One mol of ${N H}_{4} C l$ is kept in an open container and then covered with a lid. The container is now heated to $600 K$ where all ${N H}_{4} C l$ dissociated into ${N H}_{3}$ and $H C l$ . If volume of the container is $24.63$ litres. Calculate what will be the final pressure of gases inside the container. Also find whether the lid would stay or bounce off if it can withstand pressure difference of $5.5 a t m$ . Assume that outside air is at $300 K$ and $1 a t m$ pressure.

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Solution

$1 N H_{4} C l \to 1 N H_{3} + 1 H C l$
$1$ moles $N H_{4} C l$ dissolution into $2$
moles $(n)$
volume of container = volume
replaced by gas $= 24.63 m$
temperature $= 600 K$
Ans to Ideal gas equation
$P V = M R T$
$P = 24.63 = 2 \times 0.0821 \times 600$
$P = 4 a t m$
The lid would stay because pressure difference between them is less than $5.5$ atom.

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N H_{4} C l (s)

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N H_{4} C l (s)

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An air tight container having a lid with negligible mass and an area of 8 cm $^{2}$ is partially evacuated. If a 48N force is required to pull the lid off the container and the atmospheric pressure is 1.0 x 10 $^{5}$ Pa, the pressure in the container before it is opened must be: