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?
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?
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`
`b=0.025L//mol`
`V_("Real")=V_(i)-nb=50-10xx0.025=49.75L`
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