Calculate the number of moles of gas present in the container of volume $10 L$ at $300 K$ . If the manometer containing glycerin shows $5 m$ difference in level as shown diagram. calculate your answers upto two decimal point.

(Given: $d_{glycerin} = 2.72 g/mL; d_{mercury} = 13.6 g/mL Tale R = 0.08 L a t m m o l^{- 1} K^{- 1}$ )

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Solution

We know,
pressure upon a certain liquid or gas can be measured by $P = ρ g h$
Where, $ρ$ = density

So, at an equilibrium position, pressure on mercury will be equal to the pressure on glycerine.
thus, let the height of mercury column be x.
Therefore,
$ρ_{m e r c u r y} x g = ρ_{g l y c e r i n e} 5 g$
$\Rightarrow x = \frac{ρ_{g l y c e r i n e} 5}{ρ_{m e r c u r y}} = \frac{2.72 \times 5}{13.6} = 1 m$

Now,
$pressure of the gas = 1 a t m + 1 m o f H g$
$= 760 m m o f H g + 1000 m m o f H g$
$= 1760 m m o f H g$

given,
volume of the container = 10 L
Temperature = 300 K

$So moles of the gas = \frac{1760 \times 10}{760 \times 0.08 \times 300} \approx 0.96$

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Calculate the number of moles of gas present in the container of volume $10 L$ at $300 K$ . If the manometer containing glycerin shows $5 m$ difference in level as shown in the diagram. calculate your answers up to two decimal points.

(Given: $d_{glycerin} = 2.72 g/mL; d_{mercury} = 13.6 g/mL Tale R = 0.08 L a t m m o l^{- 1} K^{- 1}$ )