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Question

# The compressibility of a gas is less than unity at STP. Therefore?

A

${\mathrm{V}}_{\mathrm{m}}$ greater than $22.4\mathrm{L}$

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B

${\mathrm{V}}_{\mathrm{m}}$ less than $22.4\mathrm{L}$

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C

${\mathrm{V}}_{\mathrm{m}}$ equal to $22.4\mathrm{L}$

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D

${\mathrm{V}}_{\mathrm{m}}$ equal to $44.8\mathrm{L}$

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Solution

## The correct option is B ${\mathrm{V}}_{\mathrm{m}}$ less than $22.4\mathrm{L}$The explanation for the correct option:Option(B):$\mathrm{Vm}$ less than $22.4\mathrm{L}$Step 1: Given data:The gas is given at STP, hence at standard pressure and temperature,The value of pressure = $1\mathrm{atm}$The value of temperature = $273\mathrm{K}$The value of the gas constant = $0.0821\mathrm{L}\mathrm{atm}{\mathrm{K}}^{-1}$Step 2: Formula for compressibility factor:For an Ideal gas, $\mathrm{PV}=\mathrm{nRT}$,the compressibility factor $\mathrm{Z}=1$We know the compressibility factor (Z) is given as = $\frac{\mathrm{pV}}{\mathrm{nRT}}$Step 2: Calculating condition for the given gas:For the given gas, the compressibility factor (Z) is given as less than the unity$⇒\frac{{\mathrm{PV}}_{\mathrm{m}}}{\mathrm{nRT}}<1\phantom{\rule{0ex}{0ex}}⇒{\mathrm{PV}}_{\mathrm{m}}<\mathrm{nRT}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{V}}_{\mathrm{m}}<1\mathrm{x}0.0821\mathrm{x}273\phantom{\rule{0ex}{0ex}}⇒{\mathrm{V}}_{\mathrm{m}}<22.4\mathrm{L}$Thus, when the gas compressibility is less than unity at STP, ${\mathrm{V}}_{\mathrm{m}}$ less than $22.4\mathrm{L}$, option (B) is the correct answer.

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