Question

Average molar mass of air is $29.0gmo{l}^{-1}$. At the altitude of $x\times 10^{3}m,$ atmosphere pressure decreases from 1.00 bar to 0.50 bar. What is the value of $x?$

• A. 1
• B. 6
• C. 2
• D. 4

Answer 1

Answer: (B)

the atmospheric pressure $(p)$ at a height h above see level is related to the atmospheric pressure at level $\left(p_0\right)$ by the barometric formula.

$p=p_0{e}^{-mgh/RT}$

${\log }_e\frac{p}{p_0}={\log }_e{e}^{mgh/RT}=-\frac{mgh}{RT}$

$\therefore h=\frac{2.303RT}{mg}\log \frac{p_0}{p}$

$h=\frac{2.303\times 8.314Jmo{l}^{-1}{K}^{-1}\times 298K}{29.0\times 10^{-3}kgmo{l}^{-1}\times 9.81\times m{s}^{-2}}\log \frac{1.0}{0.5}$

$6.04\times 10^3=x\times 10^3$

Thus, $x=6$