of the atmosphere (6,371 km + 10 km), and the inner radius (r_inner) is equal to the radius of the Earth (6,371 km). Plugging these values into the above equation gives us a volume of approximately 1.083 x 10^21 m^3.

Now, we can substitute this volume into the ideal gas law equation, along with the given pressure of 1 bar, to solve for the number of moles of the atmosphere:

n = (P * V) / (RT)

Plugging in the values, we get:

n