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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) = 44.82 L`

B

`V_(m) = 22.4 L`

C

`V_(m) lt 22.4 L`

D

`V_(m) gt 22.4 L`

Text Solution

Verified by Experts

The correct Answer is:
C
More insight is obtained in the significant of `Z` if we note the following derivation:
`Z = (pV_(real))/(nRT)`
If the gas shows ideal behavior, then `V_(ideal) = nRT//p`. On putting this value of `nRT//p` in the above equation , we have
`Z = (V_(real))/(V_(i deal))`
From this relation , we can see that compressibility factor is the ratio of actual molar volume `(n = 1)` of a gas to the molar volume `(n = 1)` of it it were an ideal gas at the temperature and pressure.
If `Z lt 1` , then
`(V_(real))/(V_(i d eal)) lt 1`
or `V_(real) lt V_(i deal)`
Thus , `V_(m) lt 22.4 L`
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