What is the total mass of the atmosphere?

[EliotBry]

Homework Statement

Problem from the book "Engines, Energy and Entropy", Page 55, question 7 has me stumped. It doesn't feel like their is sufficient information to work out the mass of the air. They've given us density (as seen in the picture, if the upload works) , which is mass over volume, but unless you know the volume of mercury you can't work out the mass from that. Were only given the height of the mercury (maybe in a barometer), But without knowing the area that this height takes.

I'm truly stumped on part a), it's not complicated physics. I'm just struggling to see how you can achieve mass if you can't work out volume that the mercury takes up.

http://imgur.com/HnOoMAv

Homework Equations

Ideal gas law as in picture.

Density = mass/volume

The correct answer is 5.281×10¹⁸kg.

The Attempt at a Solution

I found the pressure of the mercury (taking T=300K at the surface for a rough estimate) as 167.85 Pa. I don't know how close that is to being correct.

I also do not understand the units of R here, what is the L stand for?

I also calculated the surface area of the Earth as 5.147×10¹⁴ m². But this is useful once I've calculated the mass of the atmosphere.

Apologies if it seems like I've not gotten far, I've just spent an hour on it running round in circles and I'm pissed off with it now.

 

[Ygggdrasil]

In a barometer, you have a column of mercury pushing against a column of atmosphere of the same size. At equilibrium, the weight of the mercury pushing at one end of the barometer is equal to the weight of the atmosphere pushing at the other end of the barometer. Assume the cross-sectional area of the barometer is 1 cm². What weight of mercury is pushing against an area of 1 cm²? What weight of atmosphere is pushing against 1 cm²?

 

[EliotBry]

Oh thankyou so much, I've managed to figure out part a) now. didn't even need the ideal gas law haha.

I feel a little embarassed now. My brain must have been a bit fried so I'll give it a rest for a bit before tackling the rest of the questions

Thankyou so much for your help.

 

[SteamKing]

The Attempt at a Solution

View attachment 98378
I found the pressure of the mercury (taking T=300K at the surface for a rough estimate) as 167.85 Pa. I don't know how close that is to being correct.

It's not clear which pressure you are talking about here. In any event, a pressure of 167.85 Pa represents a pretty good vacuum, rather than a significant pressure.

In case you didn't know it, standard atmospheric pressure is 101,325 Pa at sea level at 15° C. This is the pressure which supports a column of mercury 76 cm high.

Using the density of mercury, ρ = 13.5 g/cc, the height of the column, 76 cm, and the hydrostatic pressure law, P = ρ g h, you should be able to calculate the pressure that this column of mercury creates.

I also do not understand the units of R here, what is the L stand for?

You couldn't guess this? L stands for liters.

 

[EliotBry]

Sorry man, my brain was really frazzled earlier. L for litres, of course. I don't know what I was thinking.

 

[epenguin]

In other words, if you covered the surface of the Earth with Mercury to a depth of 760 mm, the mass of that Mercury equals the maths of the actual atmosphere?