# Pressure at a height

1. Mar 17, 2013

### Pranav-Arora

1. The problem statement, all variables and given/known data
(see attachment)

2. Relevant equations

3. 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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2. Mar 18, 2013

### ehild

Why not?

ehild

3. Mar 18, 2013

### Pranav-Arora

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 $T=293 K$.

4. Mar 18, 2013

### ehild

What is the molar mass of nitrogen gas?

ehild

5. Mar 18, 2013

### Pranav-Arora

Woops, its 28, thanks! :)

This time I get: $-1.1 \times 10^{-3}$, is this right?

6. Mar 18, 2013

### ehild

Correct.

ehild

7. Mar 18, 2013

### Pranav-Arora

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

8. Mar 18, 2013

### ehild

Try more digits.

9. Mar 18, 2013

### Pranav-Arora

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