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The compressibility factor of gases is l...

The compressibility factor of gases is less than unity at `STP`. Therefore,

A

`V_(m)` (molar volume) `gt 22.4 L`

B

`V_(m) lt 22.4 L`

C

`V_(m) = 22.4 L`

D

`V_(m) = 44.8 L`

Text Solution

Verified by Experts

The correct Answer is:
B
`Z = ((PV)_("real"))/((PV)_("idea")) = (V_("real"))/(V_("ideal"))`
For 1 mol at STP `V_("ideal") = 22.4 L`
`Z = (V_("real"))/(22.4 L) lt 1`
`:. V_("real") lt 22.4 L`
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