The Answer Is Not 0

The question is basically based on the Barometric pressure formula .

It states that

$\displaystyle p=p_0\exp\left(-\frac{Mgh}{RT}\right)$

where

$p$ is the pressure at height $h$ from sea level.

$p_0$ is the pressure at sea level.

$M$ is the molar mass of air (I used $29\,g/mol$ )

$R$ is the universal gas constant.

$T$ is the temperature.

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Let $h=10\,m$ , then we can write,

$\displaystyle p_{10}=p_{0}\exp\left(-\frac{Mgh}{RT}\right)$

$\displaystyle p_{20}=p_0\exp\left(-\frac{Mg(2h)}{RT}\right)$

To make things easier, substitute $\exp\left(-\dfrac{Mgh}{RT}\right)=x$ , i.e $p_{10}=p_0 x$ and $p_{20}=p_0x^2$ . Plugging in the expression we have to evaluate,

$\displaystyle \frac{p_0-p_0x^2}{p_0-p_0x}-2=\frac{(1-x)(1+x)}{(1-x)}-2=x-1$

$\Rightarrow x-1=\exp\left(-\frac{Mgh}{RT}\right)-1\approx \boxed{-1.11659 \times 10^{-3}}$

I used the following values:

$M=29\times 10^{-3} \,kg/mol$ , $g=9.8\,m/s^2$ , $h=10\,m$ , $R=8.314\,\,J/(mol-K)$ and $T=293\,K$ .