Question

What is the density of chlorine gas at $1.21{\text{atm}}$ and $34.9$ degree Celsius.

Answer 1

The term ideal gas alludes to a theoretical gas made out of atoms that observe a couple of rules: Ideal gas particles don't pull in or repulse one another. The lone communication between ideal gas particles would be a flexible crash upon sway with one another or a versatile impact with the dividers of the compartment.

Complete step by step answer:
We have to know that the ideal gas behavior, then the ideal gas behavior formula has to be used.
Given details,
$P = 1.21{\text{atm}}$
The unit of the temperature is converted into celsius into kelvin,
$T = 34.9^\circ C = 307.9K$
By using the following formula,
$PV = nRT$
Then, we have to rewrite the above equation,
$\dfrac{n}{V} = \dfrac{P}{{RT}}$
Where,
$R = 0.0821{\text{ L}}{\text{.atm/}}{{\text{K}}^{ - 1}}mo{l^{ - 1}}$
Applying all the values in the above equation,
$\dfrac{n}{V} = \dfrac{{1.21{\text{ atm}}}}{{(0.0821{\text{ L}}{\text{.atm/}}{{\text{K}}^{ - 1}}.mo{l^{ - 1}}) \times 307.9K}}$
Therefore,
$\dfrac{n}{V} = 0.0479{\text{mol}}{\text{.}}{{\text{L}}^{ - 1}}$
We have to know about the density in $mol.{L^{ - 1}}$. So that the above calculated value is multiplied by the molecular mass of $Cl$.
One mole of molar mass = $35.45g/mol$
Therefore, two moles of molar mass = $70.9g/mol$ .
Then,
$(70.9{\text{ g/mol) }} \times {\text{ (0}}{\text{.0479 mol/L) = 3}}{\text{.40 g/L}}$
Hence,
The density of chlorine gas = $3.40g/L$ .

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Nowadays, chlorine is used for pools, elaborate lakes and aquaria, sewage and wastewater, and different sorts of water supplies. Chlorine gas likewise is utilized as a sanitizer, Microbiota or microbicide, an algicide in food preparing frameworks, mash, and paper factory frameworks, and business and mechanical water-cooling frameworks.