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  • Need clarity, kindly explain! - States of matter - JEE Main-2

At a given temperature T , gases Ne, Ar , Xe and Kr are found to deviate from ideal gas behaviour . Their equation of state is given as p=\frac{RT}{V-b}  at T  . Here , b is the van der Waals constant . Which gas will exhibit steepest increase in the plot of Z ( compression factor ) Vs p ?

  • Option 1)

    Xe

  • Option 2)

    Kr

  • Option 3)

    Ne

  • Option 4)

    Ar

Answers (1)

best_answer

 

Compressibility Factor -

used to find deviation of a gas from ideal behaviour

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Compressibility Factor -

\dpi{100} Z=\frac{V_{real}}{V_{ideal}}

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Compressibility Factor -

For ideal gas Z=1

-

 

 

Compressibility Factor -

at  low pressures, all gases have Z less than 1.

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Compressibility Factor -

at very high pressure, all gases have Z greater than 1.

-

 

 

Compressibility Factor -

for hydrogen and helium, Z is  greater than 1 even at low pressures.

 

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we know compression \:\:factor\:\: z\:\:= 1+\frac{pb}{RT}  , at T=Constant

slop \propto \:b   ( volume of molecules )

order of radii of element Xe>Kr>Ar>Ne

b value of Xe will be higher so , the graph will be steepest .


Option 1)

Xe

Option 2)

Kr

Option 3)

Ne

Option 4)

Ar

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