Pressure at Height: Calculate Pressure Change

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Homework Statement


(see attachment)


Homework Equations





The Attempt at a Solution


I am not sure how to start with this problem. Do I need to use the barometric formula?

Barometric formula:
[tex]p=p_0exp\left(\frac{-Mgh}{RT_0}\right)[/tex]
 

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Pranav-Arora said:

Homework Statement


(see attachment)


Homework Equations





The Attempt at a Solution


I am not sure how to start with this problem. Do I need to use the barometric formula?

Barometric formula:
[tex]p=p_0exp\left(\frac{-Mgh}{RT_0}\right)[/tex]

Why not?

ehild
 
ehild said:
Why not?

ehild

I tried that but it gives me the wrong answer.
Here's my attempt:

Let h=10m, then 2h=20m.
##p_{20}=p_0 \exp\left({-\frac{Mg(2h)}{RT}}\right)##
##p_{10}=p_0 \exp\left({-\frac{Mgh}{RT}}\right)##

Let ##\exp\left({-\frac{Mgh}{RT}}\right)=x##.
[tex]\frac{p_0-p_{20}}{p_0-p_{10}}-2=\frac{1-x^2}{1-x}-2[/tex]
[tex]=(1+x)-2=x-1[/tex]

When I substitute the values of all the variable in exponential term, I get ##x-1=-5.6 \times 10^{-4}## but this is wrong. :(

I used ##M=14 \times 10^{-3} kg/mol##, ##R=8.314 J/(mol \cdot K)## and ##T=293 K##.
 
ehild said:
What is the molar mass of nitrogen gas?

ehild

Woops, its 28, thanks! :)

This time I get: ##-1.1 \times 10^{-3}##, is this right?
 
ehild said:
Correct.

ehild

But when I submit this, it is marked incorrect. :(
 
ehild said:
Try more digits.

Thanks a bunch, yes, it was a problem with significant digits. :)