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van der Waal's gas equation can be reduc...

van der Waal's gas equation can be reduced to virial eqation and virial equation (in terms of volume) is`Z=A+(B)/(V_(m))+(C)/(V_(m)^(2))+……..`
where A =first virial coefficient, B=second virial coefficient ,C = third virial coefficient. The third virial coeffdient of Hg(g) is 625 `(cm^(2)//"mol")^(2)`. What volume is available for movement of 10 moles He(g) atoms present in 50 L vessel?

A

49.75 L

B

49.25 L

C

25 L

D

50 L

Text Solution

Verified by Experts

The correct Answer is:
a
(a) `b=25cm^(3)//mol`
`b=0.025L//mol`
`V_("Real")=V_(i)-nb=50-10xx0.025=49.75L`
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Knowledge Check

  • The third virial coefficient of He gas is 4xx10^(-2)(L//"mol")^(2) , then what will be volume of 2 mole He gas at 1 atm and 273 K?

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