# Pressure at a height

(see attachment)

## 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:
$$p=p_0exp\left(\frac{-Mgh}{RT_0}\right)$$

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ehild
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(see attachment)

## 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:
$$p=p_0exp\left(\frac{-Mgh}{RT_0}\right)$$
Why not?

ehild

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##.
$$\frac{p_0-p_{20}}{p_0-p_{10}}-2=\frac{1-x^2}{1-x}-2$$
$$=(1+x)-2=x-1$$

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
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What is the molar mass of nitrogen gas?

ehild

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
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Correct.

ehild

Correct.

ehild
But when I submit this, it is marked incorrect. :(

ehild
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Try more digits.

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