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# The compressibility factor of ${ CO }_{ 2 }$ at 273 K and 100 atm pressure is 0.2005. The volume occupied by 0.2 mole of ${ CO }_{ 2 }$ gas at 100 atm and 273 K using real gas nature is $8.89\times { 10 }^{ -x }$ Litre.So the value of x is…

Hint: The compressibility factor gives us an idea about the degree to which a real gas shows deviation from the ideal gas behaviour. It is the ratio of the observed molar volume of a gas to the calculated molar volume (using ideal gas equation) of the gas at the same pressure and temperature

$Z=\cfrac { PV }{ nRT }$ (Where Z is the compressibility factor)
$\Rightarrow Z=\cfrac { P{ V }_{ real } }{ nRT }$...(1)
Since according to the ideal gas equation: $PV=nRT$
$\Rightarrow { V }_{ real }=\cfrac { nRT }{ P }$
$Z=\cfrac { { V }_{ real } }{ { V }_{ ideal } }$
We will use equation (1) to solve this question. Putting the value of n=0.2 mole, Z=0.2005, P=100 atm, R=0.0821 L atm ${ K }^{ -1 }{ mol }^{ -1 }$ and T=273 K, we get:
$0.2005=\cfrac { 1000\quad atm\times V }{ 0.2\quad mol\times 0.0821\quad L\quad atm\quad { K }^{ -1 }{ mol }^{ -1 }\times 273\quad K }$
$\Rightarrow V=8.89\times { 10 }^{ -3 }\quad L$