Question

At the top of a mountain, the thermometer reads $0^{\circ }C$ and the barometer reads $700mmHg$. At the bottom of the mountain, the thermometer reads ${30}^{\circ }C$ and, the pressure $760mmHg$. Compare the density of the air at the top of that at the bottom.

• A. $\frac{1.02}{1}$
• B. $\frac{2.04}{1}$
• C. $\frac{4.04}{1}$
• D. $\frac{6.04}{1}$

Answer 1

The correct option is A $\frac{1.02}{1}$

Solution:- (A) $\frac{1.02}{1}$

From ideal gas equation,

$PV=nRT$

$n=$ No. of moles $=\frac{\text{wt.}\left(w\right)}{\text{Mol. wt.}\left(M\right)}$

$\therefore PV=\left(\frac{w}{M}\right)RT$

$\Rightarrow PM=\left(\frac{w}{V}\right)RT$

As we know that,

$\text{Density}\left(d\right)=\frac{\text{wt.}\left(w\right)}{\text{volume}\left(V\right)}$

$\therefore PM=dRT$

For the same gas,

$d\propto \frac{P}{T}$

$\therefore \frac{d_1}{d_2}=\frac{P_1}{P_2}\times \frac{T_2}{T_1}$

Given:-

$P_1=700\text{mm Hg}$

$P_2=760\text{mm Hg}$

$T_1=0℃=273K$

$T_2=30℃=(30+273)=303K$

$\therefore \frac{d_1}{d_2}=\frac{700}{760}\times \frac{303}{273}$

$\Rightarrow \frac{d_1}{d_2}=\frac{1.02}{1}$

Hence the rtio of density of the air at the top to that at the bottom is $\frac{1.02}{1}$.