A flask containing air at $107^{o} C$ and $722 mm$ of $H g$ is cooled to $100 K$ and $760 mm$ of Hg. If density of the initial condition is $1 {g/cm}^{3}$ , then what is the final density in $(g / c m^{3})$ ?

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Solution

Given,
$Initial pressure (P_{i}) = 722 mm Final pressure (P_{f}) = 760 mm$
$Initial temperature (T_{i}) = (107 + 273) K = 380 K Final temperature (T_{f}) = 100 K$
$Initial density (d_{i}) = 1 {g /cm}^{3} Final density (d_{f}) = ?$
We know, $P M = d R T$
Where, $M = Molar mass$
Since $M$ is same here, thus we can write

$\frac{d_{i} T_{i} R}{P_{i}} = \frac{d_{f} T_{f} R}{P_{f}}$
$\Rightarrow d_{f} = (\frac{p_{f}}{p_{i}}) (\frac{T_{i}}{T_{f}}) d_{i}$

$= (\frac{760}{722}) (\frac{380}{100}) \times 1 {g/cm}^{3}$
$= 4 {g/cm}^{3}$

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A flask containing air at 107 oC and 722 mm of Hg is cooled to 100 K and 760 mm of Hg. If density of the initial condition is 1 g/cm3, then what is the final density in (g/cm3)?

A flask containing air at $107^{o} C$ and $722 mm$ of $H g$ is cooled to $100 K$ and $760 mm$ of Hg. If density of the initial condition is $1 {g/cm}^{3}$ , then what is the final density in $(g / c m^{3})$ ?